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  "body": "\n## Proof\n\nWe transport the problem to $\\mathbb{R}^m$ by a map that distorts distances by at most constant factors.\n\n1. 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 $V \\neq \\{0\\}$.\n2. 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 $|a_j(x + y)| \\le |a_j(x)| + |a_j(y)|$.\n3. By the [[equivalence of norms in finite dimension|03MC70ZRNR]], there are $c, C \u003e 0$ with $c \\|x\\|_\\infty \\le \\|x\\| \\le C \\|x\\|_\\infty$ for all $x \\in V$.\n4. 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.\n5. 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.\n6. 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 $\\alpha = c / \\sqrt{m} \u003e 0$.\n7. As $\\Lambda$ is additive, $\\alpha \\, d_2(t, s) \\le \\|\\Lambda(t) - \\Lambda(s)\\| \\le C \\, d_2(t, s)$ for all $t, s \\in \\mathbb{R}^m$.\n8. 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]].\n9. Let $K \\subseteq V$ and $L = \\Lambda^{-1}(K)$, so that $K = \\Lambda(L)$.\n10. $K$ 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]].\n11. $K$ 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 $K$.\n12. $K$ 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.\n13. By the [[Heine–Borel theorem|07PFYV65MB]] in $\\mathbb{R}^m$, $L$ is compact if and only if it is closed and bounded.\n14. By steps 10 to 13, $K$ is compact if and only if $K$ is closed and bounded.\n\n## Idea\n\nChoosing 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.\n",
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