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I am thinking about this question for a while. Is any infinite set $X$ on the line separable? I think the answer is yes, but to do so, I need to show there exists a countable subset $A$ in $X$ such that $X$ is in the closure of $A$.

Can anyone outline the proof for me? Thanks.

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By "the line", do you mean the "real line". And do you put on this set $X$ the topology given by $\mathbb R$? –  Davide Giraudo Dec 7 '11 at 15:52
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While it is not very relevant, it may still be of some interest to people reading this thread: without the axiom of choice it is consistent that there is a subset of real numbers which is not separable. –  Asaf Karagila Dec 8 '11 at 0:40
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@Asaf: in my opinion your comment is totally irrelevant but somewhat interesting. :) –  Pete L. Clark Dec 8 '11 at 2:53

2 Answers 2

Yes, assuming by "the line" you mean "the real line $\mathbb{R}$ equipped with the Euclidean topology".

Here is a broader take on this: compare with second countability. Facts:

1) Every second countable topological space is separable.
2) The converse need not hold, but every metrizable separable space (like $\mathbb{R}$!) is second countable.
3) Every subspace of a second countable space is second countable: just restrict the base.

Thus every subspace of a metrizable separable space is separable.


Added in response to the OP's comment: Hmm. In the departments with which I am familiar, students take undergraduate general topology before they take graduate real analysis. I am not completely confident in the ability of someone to succeed at the latter without having at least some mastery of the former.

Anyway, you asked me if there is a simpler explanation. Well, "simpler" is subjective, so in my opinion...no, the explanation I gave above is the one which is simplest to me. But I can explain it in a way which doesn't use second countability: here goes.

For $k \in \mathbb{Z}$ and $n \in \mathbb{Z}^+$, let $I_{n,k} = [\frac{k}{2^n},\frac{k+1}{2^n})$.

Fix $n \in \mathbb{Z}^+$ and consider the partition $\{ I_{n,k} \}_{k \in \mathbb{Z}}$ of $\mathbb{R}$. For each $k \in \mathbb{Z}$ such that $I_{n,k} \cap X$ is nonempty, choose one point $a_{n,k}$ of $X$. Let $A$ be the subset of $X$ consisting of these points $a_{n,k}$. Then $A$ is countable. Moreover, for every $x \in X$, we may construct a sequence in $A$ converging to $x$ as follows: let $x_n$ be the unique element $a_{n,k}$ $A$ lying in the same interval $I_{n,k}$ as $x$.

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I do mean the real line with This topology. I will have to learn the second countability, although it's beyond my current knowledge. Can you explain it in a it simpler way, perhaps something covered in a first graduate course in real analysis? –  Reggie Dec 7 '11 at 16:13

This is just the special case of Pete’s answer that you need for your problem, with a few more details.

There are countably many rationals, so there are countably many open intervals with rational endpoints; list them as $\{I_n:n\in\mathbb{N}\}$. Let $X$ be any subset of $\mathbb{R}$. For each $n\in\mathbb{N}$ there are two possibilities:

  1. $I_n\cap X=\varnothing$: In this case do nothing.
  2. $I_n\cap X\ne\varnothing$: Let $x_n$ be any point of $I_n\cap X$.

Now let $D=\{x_n:I_n\cap X\ne\varnothing\}$. Clearly $D$ is countable, and I claim that $D$ is a dense subset of $X$, i.e., such that $X=\operatorname{cl}_X D$.

To see this, let $x$ be any point of $X$, and let $(x-\epsilon,x+\epsilon)$ be an open interval centred at $x$. The rationals are dense in $\mathbb{R}$, so there are rationals $p\in(x-\epsilon,x)$ and $q\in(x,x+\epsilon)$. Then $(p,q)$ is an open interval with rational endpoints, so $(p,q)=I_n$ for some $n\in\mathbb{N}$. Moreover, $x\in I_n\cap X$, so $I_n\cap X\ne\varnothing$, and according to $(2)$ above there is a point $x_n\in D\cap I_n$. But $I_n\subseteq(x-\epsilon,x+\epsilon)$, so $x_n\in (x-\epsilon,x+\epsilon)\cap D$. In other words, for each $\epsilon>0$ the $\epsilon$-nbhd of $x$ contains a point of $D$, and therefore $x\in\operatorname{cl}_X D$, as desired.

To connect this with second countability: the family $\{I_n:n\in\mathbb{N}\}$ is a countable base for the Euclidean topology of $\mathbb{R}$. A base for the topology is just a collection $\mathscr{B}$ of open sets such that every open set is a union of some subcollection of $\mathscr{B}$; a space is second countable if its topology has a countable base. If $X$ is any subset of $\mathbb{R}$, $\{X\cap I_n:n\in\mathbb{N}\}$ is a countable base for the subspace topology on $X$, so second countability is hereditary: if a space is second countable, so are all of its subspaces. And the trick that I used above can pretty clearly be applied to any second countable space to get a countable dense subset $-$ which is why every second countable space is separable, and even hereditarily separable.

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$@$Brian: +1. I think this will be helpful to the OP. –  Pete L. Clark Dec 8 '11 at 0:34

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