Please could someone check my work on this exercise (from a book I am reading). Thanks!
Exercise:
Prove that $SO(n)$ and $ O(n)^- = \{ A \in O(n) \mid \det(A) = -1 \}$ are both clopen in $O(n)$.
My solution:
Since $\det: O(n) \to \{-1,1\}$ is continuous and $O(n)^- = \det^{-1} (\{-1\})$ and $SO(n) = \det^{-1}(\{1\})$ we see that they are closed. Since $SO(n) = (O^{-}(n))^c$ and $O^{-}(n) = SO^c (n)$ we see that they are open.
As a consequence, $O(n)$ is disconnected.
Alternatively, you could note that the set $U=\det^{-1} \{ ({1 \over 2}, \infty)\} \cap O(n)$ is open in $O(n)$, and $U = SO(n)$, hence $SO(n)$ is open in $O(n)$. A similar argument shows that $O(n)^-$ is open in $O(n)$. Since $O(n)$ is the disjoint union of $SO(n)$ and $O(n)^-$ we see that both $SO(n)$ and $O(n)^-$ are closed in $O(n)$.
Finally, we could note that $\det$ is continuous and $\det O(n) = \{\pm 1\}$, which is not connected. Hence $O(n)$ is not connected.
