$\def\R{\mathbf{R}}$ $\def\N{\mathbf{N}}$ Note. When I use the term "compact", I mean specifically "closed and bounded." I also only understand topological notions inside only $\R$, no other metric space/topology .

The proofs of the Heine-Borel Theorem (compact $\implies$ always finitely sub-coverable) I've seen are all indirect. They show the existence of a finite-subcover but never how to find it. My end goal is to come up with a direct proof of the Heine-Borel Theorem. Here's a conjecture I came up with that I believe is true (intuitively), but cannot prove.

Conjecture 1. Let $K \subseteq \R$ be compact, and $K'$ be the set of limit points of $K$. If $O$ is an open cover for $K'$, then $K \setminus O$ is necessarily finite.

If this conjecture is true, (which I am assuming for the moment), then this shortens our problem to finding a finite sub-cover for $K'$, since by the conjecture, of the remaining points of $K$ our sub-cover missed, we must've missed only finitely many, and can hence add back in a few more sets to cover them. Hence if $K'$ itself is finite, then it's easy to construct a finite sub-cover for $K$. The problem arises when $K'$ is infinite. I'm thinking, if $K'$ was countably infinite, then we can look at $K''$ (the set of the limit points of $K'$). If $K''$ is finite, then we can finitely sub-cover $K'$, and hence finitely sub-cover $K$. Thus if $K_0=K$ is a compact set with $K_{n+1}={K_{n}}',$ then given an open-cover of $K$, we can construct a finite sub-cover if there exists an $N \in \N$ such that $K_N$ is finite. It makes me wonder whether this is always the case for an arbitrary countable compact set, which leads me to making the following conjecture:

Conjecture 2. Let $K\subseteq\R$ be compact and countably infinite. Define $K_0 := K$, and define $K_{n+1} := {K_{n}}'$. Then there exists an $N \in \N$ such that $K_N$ is finite. In other words, as you iteratively take set of the limit points of any countable compact set, you will eventually get a finite set.

So that's good and all, but then there are compact sets where no matter how many times I iteratively take its set of limit points, it remains infinite (such as uncountable compact sets or non-empty perfect sets).

Is this line of attack worthwhile, and is it possible to extend these arguements to directly prove the Heine-Borel Theorem?

  • $\begingroup$ Your conjecture 1 is true because every infinite subset of a compact subset of $\mathbb{R}^n$ has a limit point. However I don't see how this approach can be used for uncountable case. $\endgroup$
    – freakish
    Feb 1, 2022 at 7:40

1 Answer 1


Unfortunately, you cannot constructively prove the Heine-Borel theorem. Any proof of this theorem necessarily invokes proof by contradiction or some equivalent principle at some point.

To show this, we need to come up with some model of set theory (minus the proof by contradiction rule and the axiom of choice) in which all the axioms of set theory hold but $[0,1]$ is not compact.

In particular, we can find a model known as the “effective topos” in which every function $\mathbb{N} \to \mathbb{N}$ can be computed by some algorithm. This principle is known as “Church’s Thesis”.

Starting with Church’s Thesis, we show that $[0, 1]$ is not compact. Church’s Thesis allows us, with some cleverness, to construct a continuous function $f : [0, 1] \to (0, \infty)$ such that $\inf\limits_{x \in [0, 1]} f(x) = 0$. This in turn contradicts the claim that $[0, 1]$ is compact, since $\{f^{-1}((0, 1/n)) \mid n \in \mathbb{Z}_+\}$ is an open cover with no finite sub cover. For some details of the construction of $f$, see https://projecteuclid.org/journalArticle/Download?urlId=pjm%2F1102710574&referringURL=https%3A%2F%2Fduckduckgo.com%2F&isResultClick=False.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .