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Homeproof of Heine-Borel theorem

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# proof of Heine-Borel theorem

To shorten the proofs, we will be using certain concepts and facts from general point-set topology, such as the very first assertion below. Even if you are not familiar with them, these assertions are all easily proven from the definitions (left as an exercise).

###### Compact implies closed and bounded.

Since $\mathbb{R}^{n}$ is a Hausdorff space, any compact subset $A$ has to be closed as well. That $A$ is also bounded is very easily seen by taking the open cover $\{(-k,k)^{n}\}_{{k\in\mathbb{N}}}$. Passing to a finite cover, we see that $A$ is contained in a bounded set. ∎

###### Reduction for closed and bounded implies compact.

To prove that if $A$ is closed and bounded implies it is compact, it is only necessary to prove that the rectangle $[a,b]^{n}$ is compact. For general $A$, since $A$ is bounded, it is contained in some such compact rectangle $[a,b]^{n}$. Since $A$ is closed and contained in a compact set, $A$ is also compact. ∎

###### The case $n=1$: the closed interval is compact..

Let $\mathcal{C}$ be an arbitrary cover of $[a,b]$ by open sets in $\mathbb{R}^{1}$. Define

$S=\{x\in[a,b]:\textrm{some finite subcollection of $\mathcal{C}$ covers $[a,x]% $}\}\,.$ |

Set $t=\sup S$; then $a\leq t\leq b<\infty$.

We first note that the supremum $t$ is attained: that is, there is a finite subcollection of $\mathcal{C}$ that covers $[a,t]$. Obviously we can choose one open set $U$ from $\mathcal{C}$ that covers the point $t$. This $U$ must also cover the open interval $(t-\epsilon,t+\epsilon)$ for some $\epsilon>0$. But by the definition of $t$, the interval $[a,t-\epsilon]$ has a finite subcover. When this finite subcover is put together with $U$, we obtain a finite subcover for $[a,t]$.

It is easy to see that $t$ cannot be less than $b$. For the same open set $U$ above covers the interval $(t,t+\epsilon)$, so we have a finite subcover for $[a,t+\epsilon/2]$. If $t$ is the supremum then it has go to be at the right endpoint of the interval $[a,b]$, i.e. $t=b$. ∎

For the case $n>1$, apply Tychonoff’s Theorem, in the finite case, which states:

If $X_{1},\ldots,X_{n}$ are each compact, then $X_{1}\times\cdots\times X_{n}$ in the product topology is compact.

Here we set $X_{i}$ to be the closed intervals; then the general closed rectangle in $\mathbb{R}^{n}$ is compact. It is not hard to see that the product topology for the closed rectangle is the same as its subspace topology in the product topology $\mathbb{R}^{n}=\mathbb{R}\times\cdots\times\mathbb{R}$. And it is again a standard exercise to show that the product topology on $\mathbb{R}^{n}$ is the same as the norm topology on $\mathbb{R}^{n}$ as a vector space.

This completes the proof of the Heine-Borel theorem.

# Proof by a bisection argument

There is another proof of the Heine-Borel theorem for $\mathbb{R}^{n}$ without resorting to Tychonoff’s Theorem. It goes by bisecting the rectangle along each of its sides. At the first stage, we divide up the rectangle $A$ into $2^{n}$ subrectangles. Suppose the open cover $\mathcal{C}$ of $A$ has no finite subcover. Then one of the subrectangles — call it $A_{1}$ — must have no finite subcover by $\mathcal{C}$. We can subdivide $A_{1}$ into $2^{n}$ pieces; since $A_{1}$ has no finite subcover, one of the new subrectangles of $A_{1}$ also has no finite subcover. And continue dividing to get a nested sequence of rectangles $A\supset A_{1}\supset A_{2}\supset\cdots$ whose side lengths approach zero, and possessing no finite subcover.

By the nested interval theorem, the “limit rectangle” $\bigcap_{{i=1}}^{\infty}A_{i}$ must consist of a sole point $x$, and this obviously has a finite subcover by an open set $U\in\mathcal{C}$. But $U$ must contain a small rectangle with centre $x$, which for $i$ large enough, contradicts $A_{i}$ having no finite subcover.

Of course, in both proofs of the Heine-Borel theorem, the completeness of the reals (the least upper bound property) enters in an essential way.

## Mathematics Subject Classification

54D30*no label found*

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## Comments

## Heine Borel

The proof is incomplete. It applies to the open interval ]a,b[ for which H-B is not true.

You must include consideration of the covering of both end points