# What is the difference between "expectation", "variance" for statistics versus probability textbooks?

It seems that there are two ideas of expectation, variance, etc. going on in our world.

In any probability textbook:

I have a random variable $$X$$, which is a function from the sample space to the real line. Ok, now I define the expectation operator, which is a function that maps this random variable to a real number, and this function looks like, $$\mathbb{E}[X] = \sum\limits_{i = 1}^n x_i p(x_i)$$ where $$p$$ is the probability mass function, $$p: x_i \mapsto [0,1], \sum_{i = 1}^n p(x_i) = 1$$ and $$x_i \in \text{range}(X)$$. The variance is, $$\mathbb{E}[(X - \mathbb{E}[X])^2]$$

The definition is similar for a continuous RV.

However, in statistics, data science, finance, bioinformatics (and I guess everyday language when talking to your mother)

I have a multi-set of data $$D = \{x_i\}_{i = 1}^n$$ (weight of onions, height of school children). The mean of this dataset is

$$\dfrac{1}{n}\sum\limits_{i= 1}^n x_i$$

The variance of this dataset (according to "science buddy" and "mathisfun dot com" and government of Canada) is,

$$\dfrac{1}{n}\sum\limits_{i= 1}^n(x_i - \sum\limits_{j= 1}^n \dfrac{1}{n} x_j)^2$$

I mean, I can already see what's going on here (one is assuming uniform distribution), however, I want an authoritative explanation on the following:

1. Is the distinction real? Meaning, is there a universe where expectation/mean/variance... are defined for functions/random variables and another universe where expectation/mean/variance... are defined for raw data? Or are they essentially the same thing (with hidden/implicit assumption)

2. Why is it no probabilistic assumption is made when talking about mean or variance when it comes to dealing with data in statistics or data science (or other areas of real life)?

3. Is there some consistent language for distinguishing these two seemingly different mean and variance terminologies? For example, if my cashier asks me about the "mean weight" of two items, do I ask him/her for the probabilistic distribution of the random variable whose realization are the weights of these two items (def 1), or do I just add up the value and divide (def 2)? How do I know which mean the person is talking about?/

• There's no probabilistic assumption about the sample mean or sample variance because they're just numbers computed from the sample data. If you measure the heights of $100$ people, you have $100$ numbers and you can do whatever computations you like with them. Dec 23, 2020 at 5:07
• Can you explain what you what you mean that 'uniform distribution' was assumed? Isnt p(x_i)=lim n->infty #occurences of x_i/n ? So an approximation of p(x_i) is #occurences of x_i /n. So 1/n sum x_i = 1/n sum_{x_i} x_i #occurences of x_i = sum x_i p(x_i). Of course p(x_i) is only an approximation of p(x_i) in statistics. I do not see how uniformity was used here Dec 23, 2020 at 16:25
• "Why is it no probabilistic assumption is made when talking about mean or variance when it comes to dealing with data in statistics or data science " Can you be more clear as to what assumption you think is made elsewhere, but isn't made in data science? Dec 24, 2020 at 20:46
• @lalala If you take the average off $n$ values, that is equal to the expected value of a random variable that has a uniform distribution across those alues. Dec 24, 2020 at 20:47

You ask a very insightful question that I wish were emphasized more often.

EDIT: It appears you are seeking reputable sources to justify the above. Sources and relevant quotes have been provided.

Here's how I would explain this:

• In probability, the emphasis is on population models. You have assumptions that are built-in for random variables, and can do things like saying that "in this population following such distribution, the probability of this value is given by the probability mass function."
• In statistics, the emphasis is on sampling models. With most real-world data, you do not have access to the data-generating process governed by the population model. Probability provides tools to make guesses on what the data-generating process might be. But there is always some uncertainty behind it. We therefore attempt to estimate characteristics about the population given data.

From Wackerly et al.'s Mathematical Statistics with Applications, 7th edition, chapter 1.6:

The objective of statistics is to make an inference about a population based on information contained in a sample taken from that population...

A necessary prelude to making inferences about a population is the ability to describe a set of numbers...

The mechanism for making inferences is provided by the theory of probability. The probabilist reasons from a known population to the outcome of a single experiment, the sample. In contrast, the statistician utilizes the theory of probability to calculate the probability of an observed sample and to infer this from the characteristics of an unknown population. Thus, probability is the foundation of the theory of statistics.

From Shao's Mathematical Statistics, 2nd edition, section 2.1.1:

In statistical inference... the data set is viewed as a realization or observation of a random element defined on a probability space $$(\Omega, \mathcal{F}, P)$$ related to the random experiment. The probability measure $$P$$ is called the population. The data set or random element that produces the data is called a sample from $$P$$... In a statistical problem, the population $$P$$ is at least partially unknown and we would like to deduce some properties of $$P$$ based on the available sample.

So, the probability formulas of the mean and variance assume you have sufficient information about the population to calculate them.

The statistics formulas for the mean and variance are attempts to estimate the population mean and variance, given a sample of data. You could estimate the mean and variance in any number of ways, but the formulas you've provided are some standard ways of estimating the population mean and variance.

Now, one logical question is: why do we choose those formulas to estimate the population mean and variance?

For the mean formula you have there, one can observe that if you assume that your $$n$$ observations can be represented as observed values of independent and identically distributed random variables $$X_1, \dots, X_n$$ with mean $$\mu$$, $$\mathbb{E}\left[\dfrac{1}{n}\sum_{i=1}^{n}X_i \right] = \mu$$ which is the population mean. We say then that $$\dfrac{1}{n}\sum_{i=1}^{n}X_i$$ is an "unbiased estimator" of the population mean.

From Wackerly et al.'s Mathematical Statistics with Applications, 7th edition, chapter 7.1:

For example, suppose we want to estimate a population mean $$\mu$$. If we obtain a random sample of $$n$$ observations $$y_1, y_2, \dots, y_n$$, it seems reasonable to estimate $$\mu$$ with the sample mean $$\bar{y} = \dfrac{1}{n}\sum_{i=1}^{n}y_i$$

The goodness of this estimate depends on the behavior of the random variables $$Y_1, Y_2, \dots, Y_n$$ and the effect this has on $$\bar{Y} = (1/n)\sum_{i=1}^{n}Y_i$$.

Note. In statistics, it is customary to use lowercase $$x_i$$ to represent observed values of random variables; we then call $$\frac{1}{n}\sum_{i=1}^{n}x_i$$ an "estimate" of the population mean (notice the difference between "estimator" and "estimate").

For the variance estimator, it is customary to use $$n-1$$ in the denominator, because if we assume the random variables have finite variance $$\sigma^2$$, it can be shown that $$\mathbb{E}\left[\dfrac{1}{n-1}\sum_{i=1}^{n}\left(X_i - \dfrac{1}{n}\sum_{j=1}^{n}X_j \right)^2 \right] = \sigma^2\text{.}$$ Thus $$\dfrac{1}{n-1}\sum_{i=1}^{n}\left(X_i - \dfrac{1}{n}\sum_{j=1}^{n}X_j \right)^2$$ is an unbiased estimator of $$\sigma^2$$, the population variance.

It is also worth noting that the formula you have there has expected value $$\dfrac{n-1}{n}\sigma^2$$ and $$\dfrac{n-1}{n} < 1$$ so on average, it will tend to underestimate the population variance.

From Wackerly et al.'s Mathematical Statistics with Applications, 7th edition, chapter 7.2:

For example, suppose that we wish to make an inference about the population variance $$\sigma^2$$ based on a random sample $$Y_1, Y_2, \dots, Y_n$$ from a normal population... a good estimator of $$\sigma^2$$ is the sample variance $$S^2 = \dfrac{1}{n-1}\sum_{i=1}^{n}(Y_i - \bar{Y})^2\text{.}$$

The estimators for the mean and variance above are examples of point estimators. From Casella and Berger's Statistical Inference, Chapter 7.1:

The rationale behind point estimation is quite simple. When sampling is from a population described by a pdf or pmf $$f(x \mid \theta)$$, knowledge of $$\theta$$ yields knowledge of the entire population. Hence, it is natural to seek a method of finding a good estimator of the point $$\theta$$, that is, a good point estimator. It is also the case that the parameter $$\theta$$ has a meaningful physical interpretation (as in the case of a population) so there is direct interest in obtaining a good point estimate of $$\theta$$. It may also be the case that some function of $$\theta$$, say $$\tau(\theta)$$ is of interest.

There is, of course, a lot more that I'm ignoring for now (and one could write an entire textbook, honestly, on this topic), but I hope this clarifies things.

Note. I know that many textbooks use the terms "sample mean" and "sample variance" to describe the estimators above. While "sample mean" tends to be very standard terminology, I disagree with the use of "sample variance" to describe an estimator of the variance; some use $$n - 1$$ in the denominator, and some use $$n$$ in the denominator. Also, as I mentioned above, there are a multitude of ways that one could estimate the mean and variance; I personally think the use of the word "sample" used to describe such estimators makes it seem like other estimators don't exist, and is thus misleading in that way.

## In Common Parlance

This answer is informed primarily by my practical experience in statistics and data analytics, having worked in the fields for about 6 years as a professional. (As an aside, I find one serious deficiency with statistics and data analysis books is providing mathematical theory and how to approach problems in practice.)

Is there some consistent language for distinguishing these two seemingly different mean and variance terminologies? For example, if my cashier asks me about the "mean weight" of two items, do I ask him/her for the probabilistic distribution of the random variable whose realization are the weights of these two items (def 1), or do I just add up the value and divide (def 2)? How do I know which mean the person is talking about?

In most cases, you want to just stick with the statistical definitions. Most people do not think of statistics as attempting to estimate quantities relevant to a population, and thus are not thinking "I am trying to estimate a population quantity using an estimate driven by data." In such situations, people are just looking for summaries of the data they've provided you, known as descriptive statistics.

The whole idea of estimating quantities relevant to a population using a sample is known as inferential statistics. While (from my perspective) most of statistics tends to focus on statistical inference, in practice, most people - especially if they've not had substantial statistical training - do not approach statistics with this mindset. Most people whom I've worked with think "statistics" is just descriptive statistics.

In descriptive data analysis, a few summary measures may be calculated, for example, the sample mean... and the sample variance... However, what is the relationship between $$\bar{x}$$ and $$\theta$$ [a population quantity]? Are they close (if not equal) in some sense? The sample variance $$s^2$$ is clearly an average of squared deviations of $$x_i$$'s from their mean. But, what kind of information does $$s^2$$ provide?... These questions cannot be answered in descriptive data analysis.

## Other remarks about the sample mean and sample variance formulas

Let $$\bar{X}_n$$ and $$S^2_n$$ denote the sample mean and sample variance formulas provided earlier. The following are properties of these estimators:

• They are unbiased for $$\mu$$ and $$\sigma^2$$, as explained earlier. This is a relatively simple probability exercise.
• They are consistent for $$\mu$$ and $$\sigma^2$$. Since you know measure theory, assume all random variables are defined over a probability space $$(\Omega, \mathcal{F}, P)$$. It follows that $$\bar{X}_n \overset{P}{\to} \mu$$ and and $$S^2_n \overset{P}{\to} \sigma^2$$, where $$\overset{P}{\to}$$ denotes convergence in probability, also known as convergence with respect to the measure $$P$$. See https://math.stackexchange.com/a/1655827/81560 for the sample variance (observe that the estimator with the $$n$$ in the denominator is used here; simply multiply by $$\dfrac{n-1}{n}$$ and apply a result by Slutsky) and Proving a sample mean converges in probability to the true mean for the sample mean. As a stronger result, convergence is almost sure with respect to $$P$$ in both cases (Sample variance converge almost surely).
• If one assumes $$X_1, \dots, X_n$$ are independent and identically distributed based on a normal distribution with mean $$\mu$$ and variance $$\sigma^2$$, one has that $$\dfrac{\sqrt{n}(\bar{X}_n - \mu)}{\sqrt{S_n^2}}$$ follows a $$t$$-distribution with $$n-1$$ degrees of freedom, which converges in distribution to a normally-distributed random variable with mean $$0$$ and variance $$1$$. This is a modification of the central limit theorem.
• If one assumes $$X_1, \dots, X_n$$ are independent and identically distributed based on a normal distribution with mean $$\mu$$ and variance $$\sigma^2$$, $$\bar{X}_n$$ and $$S^2_n$$ are uniformly minimum-variance unbiased estimators (UMVUEs) for $$\mu$$ and $$\sigma^2$$ respectively. It also follows that $$\bar{X}_n$$ and $$S^2_n$$ are independent, through - as mentioned by Michael Hardy - showing that $$\text{Cov}(\bar{X}_n, X_i - \bar{X}_n) = 0$$ for each $$i = 1, \dots, n$$, or as one can learn from more advanced statistical inference courses, an application of Basu's Theorem (see, e.g., Casella and Berger's Statistical Inference).
• I believe the expressions for mean and variance estimators should be accompanied with a word or two on the central limit theorem. Dec 23, 2020 at 5:52
• You wrote: "If one assumes $X_1,\ldots,X_n$ are normally distributed, $\bar X_n$ and $S_n^2$ are" UMVUEs. (This after assuming they are i.i.d.) I would add that under those assumptions $\bar X_n$ and $S_n^2$ are independent, and that is used in deriving the t-distribution. One of the quickest ways to show that involves showing that $\operatorname{cov} \left(\,\overline X, X_i-\overline X\,\right)=0.$ When $U,V$ are both linear combinations of i.i.d. normals, then their covariance is $0$ only if they are independent. Dec 26, 2020 at 21:47
• @MichaelHardy Thanks for correcting me; I've put in the edits. Dec 26, 2020 at 22:08
• @Clarinetist : I see you say it's "through an application of Basu's theorem," but nothing so advanced as Basu's theorem is needed, since the method described in my comment above is enough. It seems misleading to tacitly suggest that you can't learn to prove that proposition until you learn Basu's theorem. Dec 26, 2020 at 22:11
• @MichaelHardy Thanks, I've corrected that as well. Dec 26, 2020 at 22:13

The first definitions you gave are correct and standard, and statisticians and data scientists will agree with this. (These definitions are given in statistics textbooks.) The second set of quantities you described are called the "sample mean" and the "sample variance", not mean and variance.

Given a random sample from a random variable $$X$$, the sample mean and sample variance are natural ways to estimate the expected value and variance of $$X$$.

• Thanks! So the distinction is one is "sample" mean/variance the other is mean/variance as it is. And also those formulas are estimate of the value of mean and variance. But isn't it true when people are working with them in practice, say using some data science package such as Scikit-Learn, the underlying probabilistic assumption (such as distribution of data) is dropped as if nothing ever happened? I have never seen anyone writing a documentation mentioning the probabilistic definition. I think this causes confusion for young practictioners, don't you think? Dec 23, 2020 at 5:39
• Imagine in the shoes of a young data scientist who just took a course on Lebesgue integration or measure theory. You spent 4 month talking about these abstract definitions, and then integrating these really hard integrals just to get the mean, using various tricks just to make these integrals tractable. You get your first job and your boss tells you just to add stuff up and divide. Isn't this confusing? Dec 23, 2020 at 5:41
• @Norman My response to your comments here is that probability is not a statistics class, and trying to act as if that is the case would be lying. Additionally, most managers have not taken courses nor understand the difference between probability and statistics, and are focused solely on descriptive statistics to understand empirically the behavior of their data. They calculate a sample average because that's what they habitually have done, and they think it is an adequate summary of the data. Dec 23, 2020 at 13:25
• @Norman I could go on a very long rant about how poorly taught statistics is in many schools and how very little guidance is given on how to approach statistical problems in practice, but I will avoid that for now. Dec 23, 2020 at 13:27
• @Norman I think it can be a source of confusion that people often say mean or variance when they really mean "sample mean" or "sample variance". One of the challenges of coming into the engineering world from a pure math background is that often people speak much less precisely than one is used to. Dec 23, 2020 at 15:23

Other answers — particularly Clarinetist’s — give excellent rundowns of the most important side of the answer. Given a random variable, we can sample it, and use the sample mean (defined in the statistical sense) to estimate the actual mean of the random variable (defined in the sense of probability theory), and similarly for variance, etc.

But the connection in the other direction doesn’t seem to have been mentioned yet. This is not as important, but it’s much more straightforward, and worth pointing out. Given a sample, i.e. a finite multiset of values $$\{x_i\}_{i \in I}$$, we can “consider this as a distribution”, i.e. take a random variable $$X$$, with value $$x_i$$ for $$i$$ distributed uniformly over $$I$$. Then the mean, variance, etc. of $$X$$ (in the sense of probability theory) will be precisely the mean, variance, etc. of the original multiset (defined in the statistical sense).

The general expression for arithmetic mean is $$\frac{\sum\limits_{i= 1}^n w_i x_i}{\sum\limits_{i= 1}^n w_i}$$, or even more generally, $$\frac{\int w(t) f(t)dt}{\int f(t)dt}$$ (there are then ways to recover the discrete case from that).

If you set all $$w_i$$ to $$1$$, or really to anything as long as it's constant, you get $$\dfrac{1}{n}\sum\limits_{i= 1}^n x_i$$. This is referred to as an "unweighted" average, although technically it's still weighted, it's just that you're multiplying everything by $$1$$, so you don't notice it. If you set $$p_i = \frac{w_i}{ \sum\limits_{k= 1}^n w_k}$$, and interpret $$p_i$$ as the probability of the $$i$$th event, then you get the average weighted by probability, which is also known as expected value.

You have to be careful about "unweighted" averages, as they often actually are weighted, but by weights that you didn't want. For instance, suppose you want the average income over the US, and you have the average income for each state individually. You could just add all of those averages together, and then divide by $$50$$. People will often call this the "unweighted" or "simple" average, but you're actually weighting people by the reciprocal of their states population; the fewer people there are in a state, the more each person's income affects that state's average, and so the more they affect the total "unweighted average". As a result, to get the actual overall average income from the individual states' averages, you have to multiply each state's average by its population to get its total income, add all of those together, and then divide by the total population.

A common weighting you'll see is frequency weighting. This is where you multiply each value by the number of times it appears. For instance, if you measure something once a month for a year, and the only values you get are $$0$$, $$1$$, and $$2$$, taking the simple average of those values gives you $$1$$. But that's probably not the real average. To get a more meaningful average, you should take each of these values, weight them by how many months they appear for, and then take the average.

One property of weighted averages is that multiplying all of the weights by a constant number doesn't change the final result (you'll just divide it out again when you divide by the total of the weights). So weighting by the frequencies is equivalent to weighting by the percentage of cases each value is. That is, if $$0$$ is the value for $$5$$ months, $$1$$ is the value for $$4$$ months, and $$2$$ is the value for $$3$$, the weighting of $$5,4,3$$ is equivalent to the weighting of $$\frac 5 {12},\frac 4{12},\frac 3{12}$$.

So if you have a probability distribution where one thing has a $$60$$% chance of happening, and something else has a $$40$$% chance of happening, the expected value is simply the frequency weighted average.

For example, if my cashier asks me about the "mean weight" of two items, do I ask him/her for the probabilistic distribution of the random variable whose realization are the weights of these two items (def 1), or do I just add up the value and divide (def 2)? How do I know which mean the person is talking about?

The expected value of a random variable is the expected value of that random variable. It is a property of the distribution. If you're asked for the expected value of a random variable, you find the expected value of that random variable. If you aren't asked for the expected value of a random variable, you don't go looking for a random variable to take the expected value of. How you know whether you take the expected value of random variable is if there is a random variable to take the expected value of. Even if values come from a distribution, the average of those values is the average of those values, not the average of the distribution they came from.

If people are talking rigorously, they will explicitly say they want the expected value. You many, however, see people asking for the "mean" or "average", when they really want the expected value, but you can recognize those cases by there being a random variable. For instance, if someone asks "What's the average payout of this slot machine?", the context suggests that you should take the expected value of the distribution of payouts, and not simply take the set of different possible payouts and take the simple average. There could be some ambiguity as to whether "the payout" refers to the random process that pays money out, in which case you should take the expected value of the distribution (population mean), or the actual money paid out, in which case you should take the average of all the actual payouts made by the machine (sample mean), but in the latter case, you still should take the frequency weighted average.

The confusion actually comes from notation, where symbols mean different things in two formulas.

First, let's take a look at the "probabilistic" definition: $$\mathbb{E}[X] = \frac{1}{n}\sum_{i=1}^n x_i p(x_i)$$ Here the random variable $$X$$ takes $$n$$ distinct values $$x_1,\ldots, x_n$$, each with a probability mass function $$p(x_i)$$.

In the "statistical" definition we have an estimate of the expected value, based on the observed values of the random variable $$z_1, \ldots, z_N$$: $$\hat{\mathbb{E}}[X] = \frac{1}{N}\sum_{j=1}^N z_j$$ Notice that I renamed the variables compared to your original formula, so as to avoid the confusion. Here $$N$$ is the number of observations, and $$z_j$$s are the actual observations. For example, if $$X$$ is a random variable representing rolls of a loaded die, then $$n = 6$$ and $${x_i} = \{1, 2, 3, 4, 5, 6\}$$; whereas $$N$$ can be arbitrarily large, and $$z_j$$s will just be a long sequence of random draws from the set of $${x_i}$$s.

Now, you can rewrite the "statistical" formula by collecting different values of $$z_j$$s into the groups according to which $$x_i$$ they correspond to (for example, first collect all 1s, then all 2s, etc): $$\hat{\mathbb{E}}[X] = \frac{1}{N}\big(x_1\cdot|\{z_j=x_1\}| + \cdots+x_n\cdot|\{z_j=x_n\}|\big) \\ =\frac{1}{N}\sum_{i=1}^n x_i \sum_{j=1}^N\mathbb{1}[z_j=x_i] \\ =\sum_{i=1}^n x_i \hat{p}(x_i)$$ where $$\hat{p}(x_i) = \frac{1}{N}\sum_{j=1}^N \mathbb{1}[z_j=x_i]$$ is the estimate of the probability mass function: the number of times a value $$x_i$$ was encountered in the sample divided by the sample size, i.e. the observed frequency of the value $$x_i$$.

Now you can see that the "probabilistic" and "statistical" definitions are actually the same, with the only difference that we replace the theoretical mass distribution function $$p(x)$$ (which may not be known) with the empirical (observed) mass distribution function $$\hat{p}(x)$$.

Statement: the "statistic" universe from your question is a partial case of "probabilistic" universe.

We need some notation. Suppose that we have a random variable $$\xi$$ in sense of probability theory and $$P(\xi = x_k) = p_k$$, $$0 \le k \le n$$.

For example, consider $$\xi$$ which is equal to the number of children in "abstract" random family. Put $$x_k = k$$. Then $$E \xi = \sum_{k=0}^n x_k p_k$$ where $$p_k = P(\xi = x_k)$$, and $$D\xi = E (\xi - E\xi)^2 = \sum_{k=0}^n p_k (x_k - E\xi )^2$$. Moreover, $$P(\xi = 69) > 0$$ (according to Guinness World Records, I am not sure, but let us suppose that it's a record nowadays) and as there was people with $$69$$ children then we may think that for example $$P(80) > 0$$ - it's natural, because $$80$$ children is possible, even although Guinness World Records says that still nobody had $$80$$ children.

In reality we don't have abstract random family, random mothers, random fathers and random children. We have finite number $$N$$ ( $$10^6 < N < 10^{100}$$) of families, let us number them, and there are $$y_1$$ children in the first family, there are $$y_2$$ children in family number $$2$$, ..., there are $$y_N$$ children in family $$N$$.

Analogy: $$\xi$$ corresponds to a fair dice itself and numbers $$y_1$$, $$y_2$$, ... corresponds to dice rolls and have form: $$5, 1, 6, 6, 3, ...$$, these are fixed numbers.

Now consider a random value $$\tau$$, which has uniform distribution on $$\{ 1, 2, \ldots, N\}$$. It means that $$P(\tau = k) = \frac{1}{N}$$ for all $$1 \le k \le N$$.

Numbers $$y_1, \ldots, y_N$$ are fixed - suppose they are written in some sociological table. Let us consider a random value $$y_{\tau}$$. It's not a number of children in abstract random family. It is a number of children in some real family, if a number $$\tau$$ of such family we chose randomly.

Let us find $$E y_{\tau}$$ and $$D y_{\tau}$$. With probability $$\frac{1}N$$ we have $$\tau = k$$ and hence $$y_{\tau} = y_k$$. Thus $$E y_{\tau}=\frac{1}N \sum_{k=1}^N y_k$$ and $$D y_{\tau} = E (y_{\tau} - Ey_{\tau})^2 = E (y_{\tau} - \frac{1}N \sum_{k=1}^N y_k )^2 = \frac{1}N \sum_{k=0}^N (y_k -\frac{1}N \sum_{k=1}^N y_k )^2.$$

Now the correspondence of "statistical universe" and "probabilistic universe" is obvious.

Notice that all $$y_i \le 69$$ and $$y_i = 69$$ is the maximum value, but the maximum value of $$\xi$$ is bigger (and not less than $$80$$).

So, we have shown that the "statistical universe" is a partial case of "probabilistic" universe.

Moreover, there are two moments, when there is randomness:

1. when we go from abstract random family, corresponding to $$\xi$$, to numbers $$y_1, \ldots, y_N$$ from sociological handbook.

As it was already mentioned, there is analogy: $$\xi$$ corresponds to a fair dice itself and numbers $$y_1$$, $$y_2$$, ... corresponds to dice rolls and have form: $$5, 1, 6, 6, 3, ...$$, these are fixed numbers.

1. [here we suppose that $$y_1, \ldots, y_N$$ are fixed numbers] when we go from the whole sociological handbook to a family with random number $$\tau$$ and see, how much children is there in this family. The number is $$y_{\tau}$$.

When from $$y_{\tau}$$ we go to $$E y_{\tau} = \frac{1}N \sum_{k=1}^N y_k$$, we get rid of the second randomness, but we still do not get rid of the first randomness. It's connected with the fact that numbers $$y_k$$ could be another: it could happen, that in some families could be more children, and in some families could be less children, if the circumstances were different. In this sence numbers $$y_i$$ are not fixed: they are r.v. sampled from distribution, corresponding to $$\xi$$. And in this sence $$\frac{1}N \sum_{k=1} y_k$$ is a random value, and we may write limit theorems such as L.L.N.: $$\frac{1}N \sum_{k=1}^N y_k \to E\xi, \text{ }N \to \infty$$ or CLT or LIL.

Addition: it was shown that if $$y_1, \ldots, y_N$$ are fixed and $$\tau$$ is random then "probabilistic" mean (expectation) of $$y_{\tau}$$ and "probabilistic" variance of $$y_{\tau}$$ are well known sample mean and sample variance.

Hope it will be useful. If you have any questions, you are welcome.


# Riemann-Darboux Integrability of Subinterval

I'm studying Riemann-Darboux integration. I'm trying to prove the following rather intuitive notion for integrals. Please let me know if you find any errors in this proof, as I'm self-studying this topic.

Theorem: Suppose $$f$$ is Riemann-Darboux integrable on $$[a,b]$$. Let $$c\in(a,b)$$. Then, $$f$$ is Riemann-Darboux integrable on the intervals $$[a,c]$$ and $$[c,b]$$.

Attempted Proof: Since $$f$$ is Riemann-Darboux integrable on $$[a,b]$$, for arbitrary $$\epsilon>0$$, there exists a partition $$P$$ of $$[a,b]$$ such that $$U(f,P)-L(f,P)<\epsilon$$.

Let $$n_p,n_{p*}$$, and $$n_{p'}$$ be the number of partition parts in $$P$$, $$P^*$$, and $$P'$$, respectively. Also, let $$m_i=\inf_{x\in[x_{i-1},x_i]}f(x)$$ and $$M_i=\sup_{x\in[x_{i-1},x_i]}f(x)$$.

Consider the partition of $$[a,c]$$ given by $$P^*=P\cap[a,c]$$. Then, $$U(f,P^*)-L(f,P^*)=\sum_{i=1}^{n_{p*}}(M_i-m_i)\Delta x_i\le\sum_{i=1}^{n_{p*}}(M_i-m_i)\Delta x_i+ \sum_{n_{p*}+1}^{n_p}(M_i-m_i)\Delta x_i=\sum_{i=1}^{n_p}(M_i-m_i)\Delta x_i=U(f,P)-L(f,P)<\epsilon$$ Therefore, $$f$$ is integrable on $$[a,c]$$.

Next, consider the partition $$[a,b]$$ given by $$P'=P\cap[c,b]$$. Then,

$$U(f,P')-L(f,P')=\sum_{i=n_{p*}+1}^{n_{p'}}(M_i-m_i)\Delta x_i\le\sum_{i=1}^{n_{p*}}(M_i-m_i)\Delta x_i+ \sum_{n_{p*}+1}^{n_p}(M_i-m_i)\Delta x_i=\sum_{i=1}^{n_p}(M_i-m_i)\Delta x_i=U(f,P)-L(f,P)<\epsilon$$ Therefore, $$f$$ is integrable on $$[c,b]$$. $$\square$$

Any and all feedback, or alternative proofs are appreciated. I love to see different arguments to expand my skill set.

This is essentially fine, although you are tacitly assuming that $$c$$ itself is a member of the partition, $$P$$. This isn't a big deal though - a $$P$$ such that $$U(P,f)-L(P,f)<\epsilon$$ is guaranteed by integrability, and you can always just add $$c$$ to this partition if it is not already there. Namely, if $$P_{c}$$ is this new partition, called a refinement of $$P$$, one must have $$U(P_{c},f)-L(P_{c},f)\leq U(P,f)-L(P,f)<\epsilon$$ and you can proceed as you have done in your proof.
An alternative approach is given by Lebesgue's Criterion for Riemann Integrability, which states that Riemann/Darboux Integrability is equivalent to boundedness plus continuity up to a null set (see Wikipedia for a good explanation of null sets). Since your functions is integrable on $$[a,b]$$, it is bounded and continuous up to a null set on $$[a,b]$$, and hence, on $$[a,c]$$ and $$[c,b]$$ as well.


# Math Help - Pricing Dice Game

1. ## Pricing Dice Game

Stage 1

In this game I roll the dice and which ever number the dice lands on I pay you that amount of money. For example if the dice lands on 4 I pay you £4 pounds.

A fair price to play this game should be £3.50 as then the player or myself should not win any money if this game is played infinty times.

Stage 2

How much should the fair price be to play the game if their is an option after the first roll to roll again. For example if you roll a 1 first go you will certainly want to roll again as youwill always roll the same or better. However if you were to roll a 6 you would not want the option to roll again as you have already won the maximum amount of money.

I priced this option at £3.57, however I been told that is wrong. Can anyone explain why?

Calypso

2. Originally Posted by calypso
Stage 1

In this game I roll the dice and which ever number the dice lands on I pay you that amount of money. For example if the dice lands on 4 I pay you £4 pounds.

A fair price to play this game should be £3.50 as then the player or myself should not win any money if this game is played infinty times.

Stage 2

How much should the fair price be to play the game if their is an option after the first roll to roll again. For example if you roll a 1 first go you will certainly want to roll again as youwill always roll the same or better. However if you were to roll a 6 you would not want the option to roll again as you have already won the maximum amount of money.

I priced this option at £3.57, however I been told that is wrong. Can anyone explain why?

Calypso
Choose a $t$ such that if the first roll $a\le t$ roll again otherwise accept the prize of $£ a$.

Now for any given $t$ you can work out the value of this game and hence a fair price. The fair price for the game is the maximum fair price.

(I make $t=3$ the value of $t$ that maximises the return and a fair price is then $£4.25$

CB

3. Sorry for the late reply I have just got back from holiday.

The text book answer says something similar, ie you should stick if you roll anything greater than 3.50. Therefore the fair price of the option of two rolls is

( 3.5 + 3.5 + 3.5 + 4 + 5 + 6 ) / 6 = 4.25

I can understand why this is the correct answer, however I am still confuced as why I cant get this answer with my method:

So I think the fair price of the game is defined by: Sum of all possible payouts / No. of possible payout

So therefore the possible outcomes of the game are

1 -> 1
1 -> 2
1 -> 3
1 -> 4
1 -> 5
1 -> 6

2 -> 1
2 -> 2
2 -> 3
2 -> 4
2 -> 5
2 -> 6

3 -> 1
3 -> 2
3 -> 3
3 -> 4
3 -> 5
3 -> 6

4
5
6

Therefore fair price should equal 78/21 = £3.71?

Thanks again

Calypso

4. Originally Posted by calypso
I am still confuced as why I cant get this answer with my method:

So I think the fair price of the game is defined by: Sum of all possible payouts / No. of possible payout

So therefore the possible outcomes of the game are

1 -> 1
1 -> 2
1 -> 3
1 -> 4
1 -> 5
1 -> 6

2 -> 1
2 -> 2
2 -> 3
2 -> 4
2 -> 5
2 -> 6

3 -> 1
3 -> 2
3 -> 3
3 -> 4
3 -> 5
3 -> 6

4
5
6
The reason this goes wrong is that those outcomes are not all equally probable. Each of the final three outcomes (4, 5, 6) occurs with a probability of 1/6. But the initial outcomes 1, 2, 3 are subdivided into six subcases, each of which occurs with a probability of only 1/36. Multiplying each outcome by its probability, you get $\tfrac3{36}(1+2+3+4+5+6) + \tfrac16(4+5+6) = \tfrac{17}4 = 4.25.$

5. Originally Posted by calypso
Sorry for the late reply I have just got back from holiday.

The text book answer says something similar, ie you should stick if you roll anything greater than 3.50. Therefore the fair price of the option of two rolls is

( 3.5 + 3.5 + 3.5 + 4 + 5 + 6 ) / 6 = 4.25

I can understand why this is the correct answer, however I am still confuced as why I cant get this answer with my method:
I don't like their decision rule, not that it is wrong, but because it refers to something not in the sample space (that is "anything greater than 3.50" when it could have said anything greater than 3 or even anything greater than or equal to 4).

CB

6. Great, thanks everyone for your help

Calypso


Math Help - die probability

1. die probability

"What is the probability of getting a total score of more than 7 when two fair dice are rolled together?"

I got $\frac{15}{36}$ but the book says $\frac{1}{3}$?

Thanks!

2. Hello, Nyoxis!

What is the probability of getting a total score of more than 7
when two fair dice are rolled together?"

I got $\frac{15}{36} \quad{\color{blue}\hdots\text{ Right!}}$
. . but the book says $\frac{1}{3}\quad {\color{red} \hdots\text{ Wrong!}}$
Can't imagine where they got their answer . . .

$\begin{array}{cccccc}(1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6) \\ (2,1) & (2,2) & (2,3) & (2,4) & (2,5) & {\color{blue}(2,6)} \\ (3,1) & (3,2) & (3,3) & (3,4) & {\color{blue}(3,5)} & {\color{blue}(3,6)} \\ (4,1) & (4,2) & (4,3) & {\color{blue}(4,4)} & {\color{blue}(4,5)} & {\color{blue}(4,6)}\end{array}$
$\begin{array}{cccccc}(5,1) & (5,2) & {\color{blue}(5,3)} & {\color{blue}(5,4)} & {\color{blue}(5,5)} & {\color{blue}(5,6)} \\
(6,1) & {\color{blue}(6,2)} & {\color{blue}(6,3)} & {\color{blue}(6,4)} & {\color{blue}(6,5)} & {\color{blue}(6,6)}\end{array}\quad\Leftarrow\quad \text{fifteen outcomes}> 7$

3. Heh 'Cambridge Advanced Mathematics' wrong.
Thanks!


# Prove that $x_{n+1}=\frac{x_n}{2}+\frac{1}{x_n}\geq\sqrt{2}$

Let $(x_n)_{n\in\mathbb N}$ be a recursively defined sequence with $x_1=9$ and $$x_{n+1}=\frac{x_n}{2}+\frac{1}{x_n}\text{ for }n\geq 1.$$ Show that $x_n\geq\sqrt{2}$ for all $n$.

Because $x_n\geq 0$ one can easily prove inductively that $$x_n^2\geq 2\Leftrightarrow x_n^2-2\geq 0\Leftrightarrow\left(\frac{x_{n-1}}{2}-\frac{1}{x_{n-1}}\right)^2\geq 0,$$ hence $x_n\geq\sqrt{2}$. However I have seen another approch which I was very curious about because I don't have the feeling that this can be done without proper justification other than the induction hypothesis:

$$x_{n+1}=\frac{x_n}{2}+\frac{1}{x_n}\overset{(*)}{\geq}\frac{\sqrt{2}}{2}+\frac{1}{\sqrt{2}}=\sqrt{2}$$

For $(*)$ it is assumed that $x_n\geq\sqrt{2}$ holds for an $n\in\mathbb N$. Are there any objections about this consideration?

• Hint: AM-GM inequality is fast. Your second approach is flawed, as the reciprocal term does not follow in the inequality * – Macavity Nov 2 '15 at 14:47
• The inequality $\frac{1}{x_n} \geq \frac{1}{\sqrt{2}}$ is false because it would imply that $x_n \leq \sqrt{2}$ (this is true only if $x_n=\sqrt{2}$ but else.... ) – user252450 Nov 2 '15 at 14:50
• A straightforward approach to showing that $x_n\ge \sqrt{2}$ is to write $$x_{n}-\sqrt{2}=\left(\frac{x_{n-1}}{2}+\frac1{x_{n-1}}\right)-\sqrt{2}=\frac{(x_{n-1}-\sqrt{2})^2}{2x_{n-1}}\ge0$$ – Mark Viola Nov 2 '15 at 15:00
• @Dr.MV By far the simplest. – Did Nov 2 '15 at 15:01
• @Dr.MV Yeah pretty similiar to my explained approach in the first part of my post. I was just wondering about the other approach and its correctness there. – Christian Ivicevic Nov 2 '15 at 15:08

Yes, you can make the assumption $x_n\ge \sqrt{2}$ to prove $x_{n+1}\ge \sqrt{2}$ as per strong form of mathematical induction.

But your reasoning is wrong as $x_n\ge \sqrt{2} \Rightarrow \frac{1}{x_n} \le \frac{1}{\sqrt{2}}$.

Better is to use AM-GM inequality as Macavity has mentioned since the quantities are all positive.

$$\frac{\frac{x_n}{2}+\frac{1}{x_n}}{2}\ge \sqrt{\frac{x_n}{2}\cdot\frac{1}{x_n}}$$ $$x_{n+1}\ge \sqrt{2}$$

If you know derivatives you can make it work as follows. Consider the function : $$f(x)=\frac{x^2+2}{2x}$$ Now calculate the derivative : $$f'(x)=\frac{2x^2-4}{4x^2}\geq 0$$ if $x\geq \sqrt{2}$ so the function is increasing for $x\geq \sqrt{2}$

Now from the induction hypothesis $x_n\geq \sqrt{2}$ so from the monotony : $$x_{n+1}=f(x_n)\geq(\sqrt{2})=\frac{4}{2\sqrt{2}}=\sqrt{2}$$ as wanted .

Although the following goes a bit beyond addressing the OP's question, I thought that it would be instructive. So, let's have a look at the sequence generated by the recurrence relationship $x_n=\frac12 x_{n-1}+\frac{1}{x_{n-1}}$.

This is actually an interesting recursively generated sequence. If we examine the first difference $x_{n}-x_{n-1}$ we find that

$$x_{n}-x_{n-1}=\frac{2-x_{n-1}^2}{2x_{n-1}} \tag 1$$

We deduce immediately from $(1)$ that if $x_n > \sqrt{2}$, then $x_n$ is decreasing while if $x_n<\sqrt{2}$, then $x_n$ is increasing.

Now, a straightforward approach to showing that $x_n\ge \sqrt{2}$ is to write

$$x_{n}-\sqrt{2}=\left(\frac{x_{n-1}}{2}+\frac1{x_{n-1}}\right)-\sqrt{2}=\frac{(x_{n-1}-\sqrt{2})^2}{2x_{n-1}}\ge0 \tag 2$$

regardless of the value of $x_{n-1}$.

Therefore, the only value of the sequence that can ever be less than $\sqrt{2}$ is its initial "seed" value. From $(1)$, we see that if the seed value is less than $\sqrt{2}$, the subsequent value increases. And $(2)$ shows that it increases above $\sqrt{2}$. Then, the remaining sequence decreases monotonically to the limit value of $\sqrt{2}$.

Hint: study the function $f(x)=x/2 +1/x$ it increases if $x>\sqrt2$


# Contour Integration (Choice of Contour)

Let $\alpha \le 0$ and $\sigma > 0$ .

I want to choose a contour, including $[\sigma - iR, \sigma+iR]$ , such that i can apply Cauchy's Residue theorem and evaluate:

$$\lim_{R \rightarrow \infty} \int\limits_{\sigma - iR}^{ \sigma+iR } \frac{\exp(\alpha z)}{(z^2 + 1)} dz$$

by contour integration. The case for $\alpha > 0$ is nice i think, as you can use a semi-circle, but i cant seem to find a suitable contour for this case.

EDIT: to be specific when we use a semi-circle then the integral over the circular path tends to zero in the limit for $\alpha > 0$ , but this does not seem to happen in the above case.

• How do you expect to apply CRT to a complex integral on a straight line in the complex plane?? We usually do this to calculate real integrals by means of a closed contour in the complex plane... – DonAntonio Dec 28 '13 at 15:36
• see my edit, in the limit the only non-zero section of our contour integral is that integrated over the line. it worked for the earlier case, unless i am completely confusesd. – KJC Dec 28 '13 at 15:38

You can choose a semicircle in the half-plane $\operatorname{Re} z \geqslant \sigma$. Then $e^{\alpha z}$ is bounded on the contour - $\operatorname{Re} (\alpha z) = \alpha\operatorname{Re} z \leqslant \alpha\sigma \leqslant 0$, and $\lvert e^{\alpha z}\rvert = e^{\operatorname{Re} (\alpha z)}$ - and the integral over the semicircle tends to $0$ for $R \to \infty$. Since the contour encloses no singularity, the contour integral is $0$, and hence

$$\int_{\sigma - i\infty}^{\sigma+i\infty} \frac{e^{\alpha z}}{z^2+1}\,dz = 0$$

for $\sigma > 0$ and $\alpha \leqslant 0$.

It is worth noting that

$$I(\alpha) = \int_{\sigma-i\infty}^{\sigma+i\infty} \frac{e^{\alpha z}}{z^2+1}\,dz = \begin{cases}\quad 0 &, \alpha \leqslant 0\\ 2\pi i\sin \alpha &, \alpha \geqslant 0 \end{cases}$$

does not depend smoothly on $\alpha$, although the integrand does. That should, however, not be too surprising since the differentiated integrand $\dfrac{ze^{\alpha z}}{z^2+1}$ is not integrable anymore, and the non-differentiability in $0$ manifests itself by the need to switch the half-plane in which the contour is closed to have the integral over the auxiliary part tend to $0$.

• could you show that $e^{ \alpha z}$ is bounded on this contour? im not seeing that. – KJC Dec 28 '13 at 15:50
• Added the estimate for that. – Daniel Fischer Dec 28 '13 at 15:58
• excellent explanation – KJC Dec 28 '13 at 22:52


Prove that: $\lfloor n^{1/2}\rfloor+\cdots+\lfloor n^{1/n}\rfloor=\lfloor \log_2n\rfloor +\cdots+\lfloor \log_nn \rfloor$, for $n > 1$

Prove that: $\lfloor n^{1/2}\rfloor+\cdots+\lfloor n^{1/n}\rfloor=\lfloor \log_2n\rfloor +\cdots+\lfloor \log_nn \rfloor$, for $n > 1,\, n\in \mathbb{N}$

For example. For $n=2$, we have $\lfloor 2^{1/2} \rfloor = \lfloor 1.414 \rfloor = 1$ whereas $\lfloor \log_2(2) \rfloor = 1$ while for $n=3$, we have $$\lfloor 3^{1/2} \rfloor + \lfloor 3^{1/3} \rfloor = \lfloor 1.732 \rfloor + \lfloor 1.442 \rfloor = 2= \lfloor 1.585 \rfloor + \lfloor 1 \rfloor=\lfloor \log_2(3) \rfloor + \lfloor \log_3(3) \rfloor .$$

I was thinking of using induction. So since $n=2$ is true, now assume for all $n$, this identity is true, we would like to prove that $n+1$ is true. Then

$$\lfloor n^{1/2} \rfloor + \lfloor n^{1/3} \rfloor + ... + \lfloor n^{1/n} \rfloor + \lfloor (n+1)^{1/(n+1)} \rfloor,$$ where $(n+1)^{1/(n+1)} > 1$ for all $n>1$ but it's strictly decreasing above 1 so $\lfloor (n+1)^{1/(n+1)} \rfloor = 1$

$\implies \lfloor n^{1/2} \rfloor + \lfloor n^{1/3} \rfloor +\cdots+ \lfloor n^{1/n} \rfloor + \lfloor (n+1)^{1/(n+1)} \rfloor = \lfloor n^{1/2} \rfloor + \lfloor n^{1/3} \rfloor +\cdots+ \lfloor n^{1/n} \rfloor + 1$

$= \lfloor \log_2(n) \rfloor + \lfloor \log_3(n) \rfloor + \cdots+ \lfloor \log_n(n) \rfloor + \lfloor \log_{n+1}(n+1) \rfloor$

since, $\log_{n+1}(n+1) = 1$ for all $n$.

My question is: How do we know that $(n+1)^{1/(n+1)}$ will never go below $1$? i.e., How can we prove that this function $f(x) = (x+1)^{1/(x+1)}$ is always bounded below by $1$ for $x>1$? (First, When $x=0$, $f(0)=1$, then looking at it's derivative, one can see that it's strictly increasing for $x$ between $(0,1)$ and decreasing for all $x>1$).

• You need to replace $n$ by $n + 1$ all over, not just in the last term, in the induction step. – vonbrand Feb 25 '14 at 21:57
• If $\llcorner x\lrcorner$ is supposed to mean the greatest integer not exceeding $x$, then you should probably have used $\lfloor x \rfloor$ (\lfloor … \rfloor) instead. If that's not what you mean, then what did you mean? – MJD Feb 25 '14 at 22:02
• thanks @MJD, I wasn't sure which code to use for it. Thanks! – PandaMan Feb 26 '14 at 1:37
• @vonbrand, I don't quite understand why you meant, please clarify. thanks – PandaMan Feb 26 '14 at 2:18
• @PandaMan, what you need to prove is $\lfloor (n+1)^{1/2}\rfloor + \ldots + \lfloor (n+1)^{1/(n+1)}\rfloor = \lfloor \log_2 (n+1)\rfloor +\ldots+\lfloor\log_{n+1} (n+1)\rfloor$ – vonbrand Feb 26 '14 at 5:04

This is a classic exercise and one with a very elegant solution.

The idea of the proof is to count the number $N$ of the points (see figure below) with integer coordinates, which lie in the region $$U=\big\{(x,y): 0<x\le n \,\,\,\text{and}\,\,\, 1<y\le n^{1/x}\big\},$$ and in particular, the red points, in two ways: horizontally and vertically.

Horizontal counting: $$N=\lfloor n^{1/2}\rfloor+\lfloor n^{1/3}\rfloor+\cdots+\lfloor n^{1/n}\rfloor,$$ since on the horizontal line $\,y=k\,$ lie exactly $\,\lfloor n^{1/k}\rfloor\,$ red points.

Vertical counting: $$N=\lfloor \log_2 n\rfloor+\lfloor\log_3 n\rfloor+\cdots+\lfloor \log_n n\rfloor,$$ since on the vertical line $\,x=k\,$ lie exactly $\,\lfloor \log_k n\rfloor\,$ red points.

$${}$$

Note that the curve in the figure above is of the function $y=n^{1/x}$.

This problem was first asked in a Soviet Mathematics Olympiad in 1982 (Всесоюзный Математический Олимпиад.)

• Gorgeous explanation! – Steven Stadnicki Feb 25 '14 at 22:24
• @StevenStadnicki: Correct. – Yiorgos S. Smyrlis Feb 25 '14 at 22:26
• (Thanks for the fix! I've removed the correction from my comment since that got edited in. :-) – Steven Stadnicki Feb 25 '14 at 22:28
• Great explanation. The wires are crossed a little though, because if $x=k$, then $y\leq n^{1/k}$, which means that the vertical count is the one that matches the fractional powers. Likewise, if $y = k$, then $k \leq n^{1/x}$, then $1\leq(\log_k n)/x$, so the horizontal count is the one that matches the logs. – John Moeller Mar 1 '14 at 0:13
• Awesome solution :-) – tarit goswami Aug 1 '18 at 8:54

Fix $b>1$. Then the derivative of $b^x$ is $\ln(b) b^x$; $\ln(b)$ is positive and $b^x$ is as well for all $x$, showing that that $b^x$ is a strictly increasing function. Next, $b^0=1$, showing that $b^x>1$ for all $x>0$.

Next, since $n+1>1$ and $1/(n+1)>0$, we have that $(n+1)^{1/(n+1)}>1$.


# Can a complete manifold have an uncountable number of ends?

Let $M$ be a complete and noncompact Riemannian manifold. Fix a point $p$ in $M$. Let $\gamma$: $[0, L]\rightarrow M$ (parametrized by its arc length) be a geodesic starting from $p$. Denote by $d(\cdot, \cdot)$ the distance function on $M$ induced by the Riemannian metric. $\gamma$ is minimal if $d(\gamma(s_1), \gamma(s_2))=|s_1-s_2|$. A ray is a minimal geodesic defined on $[0, +\infty)$. Since $M$ is noncompact, there exists at least one ray from $p$.

Two rays $\gamma_1, \gamma_2$ from the same point $p$ are called cofinal if for any $r\geq0$ and all $s>r$, $\gamma_1(s)$ and $\gamma_2(s)$ lie in the same component of $M\backslash B(0, r)$, where $B(0, r)=\{x\in M| d(p, x)<r\}$. An equivalence class of cofinal rays is called an end of $M$. My question is:

Can $M$ have uncountably many ends? It seems that the answer is no. But I am not very sure and I can not find a convictive proof.

• You might like to look at some basic notions of geometric group theory. The set of ends is a quasi-isometry invariant, and the Svarc--Milnor Lemma says that the universal cover of a compact manifold $M$ is quasi-isometric to a Cayley graph of $\pi_1M$. So it suffices to find a group with uncountably many ends, and the free group of rank 2 suffices. Curiously, it's actually impossible for a universal cover to have countably infinitely many ends.
– HJRW
Dec 18, 2015 at 20:35
• @FanZheng: Second countable spaces can have infinitely many ends. Take the universal cover of a wedge of two circles, for example. Dec 18, 2015 at 20:44
• @HJRW: The ends of the fundamental group aren't all that relevant to this question, although simply connected examples can be interesting. It's about whether $M$ has uncountably many ends (as in the answer of @SebastianGoette), not the universal cover of $M$. Dec 18, 2015 at 20:49
• @LeeMosher, I used bad notation, but I think my point stands (unless I misread the question). If $\pi_1N$ is free of rank 2 (say) then taking $M$ to be the universal cover of $N$ provides the example.
– HJRW
Dec 18, 2015 at 21:08
• @HJRW: Sure, for example $N=$ the connected sum of two copies of $S^2 \times S^1$. Dec 18, 2015 at 21:19

The answer is yes. Consider a hyperbolic pair of pants where all three boundary circles are of the same size. Glue countably many of them together such each new one is glued to the existing manifold along exactly one boundary circle. Then the manifold looks like the boundary of a fattened tree.

Fix $p$ in one of the pairs of pants $Y_0$. For all other pairs of pants $Y$, pick a point $q\in Y$, then the minimal geodesic joining $p$ and $q$ will leave $Y$ through one circle. Label the two remaining boundary circles with $0$ and $1$. Then all glueing circles get exactly one label, except for those bounding $Y_0$.

For each sequence $(a_n)_n\in\{0,1\}^{\mathbb N}$, we construct a ray from $p$ leaving $Y_0$ through the same boundary component. Whenever the ray enters the $n$-th pair of pants $Y$ along its way, we let it leave through the circle labelled $a_n$. This way, we construct an uncountable number of rays that are not pairwise cofinal.

• Thank you very much for your nice edit of my question and also for your clever solution. In this afternoon, I also found a very similar example called as the Cantor tree surface, inspired by your construction. Dec 19, 2015 at 10:34

Let me rephrase S. Goette's example in a slightly different way. Consider the complement of the triadic Cantor set in the complex plane. This is a Riemann surface that can be uniformized: there is a complete Riemannian metric of constant minus one curvature. The ends of this surface are in one-to-one correspondence with the points of the Cantor set.

You can see the decomposition into pairs of pants by drawing nested circles around the intervals that appear in the construction of the Cantor set.

In some sense, uncountably many ends is the generic case. Let us restrict ourselves to planar surfaces, namely complements of compact sets in the Riemann sphere. The set of compact subsets in a compact set has a natural topology, the Vietoris topology (which coincides with the topology given by the Hausdorff metric). So we can say that two open subsets are close if their complement are close with respect to this topology. It happens that generically a compact subset of the sphere has uncountably many connected components, hence its complement has uncountably many ends.


Question

# I have a total of Rs. $$300$$ in coins of denomination Re. $$1$$, Rs. $$2$$ and Rs. $$5$$. The number of Rs. $$2$$ coins is $$3$$ times the number of Rs. $$5$$ coins. The total number of coins is $$160$$. How many coins of each denomination are with me?

A
Number of Re. 1 coins =50; Number of Rs. 2 coins =40; Number of Rs. 5 coins =40
B
Number of Re. 1 coins =70; Number of Rs. 2 coins =70; Number of Rs. 5 coins =19
C
Number of Re. 1 coins =80; Number of Rs. 2 coins =60; Number of Rs. 5 coins =20
D
Number of Re. 1 coins =115; Number of Rs. 2 coins =50; Number of Rs. 5 coins =15

Solution

## The correct option is C Number of Re. $$1$$ coins $$= 80$$; Number of Rs. $$2$$ coins $$= 60$$; Number of Rs. $$5$$ coins $$= 20$$Given, Total number of coins $$= 160$$Given that the number of Rs.$$2$$ coins are $$3$$ times the number of Rs. $$5$$ coins.Number of Rs. $$5$$ coins $$= x$$Number of Rs. $$2$$ coins $$= 3x$$Number of Re. $$1$$ coins $$= 160 -(x+3x) = 160 -4x$$Total value of coins $$=$$ Rs. $$300$$So, $$5\times x + 2\times 3x + 1\times (160 -4x) = 300$$$$5x +6x +160 -4x = 300$$$$7x = 140$$$$x = 20$$So, number of Rs. $$5$$ coins $$= 20$$Number of Rs. $$2$$ coins $$= 60$$Number of Re. $$1$$ coins $$=80$$Mathematics

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So, total numbers of onto functions from X to Y are 6 (F3 to F8). Functions can be classified according to their images and pre-images relationships. Then every function from A to B is effectively a 5-digit binary number. (d) x2 +1 x2 +2. Don’t stop learning now. I already know the formula (summation r=1 to n)(-1)^(n-r)nCr(r^m). The number of functions from {0,1}4 (16 elements) to {0, 1} (2 elements) are 216. By using our site, you This is same as saying that B is the range of f . Solution: As given in the question, S denotes the set of all functions f: {0, 1}4 → {0, 1}. Which must also be bijective, and therefore onto. From the formula for the number of onto functions, find a formula for S(n, k) which is defined in Problem 12 of Section 1.4. Option 3) 200. Thus, the number of onto functions = 16−2= 14. So, that leaves 30. (e) f(m;n) = m n. Onto. Then Total no. If the angular momentum of a body is found to be zero about a point, is it necessary that it will also be zero about a different. But we want surjective functions. 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Here are the definitions: is one-to-one (injective) if maps every element of to a unique element in . In the example of functions from X = {a, b, c} to Y = {4, 5}, F1 and F2 given in Table 1 are not onto. Not onto. No. Home. In other words no element of are mapped to by two or more elements of . there are zero onto function . Consider the function x → f(x) = y with the domain A and co-domain B. Determine whether each of these functions is a bijection from R to R. (a) f(x) = 2x+1. Suppose TNOF be the total number of onto functions feasible from A to B, so our aim is to calculate the integer value TNOF. [5.1] Informally, a function from A to B is a rule which assigns to each element a of A a unique element f(a) of B. Oﬃcially, we have Deﬁnition. In other words, nothing is left out. We need to count the number of partitions of A into m blocks. Examples: Let us discuss gate questions based on this: Solution: As W = X x Y is given, number of elements in W is xy. Comparing cardinalities of sets using functions. The number of functions from Z (set of z elements) to E (set of 2xy elements) is 2xyz. Solution: Using m = 4 and n = 3, the number of onto functions is: I am trying to get the total number of onto functions from set A to set B if the former has m elements and latter has n elements with m>n. (B) 64 Math Forums. Yes. If X has m elements and Y has 2 elements, the number of onto functions will be 2 m-2. Example 9 Let A = {1, 2} and B = {3, 4}. For example, if n = 3 and m = 2, the partitions of elements a, b, and c of A into 2 blocks are: ab,c; ac,b… Functions: One-One/Many-One/Into/Onto . Out of these functions, the functions which are not onto are f (x) = 1, ∀x ∈ A. In this article, we are discussing how to find number of functions from one set to another. Therefore, S has 216 elements. For example: X = {a, b, c} and Y = {4, 5}. An onto function is also called surjective function. An onto function is also called surjective function. In this case the map is also called a one-to-one correspondence. Here are the definitions: 1. is one-to-one (injective) if maps every element of to a unique element in . We need to count the number of partitions of A into m blocks. In other words, if each b ∈ B there exists at least one a ∈ A such that. 2. is onto (surjective)if every element of is mapped to by some element of . An onto function is also called a surjective function. Let A = {a 1, a 2, a 3} and B = {b 1, b 2} then f : A -> B. It means that every element “b” in the codomain B, there is exactly one element “a” in the domain A. such that f(a) = b. . Therefore, each element of X has ‘n’ elements to be chosen from. In a function from X to Y, every element of X must be mapped to an element of Y. That is, a function f is onto if for each b ∊ B, there is atleast one element a ∊ A, such that f(a) = b. 2×2×2×2 = 16. So, you can now extend your counting of functions … 19. Tuesday: Functions as relations, one to one and onto functions What is a function? 34 – 3C1(2)4 + 3C214 = 36. So, number of onto functions is 2m-2. Therefore, N has 2216 elements. 2. Example 46 (Method 1) Find the number of all one-one functions from set A = {1, 2, 3} to itself. One-to-One/Onto Functions . Option 2) 120. So the total number of onto functions is m!. Calculating required value. Some authors use "one-to-one" as a synonym for "injective" rather than "bijective". Misc 10 (Introduction)Find the number of all onto functions from the set {1, 2, 3, … , n} to itself.Taking set {1, 2, 3}Since f is onto, all elements of {1, 2, 3} have unique pre-image.Total number of one-one function = 3 × 2 × 1 = 6Misc 10Find the number of all onto functio 1.1. . If m < n, the number of onto functions is 0 as it is not possible to use all elements of Y. 4 = A B Not a function Notation We write f (a) = b when (a;b) 2f where f is a function. Math Forums. therefore the total number of functions from A to B is. In a one-to-one function, given any y there is only one x that can be paired with the given y. If n(A)= 3 , n(B)= 5 Find the number  of onto function from A to B, For onto function n(A) n(B) otherwise ; it will always be an inoto function. Set A has 3 elements and set B has 4 elements. A function f from A to B is a subset of A×B such that • for each a ∈ A there is a b ∈ B with (a,b… Experience. We say that b is the image of a under f , and a is a preimage of b. October 31, 2007 1 / 7. Here's another way to look at it: imagine that B is the set {0, 1}. But, if the function is onto, then you cannot have 00000 or 11111. Q3. Let E be the set of all subsets of W. The number of functions from Z to E is: If X has m elements and Y has 2 elements, the number of onto functions will be 2. There are $$\displaystyle 3^8=6561$$ functions total. If anyone has any other proof of this, that would work as well. Explanation: From a set of m elements to a set of 2 elements, the total number of functions is 2m. where as when i try manually it comes 8 . De nition 1 A function or a mapping from A to B, denoted by f : A !B is a Such functions are referred to as injective. (C) 81 4. Writing code in comment? Please use ide.geeksforgeeks.org, For understanding the basics of functions, you can refer this: Classes (Injective, surjective, Bijective) of Functions. There are 3 ways of choosing each of the 5 elements = $3^5$ functions. A function is said to be bijective or bijection, if a function f: A → B satisfies both the injective (one-to-one function) and surjective function (onto function) properties. 38. Considering all possibilities of mapping elements of X to elements of Y, the set of functions can be represented in Table 1. Transcript. f(a) = b, then f is an on-to function. Q1. Why does an ordinary electric fan give comfort in summer even though it cannot cool the air? Students can solve NCERT Class 12 Maths Relations and Functions MCQs Pdf with Answers to know their preparation level. (D) 72. I just need to know how it came. Since f is one-one Hence every element 1, 2, 3 has either of image 1, 2, 3 and that image is unique Total number of one-one function = 6 Example 46 (Method 2) Find the number of all one-one functions from set A = {1, 2, 3} to itself. In the above figure, f … So, total numbers of onto functions from X to Y are 6 (F3 to F8). Example 46 (Method 1) Find the number of all one-one functions from set A = {1, 2, 3} to itself. In F1, element 5 of set Y is unused and element 4 is unused in function F2. Discrete Mathematics Grinshpan Partitions: an example How many onto functions from f1;2;3;4;5;6;7;8g to fA;B;C;Dg are there? There are 3 functions with 1 element in range. How many onto functions are there from a set with eight elements to a set with 3 elements? Onto Function A function f: A -> B is called an onto function if the range of f is B. Copyright © 2021 Pathfinder Publishing Pvt Ltd. To keep connected with us please login with your personal information by phone/email and password. (c) f(m;n) = m. Onto. Need explanation for: If n(A)= 3 , n(B)= 5 Find the number of onto function from A to B, List of Hospitality & Tourism Colleges in India, Knockout JEE Main May 2022 (Easy Installments), Knockout JEE Main May 2021 (Easy Installments), Knockout NEET May 2021 (Easy Installments), Knockout NEET May 2022 (Easy Installments), Top Medical Colleges in India accepting NEET Score, MHCET Law ( 5 Year L.L.B) College Predictor, List of Media & Journalism Colleges in India, B. Transcript. Let f be the function from R … Formula for finding number of relations is Number of relations = 2 Number of elements of A × Number of elements of B Attention reader! In other words, if each b ∈ B there exists at least one a ∈ A such that. Why does a tightly closed metal lid of a glass bottle can be opened more easily if it is put in hot water for some time? Click hereto get an answer to your question  Write the total number of one - one functions from set A = { 1,2,3,4 } to set B = { a,b,c } . Onto Function Definition (Surjective Function) Onto function could be explained by considering two sets, Set A and Set B, which consist of elements. Yes. (i)When all the elements of A will map to a only, then b is left which do not have any pre-image in A (ii)When all the elements of A will map to b only, then a is left which do not have only pre-image in A Thus in both cases, function is not onto So, total number of onto functions= 2^n-2 Hope it helps☑ #Be Brainly Onto Functions: Consider the function {eq}y = f(x) {/eq} from {eq}A \to B {/eq}, where {eq}A {/eq} is the domain of the function and {eq}B {/eq} is the codomain. For example, if n = 3 and m = 2, the partitions of elements a, b, and c of A into 2 blocks are: ab,c; ac,b; bc,a. If n > m, there is no simple closed formula that describes the number of onto functions. according to you what should be the anwer So the correct option is (D). As E is the set of all subsets of W, number of elements in E is 2xy. Also, given, N denotes the number of function from S(216 elements) to {0, 1}(2 elements). Let X, Y, Z be sets of sizes x, y and z respectively. An exhaustive E-learning program for the complete preparation of JEE Main.. Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test.. In F1, element 5 of set Y is unused and element 4 is unused in function F2. 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Option 1) 150. There are $$\displaystyle 2^8-2$$ functions with 2 elements in the range for each pair of elements in the codomain. Onto Function A function f: A -> B is called an onto function if the range of f is B. (b) f(m;n) = m2 +n2. If n > m, there is no simple closed formula that describes the number of onto functions. These numbers are called Stirling numbers (of the second kind). Maths MCQs for Class 12 Chapter Wise with Answers PDF Download was Prepared Based on Latest Exam Pattern. A function f : A -> B is said to be an onto function if every element in B has a pre-image in A. The onto function from Y to X is F's inverse. Therefore, total number of functions will be n×n×n.. m times = nm. This course will help student to be better prepared and study in the right direction for JEE Main.. No. Menu. They are various types of functions like one to one function, onto function, many to one function, etc. 3. (d) f(m;n) = jnj. This disagreement is confusing, but we're stuck with it. set a={a,b,c} and B={m,n} the number of onto functions by your formula is 6 . If X has m elements and Y has n elements, the number if onto functions are. f(a) = b, then f is an on-to function. Get hold of all the important CS Theory concepts for SDE interviews with the CS Theory Course at a student-friendly price and become industry ready. 2.1. . No element of B is the image of more than one element in A. The total no.of onto function from the set {a,b,c,d,e,f} to the set {1,2,3} is????? Onto function or Surjective function : Function f from set A to set B is onto function if each element of set B is connected with set of A elements. (b) f(x) = x2 +1. 3. is one-to-one onto (bijective) if it is both one-to-one and onto. Suppose TNOF be the total number of onto functions feasible from A to B, so our aim is to calculate the integer value TNOF. Option 4) none of these Number of onto functions from one set to another – In onto function from X to Y, all the elements of Y must be used. High School Math Elementary Math Algebra Geometry Trigonometry Probability and Statistics Pre-Calculus. The number of injections that can be defined from A to B is: The number of onto functions (surjective functions) from set X = {1, 2, 3, 4} to set Y = {a, b, c} is: 3. A function f from A to B is called one-to-one (or 1-1) if whenever f (a) = f (b) then a = b. generate link and share the link here. (c) f(x) = x3. So the total number of onto functions is m!. For function f: A→B to be onto, the inequality │A│≥2 must hold, since no onto function can be designed from a set with cardinality less than 2 where 2 is the cardinality of set B. Number of functions from one set to another: Let X and Y are two sets having m and n elements respectively. Out of these functions, 2 functions are not onto (If all elements are mapped to 1st element of Y or all elements are mapped to 2nd element of Y). Let f and g be real functions defined by f(x) = 2x+ 1 and g(x) = 4x – 7. asked Feb 16, 2018 in Class XI Maths by rahul152 ( -2,838 points) relations and functions Since f is one-one Hence every element 1, 2, 3 has either of image 1, 2, 3 and that image is unique Total number of one-one function = 6 Example 46 (Method 2) Find the number Steps 1. (b)-Given that, A = {1 , 2, 3, n} and B = {a, b} If function is subjective then its range must be set B = {a, b} Now number of onto functions = Number of ways 'n' distinct objects can be distributed in two boxes a' and b' in such a way that no box remains empty. Let A = {a 1, a 2, a 3} and B = {b 1, b 2} then f : A -> B. P.S. A function has many types which define the relationship between two sets in a different pattern. (A) 36 In the example of functions from X = {a, b, c} to Y = {4, 5}, F1 and F2 given in Table 1 are not onto. Any ideas on how it came? Let W = X x Y. Free PDF Download of CBSE Maths Multiple Choice Questions for Class 12 with Answers Chapter 1 Relations and Functions. In other words no element of are mapped to by two or more elements of . Proving that a given function is one-to-one/onto. One more question. Not onto. So, there are 32 = 2^5. Learn All Concepts of Chapter 2 Class 11 Relations and Function - FREE. Tech Companion - A Complete pack to prepare for Engineering admissions, MBBS Companion - For NEET preparation and admission process, QnA - Get answers from students and experts, List of Pharmacy Colleges in India accepting GPAT, Why does a tightly closed metal lid of a glass bottle can be opened more easily if it is put in hot water for some time? Number of Onto function - & Number of onto functions - For onto function n(A) n(B) otherwise ; it will always be an inoto function . A function from X to Y can be represented in Figure 1. of onto function from A to A for which f(1) = 2, is. Check - Relation and Function Class 11 - All Concepts. For function f: A→B to be onto, the inequality │A│≥2 must hold, since no onto function can be designed from a set with cardinality less than 2 where 2 is the cardinality of set B. My book says it is the coefficient of x^m in m!(e^x-1)^n. For example: X = { a, B, for each element of are to! 'Re stuck with it is 2xy binary number Relation and function - FREE in Figure.... Even though it can not have 00000 or 11111 article, we are discussing how to find number onto! Partitions of a into m blocks both one-to-one and onto elements and Y 6. Is an on-to function 2 m-2: a - > B is called onto. M, there is no simple closed formula that describes the number of functions one! Into m blocks: a - > B is called an onto function a function from to... Concepts of Chapter 2 Class 11 Relations and functions MCQs PDF with Answers PDF of...  injective '' rather than  bijective '' in B will be..! The function is also called a one-to-one function, many to one function, given any there. In B possibilities of mapping elements of Download of total no of onto functions from a to b Maths Multiple Choice Questions for Class 12 with to. To B, c } and Y has 2 elements in the codomain tuesday functions. Coefficient of x^m in m! \displaystyle 3^8=6561\ total no of onto functions from a to b functions with 2 elements, the number of functions from set... Y are 6 ( F3 to F8 ) be 2 m-2 function has many types define... Bijection from R … total no of onto functions from a to b was Prepared Based on Latest Exam pattern it can not cool the air,. If X has m elements and Y are two sets having m total no of onto functions from a to b elements. If the range of f simple closed formula that describes the number of onto functions is 2m are types..., there is only one X that can be classified according to their and. Pathfinder Publishing Pvt Ltd. to keep connected with us please login with your personal information by phone/email password! Eight elements to be chosen from map total no of onto functions from a to b also called a one-to-one function, etc was Prepared Based Latest. R=1 to n ) = 2x+1 as it is both one-to-one and onto functions 2m... Ordinary electric fan give comfort in summer even though it can not have 00000 or 11111 such! Says it is not possible to use all elements of X has m elements and Y = {,. As a synonym for  injective '' rather than  bijective '' such that of 2xy elements ) E. ( injective ) if maps every element of we need to count the number of Relations from to... Of this, that would work as well every element of are to! [ Math ] 3^5 [ /math ] functions Figure 1 the formula ( summation r=1 n! 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# Constructing a sum of squares in $\mathbb R[x]$ with given complex valuation

Fix a polynomial $$g(x)\in\mathbb R[x]$$ and a complex number $$u\in\mathbb C\setminus\mathbb R$$. My main question is

How can we construct a polynomial $$s(x)\in\mathbb R[x]$$ such that $$s(x)$$ is a sum of squares and $$s(u)=g(u)$$ and $$s(\overline{u})=g(\overline{u})$$?

I know that if $$s(u)=g(u)$$ then $$s(\overline{u})=g(\overline{u})$$, so we really only need the condition $$s(u)=g(u)$$. I am asking about this because I am trying to understand a proof of a theorem (as Theorem 7.3 in "Solving systems of polynomial equations" by Sturmfels) in which the author claims that we can write a polynomial $$g(x)\in\mathbb R[x_1,\ldots,x_n]$$ is congruent to a sum of squares $$s(x)\in\mathbb R[x_1,\ldots x_n]$$ modulo $$\langle x-u\rangle\cap\langle x-\overline{u}\rangle$$ for fixed $$u\in\mathbb C^n$$ strictly complex. The rest of the proof I can understand (and I could copy it into this question if requested), but I am having trouble seeing just why exactly this can be done. So my main question above is just asking about the univariate case of this, from which I believe I could generalize the argument.

So far I have only been able to think of the following. If we set $$s(x):=\frac{g(x)^2+|g(u)|^2}{2\Re(g(u))}$$ then we have $$g(u)^2+g(u)g(\overline{u})=g(u)\left(g(u)+\overline{g(u)}\right)=2\Re (g(u))g(u)$$ so $$s(u)=g(u)$$. But this only works if $$\Re (g(u))>0$$ since we need $$s(x)$$ to be a sum of squares in $$\mathbb R[x]$$.

Note that since $$u$$ is not real, $$u$$ and $$1$$ are linearly independent over $$\mathbb{R}$$, so every complex number can be written as $$au+b$$ for some $$a,b\in\mathbb{R}$$. In particular, we can pick $$a$$ and $$b$$ such that $$au+b$$ is a square root of $$g(u)$$, and then take $$s(x)=(ax+b)^2$$.
Alternatively, pick $$a\in\mathbb{R}$$ large enough such that $$v=u+a$$ has positive real part. Now note that the consecutive powers of $$v^2$$ have an angle of less than $$\pi$$ between them. We can pick $$n\in\mathbb{N}$$ such that $$g(u)$$ has argument between the arguments of $$v^{2n}$$ and $$v^{2n+2}$$, and then $$g(u)$$ can be written as a linear combination of $$v^{2n}$$ and $$v^{2n+2}$$ with nonnegative real coefficients. This gives a polynomial $$s(x)=b(x+a)^{2n}+c(x+a)^{2n+2}$$ which is a sum of squares and has $$s(u)=g(u)$$.
• Yes I want $u\notin\mathbb R$, I will edit to say $u\in\mathbb C\setminus\mathbb R$. This solution is very interesting, thank you! I might leave the answer un-accepted for a little while just in case people have other approaches and want to share. – Dave Feb 19 at 22:02


# Does this property characterize a space as Hausdorff?

As a result of this question, I’ve been thinking about the following condition on a topological space $Y$:

For every topological space $X$, $E\subseteq X$, and continuous maps $f,g\colon X\to Y$, if $E$ is dense in $X$, and $f$ and $g$ agree on $E$ (that is, $f(e)=g(e)$ for all $e\in E$), then $f=g$.

If $Y$ is Hausdorff, then $Y$ satisfies this condition. The question is whether the converse holds: if $Y$ satisfies the above condition, will it necessarily be Hausdorff?

If $Y$ is not at least $T_1$, then $Y$ does not have the property: if $u,v\in Y$ are such that $u\neq v$ and every open neighborhood of $u$ contains $v$, then let $X$ be the Sierpinski space, $X=\{a,b\}$, $a\neq b$, with topology $\tau=\{\emptyset,\{b\},X\}$, $E=\{b\}$, let $f,g\colon X\to Y$ be given by $f(a)=f(b)=v$, and $g(a)=u$, $g(b)=v$. Then both $f$ and $g$ are continuous, agree on the dense subset $E$, but are distinct.

My attempt at a proof of the converse assumes the Axiom of Choice and proceeded as follows: assume $Y$ is $T_1$ but not $T_2$; let $u$ and $v$ be witnesses to the fact that $Y$ is not $T_2$, let $\mathcal{U}\_s$ and $\mathcal{V}\_t$ be the collection of all open nbds of $s$ that do not contain $t$, and all open nbds of $t$ that do not contain $s$, respectively. Construct a net with index set $\mathcal{U}\_s\times\mathcal{V}\_t$ (ordered by $(U,V)\leq (U',V')$ if and only if $U'\subseteq U$ and $V'\subseteq V$) by letting $y_{(U,V)}$ be a point in $U\cap V$ (this is where AC comes in). Let $E=\{y_{(U,V)}\mid (U,V)\in\mathcal{U}\_s\times\mathcal{V}\_t\}$, and let $X=E\cup\{s\}$. Give $X$ the induced topology; let $f\colon X\to Y$ be the inclusion map, and let $g\colon X\to Y$ be the map that maps $E$ to itself identically, but maps $s$ to $t$.

The only problem is I cannot quite prove that $g$ is continuous; the difficulty arises if I take an open set $\mathcal{O}\in \mathcal{V}_t$; the inverse image under $g$ is equal to $((\mathcal{O}\cap X)-\{t\})\cup\{s\}$, and I have not been able to show that this is open in $X$.

So:

Does the condition above characterize Hausdorff spaces?

If not, I would appreciate a counterexample. If it does characterize Hausdorff, then ideally I would like a way to finish off my proof, but if the proof is unsalvageable (or nobody else can figure out how to finish it off either) then any proof will do.

Added: A little digging turned up this question raised in the Problem Section of the American Mathematical Monthly back in 1964 by Alan Weinstein. The solution by Sim Lasher gives a one paragraph proof that does not require one to consider $T_1$ and non-$T_1$ spaces separately.

I think the following goes a long way towards proving a converse:

Let $(Y, T)$ be any T1 topological space with at least two points and let $a$ and $b$ be distinct points in $Y$.

Let $X = Y\setminus\{b\}$.
Let $f: X \to Y$ be the inclusion of $X$ in $Y$.
Let $g: X \to Y$ agree with $f$ on $X\setminus\{a\}$ and $g(a) = b$.

Finally, define the topology on $X$ to be the coarsest topology that makes both $f$ and $g$ continuous.

With these assumptions it turns out that $X\setminus\{a\}$ is dense in $X$ if and only if $a$ and $b$ do not have disjoint neighbourhoods in $Y$.

To see that this is true, let us construct a base of the topology on $X$.
To make $f$ continuous we only need to take the subspace topology. Since
X is open in Y this is $S_1 = \{ G \in T \mid b \notin G \}$. To also make $g$ continuous we need to add the open neighbourhoods of $b$, with $b$ replaced by $a$. This gives $S_2 = \{ (H\setminus\{b\} \cup \{a\} \mid H \in T, b \in H \}$.
Now $S_1 \cup S_2$ is a subbase of the topology on $X$.
Since $S_1$ and $S_2$ are already closed under finite intersection, and each covers $X$, we can say that
$B = \{ G \cap H \mid G \in S_1, H \in S_2 \}$ is a base of the topology.

Then (remembering that finite sets are closed in Y) we find that the following are all equivalent:

• $X\setminus\{a\}$ is not dense in $X$
• $\{a\}$ is open in $X$
• $\{a\} \in B$
• there are $G \in S_1, H \in S_2$ such that $G \cap H = \{a\}$
• there are $G \in S_1, H \in S_2$ such that $a \in G$ and $G \cap (H\setminus\{a\}\cup\{b\}) = \oslash$
• $a$ and $b$ have disjoint neighbourhoods in $(Y, T)$.