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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.

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    $\begingroup$ The problem with your argument is that these points $e$ you find may not be in $X$. To say that $X$ is separable means it CONTAINS a subset which is countable and dense in $X$. \\ You will have to find a new countable subset of $X$ which is dense in $X$, by modifying $E$ somehow. (I hope this is a useful hint.) $\endgroup$
    – GCD
    Oct 6, 2013 at 18:37
  • $\begingroup$ I don't think I follow. How could X these points not be in X, if they are in E, which is a subset of X? $\endgroup$
    – Patrick
    Oct 6, 2013 at 18:39
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    $\begingroup$ $E$ is a subset of the big space $M$, not necessarily of $X$. $X$ is also a subset of the big space $M$, but $X$ and $E$ may have no points in common. $\endgroup$
    – GCD
    Oct 6, 2013 at 18:39
  • $\begingroup$ Answered in this, but can be proved easily from definitions. $\endgroup$
    – leo
    Oct 6, 2013 at 18:41
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    $\begingroup$ $E$ and $X$ are both subsets of the same large space $M$. They need not have any points in common. For example, $M$ could be the real line, $E$ could be the rational numbers (dense in $M$), and $X$ could be the irrational numbers. $X$ is indeed separable, but in this case a countable dense subset of $X$ can't be made up of points in $E$, because none of those points are in $X$. $\endgroup$
    – GCD
    Oct 6, 2013 at 18:43

4 Answers 4

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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.

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  • $\begingroup$ I follow along until you introduce B0. That theorem is listed immediately before this question in my book. That if (M,d) is a separable metric space, then there exists a countable collection of open sets such that every open subset of M is a union of some open sets from the collection. I don't understand why any of the B's can be guaranteed to have a non empty intersection with X? $\endgroup$
    – Patrick
    Oct 6, 2013 at 19:12
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    $\begingroup$ @user99181: Because $E$ is dense in $M$. Let $x\in X$ and $n\in\Bbb Z^+$; there is some $e\in B\left(x,\frac1n\right)\cap E$. Then $x\in B\left(e,\frac1n\right)\cap X$, and $\frac1n$ is certainly rational, so $B\left(e,\frac1n\right)\in\mathscr{B}_0$. $\endgroup$ Oct 6, 2013 at 19:19
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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 $X$.

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  • $\begingroup$ For every e in E, choose a point a in X whose distance from e is, say, less than 2d(e,X). (I may not be able to get a exactly distance d(e,X) from e, but I can get close.) This is the part that threw me for a loop in the solutions I found earlier. The line above we said d(e,X)=inf{d(e,x): x in X}, then for every e, I can pick an a in X st d(e,a) is exactly d(e,X). Namely, I pick the point in X st d(e,a)=inf{(d(e,x): x in X}. $\endgroup$
    – Patrick
    Oct 6, 2013 at 19:20
  • $\begingroup$ (Sorry I just edited to fix a mistake; I'm an inveterate editor.) But nope, infimums are not always achieved. There might not be a point which actually realizes the minimum distance. For example, what is the distance from the point $0$ to the set $(0,\infty)$ in $\mathbb{R}$? The distance is zero, but there is no point in $(0,\infty)$ which actually has distance zero from $0$. $\endgroup$
    – GCD
    Oct 6, 2013 at 19:25
  • $\begingroup$ Okay, the part about the infimum and picking an a in X makes sense. Since I have this infimum of the distance, then, adding any amount > 0 to d(e,X), I guarantee the existence of an a in X st d(e,X)<= d(e,a) < d(e,X)+epsilon. Right? $\endgroup$
    – Patrick
    Oct 6, 2013 at 19:31
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    $\begingroup$ Hey @GCD, OP here, but from a different computer. Your proof makes the most sense to me...I can't follow the others. However, I'm hung up on the part where we show A is dense in X. You have d(e,a) < d(e,X) + ε/3, but I don't see why this is always true. All we have to go on is that d(e,a) < d(e,X)+1/n <= ε/3+1/n. I don't see how to guarantee 1/n <= ε/3? Suppose that the set of an's is a singleton (i.e. for all e in E, the a in X that minimized distance was the same. Then 1/n is just 1.) $\endgroup$
    – Patrick
    Oct 6, 2013 at 21:09
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    $\begingroup$ Given $e\in E$ I would like to just choose $a\in X$ associated to it so that $d(e,a) = d(e,X)$. Unfortunately this may not be possible. So I do the next best thing, which is to associated to $e$ a countable set of $a_n$'s getting arbitrarily close to the minimal distance. In the part of the proof you ask about, I simply choose $n$ large so that $1/n<\epsilon/3$; in other words so that $a_n$ comes within $\epsilon/3$ of minimizing the distance. $\endgroup$
    – GCD
    Oct 8, 2013 at 3:46
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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.

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    $\begingroup$ I should add that this relies heavily on $M$ being a metric space. In the more general setting of point set topology, separability does not imply second countability. Not even when the space is first countable as well. $\endgroup$
    – kahen
    Oct 6, 2013 at 18:54
  • $\begingroup$ I have, from a theorem in our book, that M is second countable (it doesn't use that word, but the theorem states exactly the same definition you give for second countable). Even knowing that, how can I show X is second countable? How do I know X isn't empty, or even contains an open set? $\endgroup$
    – Patrick
    Oct 6, 2013 at 19:17
  • $\begingroup$ @user99181 That doesn't matter. A $U \subseteq X$ is $(X,d)$-open if and only if there is an $(M,d)$-open $V \subseteq M$ such that $U = X \cap V$. And if $X$ is empty, then it's trivially separable. $\endgroup$
    – kahen
    Oct 6, 2013 at 19:22
  • $\begingroup$ Hello. I followed your answer, but got stuck at showing that X is also second countable. Can you hint me towards an answer? $\endgroup$
    – user370967
    Mar 4, 2018 at 21:39
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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 $$d(x,e_{(i,j)})\leq d(x,x_i) + d(x_i,e_{(i,j)}) = 2r_j \lt \epsilon.$$

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  • $\begingroup$ Could you please elaborate on the last line also? Thanks. $\endgroup$
    – MathMan
    Jul 25, 2015 at 15:31
  • $\begingroup$ @Wanderer Done. $\endgroup$
    – leo
    Jul 25, 2015 at 18:01
  • $\begingroup$ Just one small confusion. Why is $d(x_i, e_{ij}) = r_j?$ $\endgroup$
    – MathMan
    Jul 25, 2015 at 20:29
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    $\begingroup$ That's not the case, $d(x_i,e_{(i,j)})\lt r_j$. And that's because each $e_{(i,j)}$ is choose to be a point of $B(x_i,r_j)$. $\endgroup$
    – leo
    Jul 25, 2015 at 20:50

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