Clarification as to why $O(n)$ is closed?

Question

In a problem in Guillemin and Pollack they ask to show that $O(n)$ compact. The boundedness comes from the fact that the sum of each row is $1$ and as the matrix is $n \times n$ then the set is bounded. Multiple solutions then go to say that "orthogonal matrices are the inverse image of the element $I$ under the continuous map $MM^{T}$." (3rd link)

Admittedly, I don't know why this fact implies closedness. Can anyone help clarify why this is the case? Is there some theorem that I am forgetting that is being utilized?

Answer 1

Define $F: M(n,\mathbb{R})\to M(n,\mathbb{R})$ to be $M\mapsto M^TM$. Then $O(n)$ is precisely the preimage of the identity matrix $F^{-1}(\{I\})=O(n)$. Singletons are closed in $M(n,\mathbb{R})$ (it's a metric space, which is certainly sufficient). Since the preimage of closed sets under continuous maps are closed, it suffices to check that $F$ is continuous. To see this, note that each entry in the matrix $M^TM= F(M)$ is a polynomial in the matrix entries of $M$. Polynomials are obviously continuous (in fact they are smooth), so we are done.

Answer 2

The preimage of a closed set under a continuous function is itself closed.

Answer 3

The inverse image of a closed set under a continuous function is closed. This follows easily from the definition of continuity.