# Equivalence of three properties of a metric space.

Another question about the convergence notes by Dr. Pete Clark:

http://math.uga.edu/~pete/convergence.pdf

(I'm almost at the filters chapter! Getting very excited now!)

On page 15, Proposition 4.6 states that for the following three properties of a topological space $X$,

$(i)$ $X$ has a countable base.

$(ii)$ $X$ is separable.

$(iii)$ $X$ is Lindelof (every open cover admits a countable subcover).

we always have $(i)\Rightarrow (ii)$ and $(i)\Rightarrow (iii)$.

Also, we if $X$ is metrizable, we have $(iii)\Rightarrow (i)$, and thus all three are equivalent.

This last part confuses me. We establish all the implications claimed in the proof, but there seems to be a missing link in the claim that all three are equivalent: namely $(ii)\Rightarrow (iii)$.

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Actually, separable spaces aren't generally Lindelof (for an example, consider the Sorgenfrey plane). That does turn out to be the case for (pseudo)metrizable spaces, yes, but their reasoning is incorrect. It is missing precisely the link you suppose. – Cameron Buie May 15 '12 at 2:23
Dear Kyle: By the way, when you finish with the notes I'll be very grateful if you can send me one long list of corrections. – Pete L. Clark May 15 '12 at 3:10
Irrelevantly (but somewhat interesting), last week and today I am giving two lectures in our local topology and logic seminar presenting a situation in which those conditions are not equivalent at all, namely when the axiom of choice fails. See more in the paper I cited here. – Asaf Karagila May 15 '12 at 7:15

Note that a separable metric space has a countable basis. Specifically, we take a countable dense subset $S$ and take the set of balls centered at $s$ with radius $1/n$ for each $n \in N$, $s \in S$. This can be checked to be a basis. So then $(ii) \Rightarrow (i)$ is proven, which is the missing link.

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That works, too, and much more simply! – Cameron Buie May 15 '12 at 2:44

The missing implication "separable metrizable implies second countable" is rather easy to prove -- as Carl's answer shows -- but the proof should still appear in the notes.

I have uploaded a new version taking this into account. Thanks for bringing this to my attention.

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Carl’s approach is almost certainly the easiest way to patch the gap in the notes. A nice variant is to let $D=\{x_n:n\in\Bbb N\}$ be a countable dense subset of $X$, define a map $$h:X\to\Bbb R^{\Bbb N}:x\mapsto\langle d(x,x_n):n\in\Bbb N\rangle\;,$$ and prove that $h$ is s homeomorphism of $X$ onto a subspace of $\Bbb R^{\Bbb N}$: being a countable product of second countable spaces, $\Bbb R^{\Bbb N}$ is easily shown to be second countable.

It is possible to give direct proofs of $(ii)\implies(iii)$ and $(iii)\implies(ii)$, but every one that I can think of either (a) smuggles in what amounts to a proof of second countability along the way or (b) uses much higher-powered machinery.

As an example of (a), consider the following proof that $(iii)\implies(ii)$.

For each $n\in\Bbb N$ let $\mathscr{U}_n=\{B(x,2^{-n}):x\in X\}$, the set of open $2^{-n}$-balls in $X$; $\mathscr{U}_n$ is an open cover of $X$, so it has a countable subcover $\mathscr{V}_n=\{B(x_n(k),2^{-n}):k\in\Bbb N\}$. Let $$D=\{x_n(k):n,k\in\Bbb N\}\;;$$ clearly $D$ is countable. Let $W$ be any non-empty open set in $X$. Pick $x\in W$; there is an $n\in\Bbb N$ such that $B(x,2^{-n})\subseteq W$. $\mathscr{V}_n$ covers $X$, so there is some $k\in\Bbb N$ such that $x\in B(x_n(k),2^{-n})$; but then $x_n(k)\in D\cap B(x,2^{-n})\subseteq D\cap W$, and $D$ is dense in $X$. $\dashv$

With just a little more work this becomes the argument that Carl suggested to prove that $(iii)$ implies $(i)$.

As an example of (b), the fact that every metric space has a $\sigma$-discrete base almost immediately implies that every separable metric space is Lindelöf:

Let $\mathscr{B}=\bigcup\{\mathscr{B}_n:n\in\Bbb N\}$ be a base for $X$ such that each $\mathscr{B}_n$ is discrete. Suppose that $\mathscr{U}$ is an open cover of $X$ with no countable subcover. Let $\mathscr{R}\subseteq\mathscr{B}$ be a refinement of $\mathscr{U}$ covering $X$. $\mathscr{R}$ has no countable subcover, so $\mathscr{R}\cap\mathscr{B}_n$ is uncountable for some $n\in\Bbb N$. But then $\mathscr{R}\cap\mathscr{B}_n$ is an uncountable family of pairwise disjoint, non-empty open sets, and $X$ cannot be separable. $\dashv$

The reason for the difficulty in going directly between $(ii)$ and $(iii)$ is that in general separability and Lindelöfness are very far from being equivalent; in the metric setting it’s second countability that ties them together by being equivalent to each. Here are a couple of examples illustrating their independence, even in rather nice spaces.

If you retopologize $\Bbb R$ by making every $x\in\Bbb R\setminus\{0\}$ and giving $0$ a base of nbhds of the form $A\setminus C$, where $C$ is any countable subset of $\Bbb R\setminus\{0\}$, then $X$ is a very nice space that is Lindelöf but not separable.

On the other hand, a slightly more complicated retopologization of $\Bbb R$ yields a very nice space that is separable but not Lindelöf. Make each rational an isolated point. To each irrational $x$ associate a sequence $\langle q_x(k):k\in\Bbb N\rangle$ of rational numbers converging monotonically to $x$ in the usual topology; a base of the topology at $x$ consists of the sets $B_n(x)=\{x\}\cup\{q_x(k):k\ge n\}$ for $n\in\Bbb N$. Note that if $x$ and $y$ are distinct irrationals, the sets $\{q_x(k):k\in\Bbb N\}$ and $\{q_y(k):k\in\Bbb N\}$ have at most finitely many points in common; this ensures that the space is completely regular and Hausdorff. $\Bbb Q$ is a countable dense subset, so the space is separable. And $$\{B_0(x):x\in\Bbb R\setminus\Bbb Q\}\cup\Big\{\{q\}:q\in\Bbb Q\Big\}$$ is an open cover with no countable subcover, so the space is not Lindelöf.

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