I saw the following proof given of to the theorem below. I don't think the proof is correct, but I wasn't quite sure as it was given an up vote and thought I'd re post here to get some other opinions. Thanks in advance!

A metrizable Lindelöf space has a countable basis

The authors proof:

Note: This proof requires the assumption that every metrizable space with a countable dense subset has a countable basis.

Let $X$ be a metrizable Lindelhof space.

(Then as above)

For each positive integer $n$ let $\mathscr{U}_n=\left\{B\left(x,\frac1n\right):x\in X\right\}$; this is an open cover of $X$, so it has a countable subcover $\mathscr{B}_n$. Consider $\mathscr{B}=\bigcup_{n\in\Bbb Z^+}\mathscr{B}_n$.

I want to show $\mathscr{B}$ is dense in $X$.

Let $x\in X$, then let $B(x,\epsilon)$ be a basis element containing $x$. Then there exists an $n$ s.t. $x\in B(x', n)$ for some $x'$. But this implies that $x' \in B(x,\epsilon)$. So $x$ $\in \overline{\mathscr{B}}$, therefore $\overline{\mathscr{B}} = X$. Therefore $X$ has a countable basis since it contains a countable dense subset.

The original proof can be found here:

A metrizable Lindelöf space has a countable basis

My Review: -- First it seems that the set $\mathscr{B}$ is already an open covering and hence equal to all of X. Since X, the entire space, is a closed set, then $\bar{X}$ = X but X need not be countable, so $\mathscr{B}$ isn't necessarily countable either.

-- Second, the use of the open ball B(x', n) seems like it should be B(x',1/n), since thats how the author created his set he wants to verify as dense.

-- Third, I believe the set the author actually wants to verify as dense is the set $C$ = { x $\in$ $\mathscr{B}$ | x centered in an open ball = B(x,$\epsilon$) }. This set would actually be countable, and not equal to all of X in the case where X is uncountable.

-- Fourth, even considering the above point that the author is trying to prove the set $C$ as dense, I don't think the following is correct:

" Then there exists an $n$ s.t. $x\in B(x', n)$ for some $x'$. But this implies that $x' \in B(x,\epsilon)$. "

It may actually be true that C is dense, but the above phrasing implies that since x' has some neighborhood that contains x, then x' must be within epsilon of x. It seems perfectly plausible that d(x,x') $\gt$ $\epsilon$ with then n $\gt$ epsilon.

  • $\begingroup$ The countable dense subset should be $\bigcup_{n\in\mathbb N}\{x\mid B(x,\frac{1}{n})\in\mathscr R_n\}$ instead of $\bigcup_{n\in \mathbb N} \mathscr R_n$ $\endgroup$ – YuiTo Cheng Apr 21 '19 at 8:37
  • $\begingroup$ yes i noted that in my post $\endgroup$ – H_1317 Apr 21 '19 at 8:37
  • $\begingroup$ Then this proof has multiple flaws (as you have noted), but I think the main idea is ok. You can suggest edits to correct those errors you have found. $\endgroup$ – YuiTo Cheng Apr 21 '19 at 8:40
  • $\begingroup$ The proof you quoted is perfectly fine. First he defines the cover of $X$ by the open $\frac1n$-balls. This has a countable subcover $\mathcal{B}_n$. Do this for each $n \in \mathbb{N}$. We get countably many countable subcovers (so in total a countable family of open balls) and he then hints (not proves yet) that these subcovers together form a countable base. No step via separability, he directly constructs a countable base from applying the Lindelöf property repeatedly. @DanielWainfleet has filled in the missing details for you. $\endgroup$ – Henno Brandsma Apr 22 '19 at 12:05
  • 1
    $\begingroup$ Of course, $n$ must be large enough so that $\frac{1}{n}<\epsilon$. $\endgroup$ – YuiTo Cheng Apr 23 '19 at 6:50

Let $(X,d)$ be a Lindelof metric space.

For each $n\in \Bbb N$ let $B(n)=\{B_d(x,1/n):x\in X\}$). Let $C(n)$ be a countable subset of $B(n)$ with $\cup C(n)=X.$

Claim: The countable set $D=\{\emptyset\}\cup\,(\,\cup_{n\in \Bbb N}C(n)\,)$ is a base (basis) for $(X,d).$

Proof: It suffices that if $x\in U\subset X$ where $U$ is open, then there exists $\delta \in D$ such that $x\in \delta\subset U.$

There exists $n\in \Bbb N$ such that $B_d(x,1/n)\subset U.$ Then there exists $\delta=B_d(y,1/2n) \in C(2n)$ such that $x\in\delta.$ We have $d(x,y)<1/2n.$ And for any $z\in \delta$ we have $d(y,z)<1/2n.$ Therefore $$z\in \delta \implies d(x,z)\le d(x,y)+d(y,z)<$$ $$<1/2n+1/2n=1/n \implies$$ $$\implies z\in B_d(x,1/n).$$ So $x\in \delta \subset B_d(1/n)\subset U.$

  • $\begingroup$ Bonus marks available for explaining why $\emptyset \in D.$ $\endgroup$ – DanielWainfleet Apr 22 '19 at 1:53
  • $\begingroup$ Isnt that just part of the definition of a basis? $\endgroup$ – H_1317 Apr 22 '19 at 2:06
  • $\begingroup$ No $\emptyset$ is needed in any base. $\endgroup$ – Henno Brandsma Apr 22 '19 at 12:07

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