# Proof of Heine-Borel in $\mathbb R^n$

The textbook I'm using has a somewhat confusing proof of one of the directions of Heine-Borel's Theorem, namely the one that states that a closed and bounded subset of $\mathbb R^n$ is compact. It uses sequences and "Changing the centers" to 'move' the balls to $\mathbb Q^n$ and then constructs a sequence which leads to a contradiction, the latter of which I'm confused about.

Could someone give me a more coherent proof in the same style or point me in the direction of a fairly understandable proof?

-
What definition of compact do you use, convergent subsequences or finite covers? Or do you already know the equivalence between the two in the metric case? –  Plop Mar 17 '11 at 0:32
Finite covers, but we know that sequentially compact <=> compact for metric spaces. –  kevmo314 Mar 17 '11 at 0:35
In your case sequentially compact is way easier to handle, use Bolzano-Weierstrass. –  Plop Mar 17 '11 at 0:41

By Bolzano-Weierstrass, any bounded sequence $(x_k)$ in $\mathbb R^n$ has a convergent subsequence. Now try to construct a sequence sequence that does not converge, assuming a set is closed and bounded, yet there exists an open covering that has no finite subcover.

Spoiler:

Let $A \subset \mathbb R^n$ be closed and bounded, s.t. there exists an open covering which has no finite subcovering. So let $U_i$ be such an open covering. Pick an open set, which we denote by $U_0$. Let

$x_0 \in A - U_0$ , which exists by assumption.

Choose $U_1 \in (U_i)$ with $x_0 \in U_1$. This exists by assumption.

Now we proceed in that manner and choose a sequence of open sets from the covering and a sequence in $A$ that is always one step before the covering sequence.

Then the constructed sequence $(x_k)$ is bounded, as it is contained in $A$. It has a convergent subsequence, say, to $x$. Choose such a sequence and keep the label $x_k$. As $A$ is closed, $x \in A$.

By assumption, we can choose some $V$ which covers $x$. We do so at any point of the sequence, say after some fixed number of steps. We strip off all elements of the sequence which are in $V$, and construct the two sequences again in the same fashion, with $V$ being included in the sequence of open sets that we used to define the sequence $x_k$.

We repeat this. Thus we construct a bounded sequence that does not converge - because otherwise, it would be covered by an open set in the sequence we have choosen.

Hence, if $A$ is closed and bounded, every open covering has a finite subcover. $qed$

-
The first part of this proof seems to make sense and I understand that part of the construction of a finite subcovering, but I'm not sure what happens after "We strip off all the elements..." Is that necessary for the proof itself as didn't it just construct a finite subcovering? If so, could you explain that a little more? –  kevmo314 Mar 17 '11 at 0:46

The best proof I've seen is a topological one, using a few theorems about different topologies.

Namely: 1) In an order topology, $\left[ a,b \right]$ is a compact set. 2) The order topology on $\mathbb{R}$ is equivalent to the metric induced topology on $\mathbb{R}$. 3) Closed subsets of compact sets are compact.

Compact subsets of Hausdorff spaces are closed, and in the metric topology we can take $B_{n}(0)$, the ball of radius $n$ around 0, and since $\cup_{n=1}^{\infty} B_{n}(0) = \mathbb{R^{m}}$, we can choose a finite subcover, and so it is a bounded set.

That proves necessity. To prove sufficiency we simply note the following.

Let $K$ be a closed, bounded subset of $\mathbb{R}^n$. Then $K \subseteq \left[ -M, M\right]^n$ is actually compact, as $\left[ -M, M\right]$ is compact by 1), and since the order and metric topologies are equivalent, it is compact in the metric topology. By Tychonoff's Theorem(not even necessary, finite products of compact sets are compact), we have $\left[ -M, M\right]^n$ is also compact. Closed subsets of compact sets are compact, and $K$ is a closed subset of $\left[-M,M\right]$, so $K$ is compact, proving sufficiency.

-

One can use the following lemma to prove it: Let $(\mathbb{R}^n, d)$ be a Euclidean space with either the Euclidean metric or the taxicab metric. Then every bounded sequence in $(\mathbb{R}^n, d)$ has at least one convergent subsequence.

-