Proof of the Heine–Borel theorem in finite-dimensional normed spaces

Proof

We transport the problem to $\mathbb{R}^m$ by a map that distorts distances by at most constant factors.

1. Let $\mathbb{K}$ be $\mathbb{R}$ or $\mathbb{C}$. If $V = \{0\}$, its subsets are $\varnothing$ and $V$. Both are finite, so compact, and both are closed and bounded. Let $V \neq \{0\}$.
2. Take a basis $e_1, \ldots, e_n$ of the finite-dimensional space $V$, with coordinates $a_j$ as in this result, and let $\|x\|_\infty = \max_j |a_j(x)|$. It is a norm: it vanishes only at $0$, it is homogeneous, and $|a_j(x + y)| \le |a_j(x)| + |a_j(y)|$.
3. By the equivalence of norms in finite dimension, there are $c, C > 0$ with $c \|x\|_\infty \le \|x\| \le C \|x\|_\infty$ for all $x \in V$.
4. Let $m = n$ if $\mathbb{K} = \mathbb{R}$ and $m = 2n$ if $\mathbb{K} = \mathbb{C}$. Let $\Lambda \colon \mathbb{R}^m \to V$ be $\Lambda(t) = \sum_j t_j e_j$, or $\Lambda(t) = \sum_j (t_{2j-1} + \mathrm{i} \, t_{2j}) e_j$ in the complex case. It is an additive bijection.
5. Let $\|t\|_2$ be the Euclidean length, so that $d_2(t, s) = \|t - s\|_2$ is the Euclidean metric. Each coordinate $a_j(\Lambda(t))$ is $t_j$, or $t_{2j-1} + \mathrm{i} \, t_{2j}$. Its modulus is at most $\|t\|_2$, and at least each $|t_k|$ that appears in it.
6. Hence $\|t\|_2 / \sqrt{m} \le \max_k |t_k| \le \|\Lambda(t)\|_\infty \le \|t\|_2$. With step 3, $\alpha \|t\|_2 \le \|\Lambda(t)\| \le C \|t\|_2$, where $\alpha = c / \sqrt{m} > 0$.
7. As $\Lambda$ is additive, $\alpha \, d_2(t, s) \le \|\Lambda(t) - \Lambda(s)\| \le C \, d_2(t, s)$ for all $t, s \in \mathbb{R}^m$.
8. So $\Lambda$ and $\Lambda^{-1}$ are Lipschitz, with constants $C$ and $1 / \alpha$. They satisfy the epsilon-delta condition, so they are continuous by continuity via open sets, and $\Lambda$ is a homeomorphism.
9. Let $K \subseteq V$ and $L = \Lambda^{-1}(K)$, so that $K = \Lambda(L)$.
10. $K$ is closed if and only if $L$ is closed. If $K$ is closed, $L$ is the preimage of $K$ under $\Lambda$. If $L$ is closed, $K$ is the preimage of $L$ under $\Lambda^{-1}$. Use continuity via closed sets.
11. $K$ is bounded if and only if $L$ is bounded. By step 7, distances in $K$ are at most $C$ times distances in $L$. Distances in $L$ are at most $1 / \alpha$ times distances in $K$.
12. $K$ is compact if and only if $L$ is compact. Each is the image of the other under a continuous map, so continuous image of a compact set is compact applies both ways.
13. By the Heine–Borel theorem in $\mathbb{R}^m$, $L$ is compact if and only if it is closed and bounded.
14. By steps 10 to 13, $K$ is compact if and only if $K$ is closed and bounded.