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

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!

Solution 1:

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.

Solution 2:

I think it's worth pointing out that, although second-countability implies Lindelöf requires some choice, the Heine-Borel theorem itself doesn't. To prove that a closed and bounded subset of $\mathbb{R}$ is compact, consider first the case $[0,1]$.

Given an open cover of $[0,1]$, consider the set $S = \{x \in [0,1]\mid\text{there is a finite subcover of }[0,x]\}$. Then $0 \in S$, $S$ is open, and the least upper bound of $S$ is an element of $S$ (check each of these facts). It follows that $1 \in S$ is the least upper bound, so there is a finite subcover.

To prove that any closed and bounded set is compact, realize that it is a closed subset of $[-a,a] \cong [0,1]$ for sufficiently large $a$.