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 $$E_j = \{x \in \mathbb{R}: \text{$x$ is the $j$th coordinate of $y$ for some $y \in E$}\}.$$ 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.
Best Answer
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 $E$.