Show that the set of all $n \times n$ orthogonal matrices, $O(n)$, is a compact subset of $\mbox{GL} (n,\mathbb R)$

Hints:

• $\mathcal O(n)$ is the continuous inverse image of a closed set…

• There exists a rather obvious bound for every element in $\mathcal O(n)$ and thus this set is bounded…

Now Heine-Borel and we're done.

Added: Thanks to the comments by Berci I think some confusion may happen here, since the orthogonal group is not the inverse image of $\,\{1,-1\}\,$ under the determinant map in the whole $\,GL(n,\Bbb R)\,$ (as there are matrices with determinant $\,\pm 1\,$ which are not orthogonal, of course).

So let us try the following approach: the map $\,T:GL(n,\Bbb R)\to GL(n,\Bbb R)\,$ defined by $\,T(A):=AA^t\,$ is continuous since every entry in the matrix $\,AA^t\,$ is a polynomial on the entries of $\,A\,$ ,and now we can see that

$$\mathcal O(n)=T^{-1}(\{I\})$$

and, of course, $\,\{I\}\,$ is a closed set, so $\,\mathcal O(n)\,$ is closed.

In fact, you can say more:

Using the canonical action of $SO(n)$ on $\mathbb{S}^{n-1}$, you can show that $SO(n)$ is homeomorphic to $\mathbb{S}^{n-1}$. But you have the exact sequence $1 \to SO(n) \to O(n) \to \mathbb{Z}_2 \to 1$, so $[O(n):SO(n)]=2$ and $O(n)$ is homeomorphic to $\mathbb{S}^{n-1} \coprod \mathbb{S}^{n-1}$.