# Check on a proof I saw on another thread: Metrizable Lindelöf spaces have a countable basis

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.

• The countable dense subset should be $\bigcup_{n\in\mathbb N}\{x\mid B(x,\frac{1}{n})\in\mathscr R_n\}$ instead of $\bigcup_{n\in \mathbb N} \mathscr R_n$ – YuiTo Cheng Apr 21 '19 at 8:37
• yes i noted that in my post – H_1317 Apr 21 '19 at 8:37
• 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. – YuiTo Cheng Apr 21 '19 at 8:40
• 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. – Henno Brandsma Apr 22 '19 at 12:05
• Of course, $n$ must be large enough so that $\frac{1}{n}<\epsilon$. – 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 $$x\in \delta \subset B_d(1/n)\subset U.$$

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