Let $ M^{n \times n}( \mathbb{R})$ denote the set of $n \times n$ matrices with real coefficients and $S_O$ the set of orthogonal $n \times n$ matrices with real coefficients.
I want to prove that $S_O$ is compact in $ M^{n \times n}( \mathbb{R})$. I know this may appear to be a duplicate of other similar posts, but I have the following problem:
Most of the proofs I have come across seem to use the Heine-Borel property, meaning that one shows that $S_O$ is closed and bounded, and therefore also compact. But as far as I am aware, the Heine-Borel Theorem holds only for subsets of $\mathbb{R}^n$. I do realize that some metric spaces/vector spaces/topological spaces have the so called Heine-Borel property and the Theorem can therefore be applied, but how do I know/prove that the space $ M^{n \times n}( \mathbb{R})$ has this property?
Could it also be that I am missing out on something very obvious? To me $S_O$ is a subset of $ M^{n \times n}( \mathbb{R})$ and not $\mathbb{R}^n$, so using the Theorem does seem a little bit off to me.
I really hope someone out there has some kind of useful input regarding this dilemma.
Thanks.
What topology/metric do you put on $M^{n \times n}(\mathbb{R})$ besides the obvious topology/metric given by the identification of $M^{n \times n}(\mathbb{R})$ with $\mathbb{R}^{n \times n}$.
If you do give it the obvious choice of a metric, then Heine-Borel theorem clearly applies.
Then since all norms a finite dimensional vector space are equivalent, for any norm, we get the Heine Borel theorem, since closed and bounded in one norm implies closed and bounded in the Euclidean norm, for which we already have the Heine Borel theorem.