The set of traces of orthogonal matrices is compact 
Is the following set compact: $$M = \{ \operatorname{Tr}(A) : A \in M(n,\mathbb R) \text{ is orthogonal}\}$$ where $\operatorname{Tr}(A) $ denotes the trace of $A$?

In order to be compact $M$ has to be closed and bounded.
$\|A\|=\sqrt {\sum_{i,j} {a_{ij}}^2}=\sqrt n $ and hence bounded. So $\operatorname{Tr}(A)<\sqrt n $. Hence $M$ is bounded.
Now we have to prove that $M$ is closed. Let $\operatorname{Tr}(A_n)$ be a sequence of matrices converging to $\operatorname{Tr}(A)$ where $A_n$ is a sequence of orthogonal matrices. The only thing remaining to show is that $A$ is orthogonal.

Hint: To show that $A$ is orthogonal, consider $\lim_{n \to \infty} \left\| A_n^TA_n - I \right\|$, noting that the function
$$
f(X) = \left\| X^TX - I \right\|
$$
is continuous.
