Heine-Borel Theorem/Real Line

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 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 $z \in G$ 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 do is 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 the same finite subset of $\mathcal U$ that $\left[{a \,. \, . \, x}\right]$ can.

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 $\left[{a \,. \, . \, x}\right]$.

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

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

This means that $\left[{a \,. \, . \, x}\right]$ is covered by one set in $\mathcal U$.

So all such $x$ 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 $c > b$ by leastness of $g$.

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 $g \ge a$ 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 assume that $\epsilon < g - a$.

By leastness of $g$, there is 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 $\left[{a \,. \, . \, g + \frac \epsilon 2}\right]$ is covered by $\left\{{U_1, U_2, \ldots, U_r, U_0}\right\}$.

So $g + \frac \epsilon 2$ 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.

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.

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.