Axiom of Choice – Does Proving Second Countable Implies Lindelöf Require It?

axiom-of-choicegeneral-topologysecond-countableset-theory

Background:

This question came up in my homework (but was not a homework problem). The problem was proving one direction of the Heine-Borel theorem. As with all proofs of compactness, one begins with, "Suppose $A$ is closed and bounded, and $\mathcal{U}$ is an open cover …" My proof, which I believe is typical, constructed a convergent sequence of points and derived a contradiction. The problem: to construct a sequence, one needs the cover to be countable.

So I invoked that second-countability implies Lindelöf. This is not controversial, of course, but I included a little argument along the lines of the answers in this question: Every open cover has a countable subcover (Lindelöf's lemma).

In particular, for each $x\in A$ there is some $U$ in $\mathcal{U}$ with $x\in U$; second-countability implies for each $x\in A$ there is some basis element $Q$ with $x\in Q\subset U$; the set of all such $Q$ covers $A$ and each $Q\subset U$ for at least one $U\in\mathcal{U}$, so picking one $U\supset Q$ for each $Q$ gives a countable cover of $A$.

I noticed that the answers in that thread included similar phrases: "Now for each $B\in\mathcal{B}U$ choose some $U(B)\in U$ such that $B\subseteq U$", "for each element $O$ of $\Sigma$ choose an element $U$ of $\Omega$ containing it"

My argument and the arguments suggested in the thread linked both seem to assume the Axiom of Choice. I'm not sure that I understand the axiom of choice, though. So, two questions:

1) Are we actually using the axiom of choice in these arguments?

2) Does (second countable) $\Rightarrow$ (Lindelöf) require the axiom of choice?

Thanks!

Best Answer

The results requires some choice. It is consistent that $\mathbb R$ is not Lindelöf but still second countable (without the axiom of choice).

The axiom of choice for countable families of subsets of $\mathbb R$ is in fact equivalent that second countable implies Lindelöf, and to a few other interesting assertions:

H. Herrlich and G. E. Strecker, When is $\mathbb{N}$ Lindelöf?, Comment. Math. Univ. Carolinae 38,3 (1997), 553-556.