---
tag: G0485GZA7Y
kind: proof
name: Proof of the Heine–Borel theorem in finite-dimensional normed spaces
msc: [46B20]
proves: 0PHT6NBYPB
uses: [07PFYV65MB, 03MC70ZRNR, 0J3578562S, 018W3V3TCT, 06MRNP4BNT, 059THXBRBB, 0E1MMCAY5B, 0TMZ21WKCN, 0YF82GBTND, 0B8GPXFSP7, 0K9AAJ8G51, 00R0X1WGPX, 0G5AV9Q4RR, 0P953MG05E, 05GZ11PWRY]
sources:
  - url: https://en.wikipedia.org/wiki/Heine%E2%80%93Borel_theorem
    locator: In the theory of topological vector 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|00R0X1WGPX]], and both are [[closed|0G5AV9Q4RR]] and [[bounded|0P953MG05E]]. Let $V \neq \{0\}$.
2. Take a [[basis|06MRNP4BNT]] $e_1, \ldots, e_n$ of the [[finite-dimensional|018W3V3TCT]] space $V$, with coordinates $a_j$ as in [[this result|059THXBRBB]], 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|03MC70ZRNR]], 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|0E1MMCAY5B]]. 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|0TMZ21WKCN]], with constants $C$ and $1 / \alpha$. They satisfy the epsilon-delta condition, so they are continuous by [[continuity via open sets|0YF82GBTND]], and $\Lambda$ is a [[homeomorphism|05GZ11PWRY]].
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|0B8GPXFSP7]].
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|0K9AAJ8G51]] applies both ways.
13. By the [[Heine–Borel theorem|07PFYV65MB]] 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.

## Idea

Choosing a basis identifies $V$ with $\mathbb{R}^n$ or $\mathbb{R}^{2n}$. By the equivalence of norms, this identification changes distances by at most fixed factors. Such a map preserves closed sets, bounded sets and compact sets, so the classical Heine–Borel theorem carries over.