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Show that a metric space has a countable base if and only if each open cover has a countable subcover.

Proof

$\Rightarrow$ using Lindöf's lemma

But how does one show $\Leftarrow$?

Consider $\{ U_i\}_{i\in I}$ to be the open cover, then there is a countable subcover $\{ U_j\}_{j\in J}$ where $J\subseteq I$ countable.

But I can't use this open cover as a base. How should I manipulate this making sure it fits?

Example

Consider $]0,+\infty[$ and the usual metric, then $\{ {]0,3[},{]2,5[, {]4,7[}, \ldots} \}$ is a countable cover. This is not a countable base though. $]1,2[$ cannot be constructed as a union of base sets.

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    $\begingroup$ Consider the cover by balls of radius $1/n$, this has a countable subcover. Now take the union of all these over all $n$. $\endgroup$ – JHance Jun 6 '15 at 21:45
  • $\begingroup$ Btw, Lindelöf, not Lindöf. Whence the name "Lindelöf space" naturally :). $\endgroup$ – MickG Jun 7 '15 at 7:52
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In more technical terms, "has a countable base" = "is second-countable", and "every open cover has a countable subcover" = "the space is Lindelöf". So you are asking how to prove that a Lindelöf metric space is second countable. I found a proof of the double implication here. What it does to prove the implication you're interested in is this. Consider $C_k=\{B_{\frac1k}(x):x\in X\}$, $X$ being our Lindelöf metric space, that is for each $k$ we consider the set (or family) $C_k$ of all the balls of radius $\frac1k$ with center in $X$. For each of these, we have a countable subcover $S_k$. Let $\mathcal{B}=\bigcup_kS_k$. This set is a countable base. That it be countable is evident since it is a countable union of countable subcovers. Take an open set $U\subseteq X$ and $x\in U$. It is clear that we can find $k\in\mathbb{N}$ such that a ball in $S_k$ is contained in $U$ and $x$ is in that ball. Indeed, let $r:=\sup\{\epsilon\in\mathbb{R}:B_\epsilon(x)\subseteq U\}$. Choose $\ell$ such that $\frac1k<\frac r3$. We will surely find an element of $S_\ell$ containing $x$, since $S_\ell$ is a subcover. This element will necessarily be contained in $B_r(x)$, since if $y$ is in that ball, $d(x,y)\leq d(x,z)+d(z,y)<\frac r3+\frac r3$, $z$ being that ball's center. Thus, any open set can be written as a union of elements of $\mathcal{B}$, proving it is a base.

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