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

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

This means that if $$z \in G$$ is $$\mbox{good}$$, 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 $$\mbox{good}$$.

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

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 $$\mbox{good}$$.

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 $$\mbox{good}$$ such that $$c > b$$.

From our initial observation that if $$x$$ is $$\mbox{good}$$, and $$a \le y \le x$$, then $$y$$ is $$\mbox{good}$$, it follows that $$b$$ is $$\mbox{good}$$, 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 $$\mbox{good}$$ such that $$c > b$$ by leastness of $$g$$.

Again, from our initial observation, it follows that $$b$$ is $$\mbox{good}$$, 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 $$\mbox{good}$$ $$c$$ 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 $$\mbox{good}$$, 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.