---- > [!theorem] Theorem. ([[Heine-Borel theorem]]) > A [[subspace topology|subspace]] $A$ of $\mathbb{R}^{n}$ is [[compact]] if and only if it is [[closed set|closed]] and is [[bounded set|bounded]] in the [[Euclidean metric]] $d$ or the [[sup Metric|sup metric]] $\rho$. > [!generalization] > - [[for metric spaces, compact iff complete and totally bounded]] ^generalization > [!proof]- Proof. ([[Heine-Borel theorem]]) > It suffices to consider only the [[metric]] $\rho$, because the inequality $\rho(x,y) \leq d(x,y) \leq \sqrt{ n } \rho(x,y)$ > implies that $A$ is [[bounded set|bounded]] under $d$ iff it is [[bounded set|bounded]] under $\rho$. > > $\to$. Suppose $A$ is [[compact]]. Then by [[every compact subspace of a Hausdorff space is closed]], $A$ is [[closed set|closed]]. Now, the set collection $\{ B_{\rho}(\b 0, m) : m \in \mathbb{N}\}$ > [[cover|covers]] $A$. Since $A$ is [[compact]] it has a finite subcover. The ball with largest radius $m$ comprising that subcover witnesses that $A$ is bounded. > > > $\leftarrow.$ Conversely, suppose that $A$ is closed and bounded in $X$. Since $A$ is bounded, we can fit a closed (sup-)cube around $A$; by [[closed intervals are compact]] and [[products preserve compactness]] this cube is [[compact]]. Since $A$ is a [[closed set]] living in a compact set, we are done by [[closed subspaces of compact spaces are compact]]. > > > (Completeness of $\mathbb{R}$ was invoked when claiming that closed intervals are compact.) ---- #### ----- #### References > [!backlink] > ```dataview TABLE rows.file.link as "Further Reading" FROM [[]] FLATTEN file.tags GROUP BY file.tags as Tag > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```