# The Unapologetic Mathematician

## The Heine-Borel Theorem

We’ve talked about compact subspaces, particularly of compact spaces and Hausdorff spaces (and, of course, compact Hausdorff spaces). So how can we use this to understand the space $\mathbb{R}$ of real numbers, or higher-dimensional versions like $\mathbb{R}^n$?

First off, $\mathbb{R}$ is Hausdorff, which should be straightforward to prove. Unfortunately, it’s not compact. To see this, consider the open sets of the form $(-x,x)$ for all positive real numbers $x$. Given any real number $y$ we can find an $x$ with $|y|, so $y\in(-x,x)$. Therefore the collection of these open intervals covers $\mathbb{R}$. But if we take any finite number of them, one will be the biggest, and so we must miss some real numbers. This open cover does not have a finite subcover, and $\mathbb{R}$ is not compact. We can similarly show that $\mathbb{R}^n$ is Hausdorff, but not compact.

So, since $\mathbb{R}^n$ is Hausdorff, any compact subset of $\mathbb{R}^n$ must be closed. But not every closed subset is compact. What else does compactness imply? Well, we can take the proof that $\mathbb{R}^n$ isn’t compact and adapt it to any subset $A\subseteq\mathbb{R}^n$. We take the collection of all open “cubes” $(-x,x)^n$ consisting of $n$-tuples of real numbers, each of which is between $-x$ and $x$, and we form open subsets of $A$ by the intersections $U_x=(-x,x)^n\cap A$. Now the only way for there to be a finite subcover of this open cover of $A$ is for there to be some $x$ so that $U_x=A$. That is, every component of every point of $A$ has absolute value less than $x$, and so we say that $A$ is “bounded”.

We see now that every compact subset of $\mathbb{R}^n$ is closed and bounded. It turns out that being closed and bounded is not only necessary for compactness, but they’re also sufficient! To see this, we’ll show that the closed cube $\left[-x,x\right]^n$ is compact. Then a bounded set $A$ is contained in some such cube, and a closed subset of a compact space is compact. This is the Heine-Borel theorem.

In the $n=1$ case, we just need to see that the interval $\left[-x,x\right]$ is compact. Take an open cover $\{U_i\}$ of this interval, and define the set $S$ to consist of all $y\in\left[-x,x\right]$ so that a finite collection of the $U_i$ cover $\left[-x,y\right]$. Then define $t$ to be the least upper bound of $S$. Basically, $t$ is as far along the interval as we can get with a finite number of sets, and we’re hoping to show that $t=x$. Clearly it can’t go past $x$, since $S\subseteq\left[-x,x\right]$. But can it be less than $x$?

In fact it can’t, because if it were, then we can find some open set $U$ from the cover that contains $t$. As an open neighborhood of $t$, the set $U$ contains some interval $(t-\epsilon,t+\epsilon)$. Then $t-\epsilon$ must be in $S$, and so there is some finite collection of the $U_i$ which covers $\left[-x,t-\epsilon\right]$. But then we can just add in $U$ to get a finite collection of the $U_i$ which covers $\left[-x,t+\frac{\epsilon}{2}\right]$, and this contradicts the fact that $t$ is the supremum of $S$. Thus $t=x$ and there is a finite subcover of $\left[-x,x\right]$, making this closed interval compact!

Now Tychonoff’s Theorem tells us that products of closed intervals are also compact. In particular, the closed cube $\left[-x,x\right]^n\subseteq\mathbb{R}^n$ is compact. And since any closed and bounded set is contained in some such cube, it will be compact as a closed subspace of a compact space. Incidentally, since $n$ is finite, we don’t need to wave the Zorn talisman to get this invocation of the Tychonoff magic to work.

As a special case, we can look back at the one-dimensional case to see that a compact, connected space must be a closed interval $\left[a,b\right]$. Then we know that the image of a connected space is connected, and that the image of a compact space is compact, so the image of a closed interval under a continuous function $f:\mathbb{R}\rightarrow\mathbb{R}$ is another closed interval.

The fact that this image is an interval gave us the intermediate value theorem. The fact that it’s closed now gives us the extreme value theorem: a continuous, real-valued function $f$ on a closed interval $\left[a,b\right]$ attains a maximum and a minimum. That is, there is some $c\in\left[a,b\right]$ so that $f(c)\geq f(x)$ for all $x\in\left[a,b\right]$, and similarly there is some $d\in\left[a,b\right]$ so that $f(d)\leq f(x)$ for all $x\in\left[a,b\right]$.

January 18, 2008 -

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5. [...] corresponding , and so we find that is a closed interval covered by the open intervals . But the Heine-Borel theorem says that is compact, and so we can find a finite collection of the which cover . Renumbering the [...]

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6. [...] linear operators — is continuous by our assumption, the function is continuous as well. The extreme value theorem tells us that since is compact this continuous function must attain a maximum, which we call [...]

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