Heine-Borel Theorem

Theorem

If $F$ is a closed, bounded set of real numbers, then every open cover of $F$ has a finite subcover.

General Result

Any closed, bounded subspace of $\R^n$ is compact.

Proof

If $F$ is a closed, bounded set of real numbers, then it is the union of a finite collection of closed real intervals.

It is therefore a subset of a closed real interval that extends over the entire set $F$, and it follows we need only prove this theorem for a single closed real interval.

Let $\left[{a\,.\,.\,b}\right]$ be a closed real interval.

Let $\mathcal U$ be any open cover of $\left[{a\,.\,.\,b}\right]$.

Let $G = \left\{{ x \in \R: x \ge a, \left[{a\,.\,.\,x}\right] \text{ is covered by a finite subset of } \mathcal U }\right\}$.

Let us call points in $G$ $\text{good}$ (for $\mathcal U$).

This means that if $z$ is $\text{good}$, then $\left[{a\,.\,.\,z}\right]$ has that finite subcover we are trying to demonstrate for the whole of $\left[{a\,.\,.\,b}\right]$.

That is, what we want to show that $b$ is $\text{good}$.

Now, if $x$ is $\text{good}$, and $a \le y \le x$, then $y$ is $\text{good}$.

This is because $\left[{a\,.\,.\,y}\right] \subseteq \left[{a\,.\,.\,x}\right]$, and so $\left[{a\,.\,.\,y}\right]$ can be covered with any finite subset of $\mathcal U$ that covers $\left[{a\,.\,.\,x}\right]$, and we know that at least one such subset exists.

Now we show that $G \ne \varnothing$.

At the same time we show that $G \supseteq \left[{a\,.\,.\,a + \delta}\right]$ for some $\delta > 0$.

First, we see that $a$ must belong to some $U \in \mathcal U$, since $\mathcal U$ covers $\left[{a\,.\,.\,x}\right]$. Fix this $U$.

Since $U$ is open, $\left[{a\,.\,.\,a + \delta}\right) \subseteq U$ for some $\delta > 0$.

Hence $\left[{a\,.\,.\,x}\right] \subseteq U$ for all $x \in \left[{a\,.\,.\,a + \delta}\right)$.

It follows that all these $x$ are $\text{good}$.

Now the non-empty $G$ is either bounded above or it is not.

• Suppose $G$ is not bounded above.

Then there is some $c$ which is $\text{good}$ such that $c > b$.

From our initial observation that if $x$ is $\text{good}$, and $a \le y \le x$, then $y$ is $\text{good}$, it follows that $b$ is $\text{good}$, and hence the result.

• Now suppose $G$ is bounded above.

By the Continuum Property, $G$ admits a supremum in $\R$.

So let $g = \sup G$.

If $g > b$, then there is some $c$ which is $\text{good}$ such that $c > b$ as $g$ is the least upper bound for $G$.

Again, from our initial observation, it follows that $b$ is $\text{good}$, and hence the result.

So, let us try and get a contradiction, and assume that $g \le b$.

Note that $g > a$ as we have already seen that $\left[{a\,.\,.\,a + \delta}\right) \subseteq G$ for some $\delta > 0$.

Now since $g \in \left[{a . . b}\right]$, $g$ must belong to some $U_0 \in \mathcal U$.

Since $U_0$ is open, there exists some neighborhood $N_\epsilon \left({g}\right)$ of $g$ such that $U_0 \supseteq N_\epsilon \left({g}\right)$.

Since $g > a$, we can arrange that $\epsilon < g - a$.

As $g$ is the least upper bound, there must be a $\text{good}$ $c$ such that $c > g - \epsilon$.

This means $\left[{a,.\,.\,c}\right]$ is covered by a finite subset of $\mathcal U$, say $\left\{{U_1, U_2, \ldots, U_r}\right\}$.

Then $\left[{a\,.\,.\,g + \dfrac {\epsilon} {2}}\right]$ is covered by $\left\{{U_1, U_2, \ldots, U_r, U_0}\right\}$.

So $g + \dfrac {\epsilon} {2}$ is $\text{good}$, contradicting the fact that $g$ is an upper bound for $G$.

This contradiction implies that $g > b$, and the proof is complete.

$\blacksquare$

Proof of General Result

It holds for $n = 1$, as follows.

Suppose $C$ is a closed, bounded subspace of $\R$.

Then $C \subseteq \left[{a\,.\,.\,b}\right]$ for some $a, b \in \R$.

Moreover, $C$ is closed in $\left[{a . . b}\right]$ by the corollary of Closed Sets in Topological Subspace.

Hence $C$ is compact, by Closed Subspace of Compact Space is Compact.

Now suppose $C \subseteq \R^n$ is closed and bounded.

Since $C$ is bounded, $C \subseteq \left[{a\,.\,.\,b}\right] \times \left[{a\,.\,.\,b}\right] \times \cdots \times \left[{a\,.\,.\,b}\right] = B$ for some $a, b \in \R$.

Now $B$ is compact by Topological Product of Compact Spaces.

Also, $C$ is closed in $B$ by the corollary of Closed Sets in Topological Subspace.

Hence $C$ is compact, by Closed Subspace of Compact Space is Compact.

$\blacksquare$

Note

This does not apply in the general metric space.

A trivial example is $\left({0 . . 1}\right)$ as a subspace of itself.

It is closed and bounded but not compact.

Source of Name

This entry was named for Heinrich Eduard Heine and Émile Borel.

The theorem is sometimes called the Borel-Lebesgue Theorem, for Émile Borel and Henri Léon Lebesgue.

Sources

• Elon Lages Lima: Análise Real 1 (1989): Chapter 5, Theorem 10