Is $O(k)\times O(n-k)$ closed in $SO(n)$?

Question

Let $O(m)$ denote the group of orthogonal matrices under multiplication, and let $SO(m)$ be the special orthogonal group over $\mathbb{R}$. Let \begin{equation*} (O(k)\times O(n-k))\cap SO(n):=\left\{A=\begin{pmatrix} B & 0\\ 0 & C \end{pmatrix} \in \mathbb{R}^{n\times n} \mid B\in O( k) ,\ C\in O( n-k) ,\ \det( B)\det( C) =1\right\}. \end{equation*}

I want to prove/disprove $O(k)\times O(n-k)$, $k=1,\dots,n-1$ is closed in $SO(n)$.

I do not really know how.

Even for $k=1$, I am not too sure. For $k=1$, we have \begin{equation*} \left\{\begin{pmatrix} b & 0\\ 0 & C \end{pmatrix} \in \mathbb{R}^{n\times n} \mid b\in O( 1) ,\ C\in O( n-1) ,\ b\det( C) =1\right\} =H_{1} \cup H_{-1} , \end{equation*} where for $b=1,-1$ we let \begin{equation*} H_{b} :=\left\{\begin{pmatrix} b & 0\\ 0 & C \end{pmatrix} \in \mathbb{R}^{n\times n} \mid \ C\in O( n-1) ,\ \det( C) =b\right\} . \end{equation*} I want to say this is essentially the union of $SO(n-1)$ and thus closed, but I am not really comfortable with seeing $SO(n-1)$ as a subset of $SO(n)$ incorporating the topological consistency.

Answer 1

Since each orthogonal group is compact, $O(k)\times O(n-k)$ is compact and therefore a closed subset of $\mathbb{R}^{n\times n}$. And your set is a closed subset of this one, because it's the set of thos elements whose determinant is $1$. Since $\det$ is continuous and $\{1\}$ is closed, this is again a closed set.