$O(n,\mathbb R)$ of all orthogonal matrices is a closed subset of $M(n,\mathbb R).$ 
Let $M(n,\mathbb R)$ be endowed with the norm $(a_{ij})_{n\times n}\mapsto\sqrt{\sum_{i,j}|a_{ij}|^2}.$ Then the set $O(n,\mathbb R)$ of all orthogonal matrices is a closed subset of $M(n,\mathbb R).$

My Attempt: Let $f:M(n,\mathbb R)\to M(n,\mathbb R):A\mapsto AA^t.$ Choose a sequence $\{A_k=(a^k_{ij})\}\subset M(n,\mathbb R)$ such that $A_k\to A=(a_{ij})$ for chosen $A\in M(n,\mathbb R).$ Then $\forall~i,j,$ $a_{ij}^k\to a_{ij}$ in $\mathbb R.$
Now $A_kA_k^t=(\sum_{p=1}^n a_{ip}^ka_{jp}^k)~\forall~k\in\mathbb Z^+.$
Choose $i,j\in\{1,2,...,n\}.$ Then for $p=1,2,...,n;~a_{ip}^k\to a_{ip},~a_{jp}^k\to a_{jp}$ in $\mathbb R\implies \sum_{p=1}^n a_{ip}^ka_{jp}^k\to \sum_{p=1}^n a_{ip}a_{jp}$ in $\mathbb R.$ 
So $(\sum_{p=1}^n a_{ip}^ka_{jp}^k)\to (\sum_{p=1}^n a_{ip}a_{jp})\implies A_kA_k^t\to AA^t.$
So $f$ is continuous on $M(n,\mathbb R).$
Now $O(n,\mathbb R)=f^{-1}(\{I\}).$ The singleton set $\{I\}$ being closed in $M(n,\mathbb R),$ $O(n,\mathbb R)$ is closed in $M(n,\mathbb R).$

I'm not absolutely sure about the steps. Is't a correct attempt?


It would be quicker to observe that $f$ is a vector of polynomials in the natural coordinates, and polynomials are continuous, so $f$ is continuous.
