# Giving a sequential proof of the Heine-Borel theorem

I'm currently studying the section on metric spaces from Terence Tao's Analysis II, and I'm having difficulty proving the Heine-Borel theorem, whose proof he left as an exercise.

The precise statement is this:

Let $$(\mathbb{R}^n, d)$$ be a Euclidean space with either the Euclidean metric, the taxicab metric, or the sup norm metric. Let $$E$$ be a subset of $$\mathbb{R}^n$$. Then $$E$$ is compact if and only if it is closed and bounded.

The definition of compactness isn't formulated as "every open cover has a finite subcover," rather:

A metric space $$(X, d)$$ is said to be compact if every sequence in $$(X, d)$$ has at least one convergent subsequence. A subset $$Y$$ of a metric space $$X$$ is said to be compact if the subspace $$(Y, d|_{Y \times Y})$$ is compact.

He leaves two hints: the first is to use the Heine-Borel theorem for the real line, which I already proved (that is, a subset of $$\mathbb{R}$$ is compact if and only if it is closed and bounded). The other is to use the equivalence of the Euclidean, taxicab, and sup norm metrics, as well as the equivalence of convergence in these metrics with componentwise convergence. To be more precise, a sequence in $$\mathbb{R}^n$$ converges to a point with respect to the Euclidean, taxicab, or sup norm metric if and only if each of its components converge to the respective components of that point.

I have already shown one direction, which is true in any metric space: if a set is compact, then it is also closed and bounded.

However, I can't seem to figure out how to prove the converse; that is, if $$E$$ is closed and bounded, then every sequence in $$E$$ has a convergent subsequence. My idea was to do this: for each $$1 \leq j \leq n$$, let $$E_j$$ be the set $$E_j = \{x \in \mathbb{R}: \text{x is the jth coordinate of y for some y \in E}\}.$$ The boundedness of each $$E_j$$ follows from the boundedness of $$E$$. Then, if I can prove that each $$E_j$$ is closed, I can then use the Heine-Borel theorem on the real line to keep constructing subsequences that converge in each component, until I get a subsequence for which all the components converge. Then I'm finished. However, I can't seem to prove that $$E_j$$ is closed.

I'm not sure if I'm going down the right path, but this seems like it should be a relatively straightforward problem, since it seems like I should be able to easily use componentwise convergence and the Heine-Borel theorem for the real line to prove the result. Any help would be greatly appreciated.

• Your proof will work. Note that $E_j^c=\pi_j(E^c)$ is open (in $\mathbb R$) so $E_j$ is closed (in $\mathbb R)$. – Matematleta Aug 8 '19 at 4:36
• @Matematleta What does the $\pi_j$ in $\pi_j(E^c)$ denote? – Emory Sun Aug 8 '19 at 4:37
• the projection on the $j^{th}$ coordinate, which is an open map. – Matematleta Aug 8 '19 at 4:38
• @Matematleta How can I prove that such a map preserves openness using elementary methods? The text has not yet begun to discuss functions on metric spaces, much less functions that preserve topological structure. Edit: Never mind, I'll try to fill in that detail by myself. Thanks for answering! – Emory Sun Aug 8 '19 at 4:39
• As a side note, this theorem is normally called the Bolzano-Weierstrass theorem, not the Heine-Borel theorem. Heine-Borel refers instead to the analogous result with the open cover definition of compactness. – Eric Wofsey Aug 8 '19 at 5:11

You don't need $$E_j$$ to be closed: just take its closure $$F_j$$ (or just any closed bounded interval containing it). Then $$F_j$$ is closed and bounded, and you can apply your argument to find a subsequence which converges in $$F_j$$ on each coordinate. Then, because $$E$$ is closed in all of $$\mathbb{R}^n$$, the coordinatewise limit of this subsequence must actually be in $$E$$.

• Wow, I never thought of that. Thanks! – Emory Sun Aug 8 '19 at 6:00
• Suppose $(x_{1}^{(m_{1}(k))})_{k=0}^{\infty},\dots, (x_{n}^{(m_{n}(k))})_{k=0}^{\infty}$ are the convergent subsequences. How does one even know that the n-tuple $(x_{1}^{(m_{1}(k))},\dots, x_{n}^{(m_{n}(k))})$ lies in $E$ (note that the functions $m_{1},\dots, m_{n}$ are all different)? – Karthik Kannan Sep 16 '19 at 21:50
• @KarthikKannan: It might not be. You have to instead pick the subsequences in order so you can use the same subsequence on each coordinate. In other words, first choose a subsequence that converges on the first coordinate, then choose a subsequence of that subsequence (not of the full original sequence!) that converges on the second coordinate, and so on. – Eric Wofsey Sep 16 '19 at 22:06
• That is really smart! Thanks! – Karthik Kannan Sep 16 '19 at 22:15

Aim for a simpler result. Namely if $$S$$ is a sequence in E, it has a subsequence that will converge in the $$i$$th index.

Now we start with $$S = S_0$$. Construct a subsequence $$S_1$$ of that that converges in the first index. Then a subsequence $$S_2$$ of $$S_1$$ that converges in the second index (note, we only need the compactness of E and the previous lemma to show this). Keep going until we have a subsequence $$S_n$$ of $$S_0\ldots S_{n-1}$$ that converges in every index. Verify that that subsequence converges to a point.

• Yes, that was my plan. But it requires me to first show that the component wise projection of $E$ is compact first, which is what I am currently attempting to resolve. – Emory Sun Aug 8 '19 at 4:55
• @EmorySun I would just let $x$ be the infimum of the set of real numbers $y$ such that $S$ is below $y$ in the $i$th component an infinite number of times. And then I would pick an element of $S$ that is below $x+1$, the next one below $x+\frac{1}{2}$, the next one below $x+\frac{1}{3}$ and so on. That subsequence will converge to $x$. – btilly Aug 8 '19 at 5:01