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I was thinking a bit about the proof of the fact that $\ell^{\infty}$ is not separable. And from the proof I saw, which uses a subspace which is discrete and uncountable, I thought I can prove it for any space which has the property.

Let's take $M$ a metric space with a discrete subspace $X$. Let $D$ be a dense subset of $M$. For every $x\in X$, we consider the ball $B_x=B_{\frac{1}{3}}(x)$. Then since $D$ is dense, for every such $x$, $B_x\cap D\neq \varnothing$, so let's take an arbitrary fixed element $a(x)\in B_x\cap D$ for each $x$.

Since $x\neq y\Rightarrow B_x\cap B_y=\varnothing$ (because $d(x,y)=1$), so $(B_x\cap D)\cap (B_y\cap D)=\varnothing\Rightarrow a(x)\neq a(y)$, so $x\mapsto a(x)$ is injective, thus $D$ is uncountable and $M$ is not separable.

Is that right?

Edit: Now I see that this can be solved using this since the subset $D$ would not be separable.

Nell
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  • By a discrete subset $S$, do you mean $S$ with subspace topology is discrete, or $S$ with subspace metric has discrete metric? – edm Aug 18 '17 at 05:10

2 Answers2

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It's correct. The underlying topological argument is as follows. If you can find a collection of pairwise disjoint nonempty open sets in $X$, then any dense subset of $X$ must have cardinality at least as large as that of the collection.

While the argument used in your edit will work in metric spaces, it fails in general for topological spaces.

John Griffin
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  • Could you provide an example of a top. space that it fails in? – Duncan Ramage Aug 18 '17 at 04:30
  • John, now that I see it, the proposition seems very general in the sense that in a topological space we must only guarantee the existence of an uncountable family of that kind (And I think that does not imply metrizable). – Nell Aug 18 '17 at 04:32
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    @DuncanRamage The classic example is the Sorgenfrey plane. It is separable because $\mathbb{Q}\times\mathbb{Q}$ is dense, but it has an uncountable discrete subspace (the antidiagonal) which is thus nonseparable. You can also cook up an example by taking an uncountable discrete space and adjoining an element whose only neighborhood is the whole space. – John Griffin Aug 18 '17 at 04:33
  • @Nell The main generality that is gained is that there is no mention of a metric. But in reality its pretty much the same as what you've provided. – John Griffin Aug 18 '17 at 04:36
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    That example is very interesting. – Nell Aug 18 '17 at 04:37
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    @JohnGriffin, thank you! – Duncan Ramage Aug 18 '17 at 04:42
  • Is that collection of pairwise disjoint nonempty open sets constructed using the uncountable discrete subset? – edm Aug 18 '17 at 05:01
  • @edm Precisely! If you have an uncountable discrete metric space, then the balls of radii $1/2$ around each point of the subset will be as desired. If we just assume topologocally discrete, we can proceed as in orangeskid's answer. – John Griffin Aug 18 '17 at 05:06
  • I think your argument could contain error for a general topological space, specifically the construction of that collection of open sets. For example, consider three-point space ${x,y,z}$ with topology ${\varnothing,{y},{x,y},{y,z},{x,y,z}}$. For the discrete subset ${x,z}$, you can only construct the collection of open sets ${{x,y},{y,z}}$ which is not pairwise disjoint. – edm Aug 18 '17 at 05:17
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    @edm Oh sorry I thought you meant in the particular case of metric spaces (as in OP with $\ell^\infty$). As you show, in general we can't construct the open sets from a discrete space. This actually shows from a different angle why the argument in the edit of the OP won't work in general. There are separable spaces with uncountable discrete subspaces, which means the open sets obtained from the discrete subspace cannot be pairwise disjoint. – John Griffin Aug 18 '17 at 05:22
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All is good, just want to point out the definition of a discrete subset $X$. $X$ is discrete if for every $x$ in $X$ there exists an open subset $U_x$ of $M$ such that $U_x \cap X= \{x\}$. If $M$ is a metric space that means for every $x$ in $X$ there exists an open ball $B_x \colon = B(x, r_x)$ so that $B_x \cap X = \{x\}$. Now, apriori, the balls $B_x$ might intersect, but a minor adjustment will do: the balls $B'_x\colon = B(x, \frac{r_x}{2})$ are pairwise disjoint. Indeed, assume that $B'_x \cap B'_y \ne \emptyset$. Then $d(x,y)< \frac{r_x+ r_y}{2}$. If $r_y \le r_x$ that would imply $d(x,y) < r_x$, contradiction.

orangeskid
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