Heine–Borel theorem: Difference between revisions
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In [[mathematics]], the '''Heine-Borel theorem''' characterises the [[compact space|compact]] [[subset]]s of the [[real number]]s. | In [[mathematics]], the '''Heine-Borel theorem''' characterises the [[compact space|compact]] [[subset]]s of the [[real number]]s. | ||
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A Euclidean space of fixed finite dimension ''n'' also forms a [[metric space]] with the Euclidean distance as metric. As a [[topological space]], the same statement holds: a subset is compact if and only if it is [[closed set|closed]] and [[bounded set|bounded]]. | A Euclidean space of fixed finite dimension ''n'' also forms a [[metric space]] with the Euclidean distance as metric. As a [[topological space]], the same statement holds: a subset is compact if and only if it is [[closed set|closed]] and [[bounded set|bounded]]. | ||
==Discussion== | |||
The theorem makes two assertions. Firstly, that a compact subset of '''R''' is closed and bounded. A compact subset of any [[Hausdorff space]] is closed. The metric is a continuous function on the compact set, and a continuous function on a compact set is bounded. | |||
The second and major part of the theorem is that a closed bounded subset of '''R''' is compact. We may reduce to the case of a closed interval, since a closed subset of a compact space is compact. | |||
One proof in this case follows directly from the definition of compactness is terms of [[open cover]]s. Consider an open cover ''U''<sub>λ</sub>. Let ''S'' be the subset of the closed interval [''a'',''b''] consisting of all ''x'' such that the interval [''a'',''x''] has a finite subcover. The set ''S'' is non-empty, since ''a'' is in ''S''. If ''b'' were not in ''S'', consider the [[supremum]] ''s'' of ''S'', and show that there is another ''t'' between ''s'' and ''b'' which is also in ''S''. This contradiction shows that ''b'' is in ''S'', which establishes the result. | |||
A second proof relies on the [[Bolzano-Weierstrass theorem]] to show that a closed interval is [[Sequentially compact space|sequentially compact]]. This already shows that it is [[Countably compact space|countably compact]]. But '''R''' is [[separable space|separable]] since the [[rational number]]s '''Q''' form a [[countability|countable]] [[dense set]], and this applies to any interval as well. Hence countable compactness implies compactness. | |||
Finally we note that a finite product of compact spaces is compact, and a closed bounded subset of '''R'''<sup>''n''</sup> is a closed subset of a "closed box", that is, a finite product of closed bounded intervals.[[Category:Suggestion Bot Tag]] | |||
Latest revision as of 16:00, 26 August 2024
In mathematics, the Heine-Borel theorem characterises the compact subsets of the real numbers.
The real numbers form a metric space with the usual distance as metric. As a topological space, a subset is compact if and only if it is closed and bounded.
A Euclidean space of fixed finite dimension n also forms a metric space with the Euclidean distance as metric. As a topological space, the same statement holds: a subset is compact if and only if it is closed and bounded.
Discussion
The theorem makes two assertions. Firstly, that a compact subset of R is closed and bounded. A compact subset of any Hausdorff space is closed. The metric is a continuous function on the compact set, and a continuous function on a compact set is bounded.
The second and major part of the theorem is that a closed bounded subset of R is compact. We may reduce to the case of a closed interval, since a closed subset of a compact space is compact.
One proof in this case follows directly from the definition of compactness is terms of open covers. Consider an open cover Uλ. Let S be the subset of the closed interval [a,b] consisting of all x such that the interval [a,x] has a finite subcover. The set S is non-empty, since a is in S. If b were not in S, consider the supremum s of S, and show that there is another t between s and b which is also in S. This contradiction shows that b is in S, which establishes the result.
A second proof relies on the Bolzano-Weierstrass theorem to show that a closed interval is sequentially compact. This already shows that it is countably compact. But R is separable since the rational numbers Q form a countable dense set, and this applies to any interval as well. Hence countable compactness implies compactness.
Finally we note that a finite product of compact spaces is compact, and a closed bounded subset of Rn is a closed subset of a "closed box", that is, a finite product of closed bounded intervals.