= Heine-Borel theorem
{c}
{title2=$K\subseteq\mathbb R^d:\quad K\text{ compact}\iff K\text{ closed and bounded}$}
{wiki=Heine–Borel_theorem}
For every finite $d\geq1$, a <subset> $K\subseteq\mathbb R^d$ of <Euclidean space> is <compact> if and only if it is a <closed set> and a <bounded set>. Here <closed> means <closed> in the ambient <Euclidean space>, not merely in $K$ itself, and the <bounded set> condition uses the <Euclidean norm>. The <empty set> is included: it is a <closed set>, a <bounded set>, and <compact>. https://www.jirka.org/ra/html/sec_metcompact.html[Jiří Lebl, Basic Analysis, Theorem 7.4.14]{ref}
This makes an apparently infinite test practical. To verify <compactness> of a <closed interval> $[a,b]$ with finite endpoints, we need not examine every <open cover>: it is <closed> in the <real line> and a <bounded set>. By contrast, the <open interval> $(0,1)$ is a <bounded set> but is not <closed> in the <real line>, as the missing-endpoint figure in <compact space> illustrates. The whole <real line> is <closed> but is not a <bounded set>, so it too fails <compactness>. These examples show why both hypotheses are needed.
The same criterion holds in any real or complex <finite-dimensional vector space> equipped with a <norm>: all <norms> there are <equivalent norms>, so they give the same <open sets>, <closed sets>, and <bounded sets>. Since <compactness> depends only on the <topology>, choosing a different <norm> does not change the conclusion. Without an appropriate <topology>, a bare <vector space> alone has no notion of <compactness>.
The restriction to a <finite-dimensional vector space> is essential. In the <l2 sequence space>, an <infinite-dimensional vector space> that is also a <Hilbert space>, let $e_n$ be the <vectors> of its standard <orthonormal basis>. All lie in the <closed unit ball> $B=\{x:\lVert x\rVert\leq1\}$, which is a <closed set> and a <bounded set> in the <norm topology>. For distinct indices, their <inner product> is zero, giving
$$
\lVert e_n-e_m\rVert^2=\lVert e_n\rVert^2+\lVert e_m\rVert^2-2\operatorname{Re}\langle e_n,e_m\rangle=2.
$$
Thus distinct terms remain distance $\sqrt2$ apart. No <subsequence> is a <Cauchy sequence>, hence none is a <convergent sequence>. The equivalence of <compactness> and <sequential compactness> in a <metric space> proves that this <closed unit ball> is not <compact>. The conclusion concerns the <norm topology>.
A short proof outline in <Euclidean space> explains the role of a <finite-dimensional vector space>. For a <sequence> in a <bounded set>, apply the <Bolzano-Weierstrass theorem> to each of the <Cartesian coordinates> in turn, retaining a <subsequence> at each of the finitely many steps. The final <subsequence> is a <convergent sequence> in all <Cartesian coordinates>, and a <closed set> contains its <limit of a sequence>. This yields <sequential compactness>, hence <compactness>. Conversely, a <compact subset> of a <metric space> is a <closed set> and a <bounded set>. https://www.jirka.org/ra/html/sec_metcompact.html[Lebl, Basic Analysis, proof of Theorem 7.4.14]{ref}
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