Prove that a subset of a separable set is itself separable

Question

The problem statement, all variables and given/known data:
Show that if $X$ is a subset of $M$ and $(M,d)$ is separable, then $(X,d)$ is separable. [This may be a little bit trickier than it looks - $E$ may be a countable dense subset of $M$ with $X\cap E = \varnothing$.] Definitions Per our book:

A metric space $(M,d)$ is separable if there exists a countable dense $E$ contained in $M$.

$E$ contained in $M$ is dense if $\forall m\in M$, $\forall ε>0\in\mathbb R$, $\exists e\in E$ s.t. $d(m,e) < ε$

The attempt at a solution

My best attempt was doomed from the start, because I don't quite understand the hint. My thought process went as follows:

Since $X$ is a subset of $M$, $\forall x\in X, x\in M$. Thus, since $E$ is dense in $M$,

$\forall x\in X$, and ε > 0, $\exists e\in E$ st $d(x,e)<ε $.

At this point, I was done, because the set of $e$'s satisfying the above, is a subset of $E$, a countable set. So a subset of a countable set is dense in $X$, and $X$ is separable. This is incorrect, but I cannot see why. Any help clearing up the confusion would be greatly appreciated. Thanks!

Edit: I wish I could upvote all of you for your help! I really appreciate the speedy replies and attempts to make this information clear to me.

Answer 1

Note that it’s very important here that you’re working in a metric space, because the statement isn’t true in topological spaces in general.

Since $X$ is a subset of $M$, $\forall x\in X, x\in M$. Thus, since $M$ is dense in $E$,

$\forall x\in X$, and $\epsilon > 0$, $\exists e\in E$ st $d(x,e)< \epsilon $.

‘Since $M$ is dense in $E$’ doesn’t make sense. First, you never defined $E$. I can guess that it’s supposed to be a countable dense subset of $M$, but then your statement is just backwards: $E$ is dense in $M$. In any case, finding points of $E$ near $x$ doesn’t help to show that $X$ is separable: to show that you must find a countable subset of $X$ that is dense in $X$, and $E$ might be completely disjoint from $X$. For a concrete example of this possibility, let $M=\Bbb R$ with the usual metric, and let $X$ be the set of irrational numbers. We know that $\Bbb R$ is separable, because $\Bbb Q$ is a countable dense subset of $\Bbb R$. The theorem that you’re to prove says that $X$ is also separable, i.e., that there is some countable set $D$ of irrational numbers that is dense in $X$, but $D$ certainly can’t be $\Bbb Q$, our familiar countable dense set of reals: no member of $\Bbb Q$ is even in $X$.

One way to prove the theorem is to show that if $E$ is a countable dense subset of $M$, then $$\mathscr{B}=\{B(e,r):e\in E\text{ and }0<r\in\Bbb Q\}$$ is a base for $M$, meaning that if $x\in M$, and $U$ is an open set containing $x$, then there is some $B(e,r)\in\mathscr{B}$ such that $x\in B(e,r)\subseteq U$. Note that $\mathscr{B}$ is a countable family of open balls. Then let $$\mathscr{B}_0=\{B\cap X:B\in\mathscr{B}\text{ and }B\cap X\ne\varnothing\}\;,$$ and show that $\mathscr{B}_0$ is a base for $X$. Since $\mathscr{B}$ is countable, so is $\mathscr{B}_0$. Finally, for each $B\in\mathscr{B}_0$ pick one point $x_B\in B$, and let $D=\{x_B:B\in\mathscr{B}_0\}$; $D$ is countable, and it’s not too hard to show that it’s dense in $X$ and hence that $X$ is separable.

Answer 2

EDIT: Fixed a mistake.

Here is a slightly different answer, though of course basically equivalent to the other two.

$X$ is a subset of $M$. $E$ is a countable dense subset of $M$. We would like to find a countable dense subset of $X$.

The problem, as hashed out in the comments above, is that we would like to use $E$ as our countable dense subset of $X$, but the points of $E$ may not actually belong to $X$.

So one thing we can do is this: Every point $e$ in $E$ has some distance $$ d(e, X) = \inf\{ d(e, x) : x\in X\} $$ i.e. how far it is from the set $X$.

For every $e$ in $E$, choose points $a_n$ in $X$ whose distance from $e$ is, say, less than $d(e,X) + 1/n$. (I may not be able to get $a$ exactly distance $d(e,X)$ from $e$, but I can get close.)

Now, let $A$ be the set of all $a_n$ that I made, countably many from each $e$ in $E$. (Okay, some $a_n$ may belong to multiple $e$'s, but so what.) $A$ is in $X$, by definition, and $A$ is countable, as $E$ was (countable union of countable sets is countable).

So I just have to show that $A$ is dense in $X$. Let $x$ be any point of $X$, and take any $\epsilon>0$. Then there is some $e\in E$ within $\epsilon/3$ of $x$, i.e. $d(x,e)<\epsilon/3$. That means $d(e, X)\leq \epsilon/3$, so there is some $a\in A$ such that $$ d(e,a) < d(e,X) + \epsilon/3 \leq 2\epsilon/3. $$

So $$ d(x,a) \leq d(x,e) + d(e,a) < \epsilon/3 + 2\epsilon/3 = \epsilon. $$

So $A$ is dense in $X$.

Answer 3

Hint (for a different proof strategy): Show that $(M,d)$ is second countable, i.e. there is a countable collection $\mathcal B = \{U_n : n \in \omega\}$ of open sets of $M$ such that any open set $U$ in $M$ can be written as a union of elements of $\mathcal B$. Then show that that $X$ also is second countable. And finally show that second countable implies separable.

Answer 4

This answered here which is the approach in the Brian M. Scott's answer.

Since you pointed out that you are not used to the terms used, and that this came from a course of math for economists, I'll try to give an answer using the definitions that you provided.

First, as an advice, when you are asked to proof something, at least at the beginning, it is better to look at the definitions and write down what you have to prove for the object you are supposed to prove things about.

In this case you have to prove that $(X,d)$ is separable.

What is $(X,d)$? It is the metric space formed by the subset $X$ of $M$, in which we measure distances just as we do in $M$.

What you have to prove about $(X,d)$? You have to prove that there exist a countable subset $E$ of $X$ such that $E$ is dense in $X$.

But you know that $(M,d)$ is separable, so there exist a countable subset $D$ of $M$, say $$D = \{x_1,x_2,x_3,\ldots\},$$ which is dense in $M$.

Now, let's enumerate the positive rationals as $$\Bbb Q \cap (0,\infty) = \{r_1,r_2,r_3,\ldots\}.$$

Define $$\Delta = \{(i,j) : B(x_i,r_j)\cap X\neq\emptyset\}.$$ $\Delta$ is not empty because otherwise $D$ is not dense in $M$.

Then, for each $(i,j)\in\Delta$ there is an $e_{(i,j)}\in B(x_i,r_j)\cap X$. Define then $$E = \{e_{(i,j)}: (i,j)\in \Delta\}.$$

This $E$ is a countable dense subset of $X$, so we are done.

If this last statement is not clear, let me know it and I'll elaborate on it.

Edit Indeed, Let $x\in X$ and $\epsilon\gt 0$. There's an $r_j$ such that $r_j\lt \epsilon/2$. Since $D$ is dense in $M$ there exist some $x_i\in D$ such that $d(x,x_i)\lt r_j$. So both $x$ and $e_{(i,j)}$ are elements of the ball $B(x_i,r_j)$, therefore $$d(x,e_{(i,j)})\leq d(x,x_i) + d(x_i,e_{(i,j)}) = 2r_j \lt \epsilon.$$