Heine-Borel Theorem ($\mathbb{R}^k$) (in ZF)

Question

Heine-Borel Theorem; If $E \subset \mathbb{R}^k$, then $E$ is compact iff $E$ is closed and bounded.

I have proved 'closed and bounded⇒compact' and 'compact⇒bounded'. (There exists $r\in \mathbb{R}$ such that for every $x\in E$, $|x|<r$)

The proof in Rudin PMA p.40 uses 'countable axiom of choice'

I have googled it and found some proofs, but they all used some weaker form of AC.

Please help me how to show that $compact⇒closed$ in ZF..

Answer 1

We will prove that compact implies closed by contraposition.

Suppose that $E$ is not closed. Then there is some $x\notin E$ such that every neighbourhood of $x$ has a nonempty intersection with $E$. In particular, the collection $E_n:=E\setminus \overline {B(x,1/n)}$ (where $\overline {B(x,1/n)}$ is the closed ball centered at $x$ with radius $1/n$) is an infinite, nondecreasing open cover of $E$ (because for any $p\in E$ and $n>1/d(x,p)$ we have $p\in E_n$).

It is enough to show that $E_n$ does not stabilize. But if it did, then we would have for some $N<\omega$ that $E_N=E$, so $B(x,1/N)$ would be disjoint from $E$, so $x$ would not belong to the closure of $E$, so we're done.

This argument should work in an arbitrary Hausdorff space, though without countable character the cover will not be a sequence, but a directed set.

Answer 2

Any compact subset of a Hausdorff space is closed.

Proof: Let $A \subset X$ be compact. We show that $X -A$ is open. Let $x \in X - A$. Since $X$ is Hausdorff, for every $a \in A$, there are disjoint open sets $U_a$ and $V_a$ such that $x \in U_a$ and $a \in V_a.$ So $\{V_a\}_{a \in A}$ is an open cover which has a finite subcover $\{V_{a_1},...,V_{a_n}\}$. Let $V =\bigcup\limits_{i=1}^n V_{a_i}$ and $U = \bigcap\limits_{i=1}^n U_{a_i}$. Clearly, $U \cap V = \emptyset$, so $x \in U \subset X -A$. Hence, $X - A$ is open.