Proving the limit of any convergent sequence of orthogonal matrices is again an orthogonal matrix

Question

I would like to show the set $O_n(\Bbb R)$ of orthogonal $n\times n$ matrices is closed in $M_n(\Bbb R)$ by showing every convergent sequence in $O_n(\Bbb R)$ has a limit in $O_n(\Bbb R)$ and not associating matrices as linear operators like it is done here or concluding $O_n(\Bbb R)$ is the preimage of a closed set under a continuous function, but using only sequences in the most elementary way possible.

My attempt:

We can identify $M_n$ with $\Bbb R^{n+\ldots+n}=\Bbb R^{n\cdot n}$ viewing $A\in M_n(\Bbb R), A=(a_{ij}), i,j\in\{1,\ldots,n\}$ as $$A=(a_{11},\ldots,a_{1n},\ldots,a_{j1},\ldots,a_{jn},\ldots,a_{n1},\ldots,a_{nn})\tag 1$$ in order to apply the following proposition:

Let $a_k=(a_k^1,\ldots,a_k^n)\in\Bbb R^n.$ It holds: $$\lim_{k\to\infty}a_k=L=(\ell_1,\ldots,\ell_n)\iff\lim_{k\to\infty} a_k^i=\ell_i,\forall i\in\{1,\ldots,n\}.$$

Let $M\in O_n, M=(m_{ij})$. Then the condition $MM^\tau=I$ gives us $$\Sigma_{ij}:=\sum_{\ell=1}^nm_{i\ell}m_{j\ell}=\begin{cases}1,&\text{ if } i=j,\\0,&\text{ if } i\ne j\end{cases}, i,j\in\{1,\ldots,n\}.$$

Let's rewrite $$O_n=\{M\in M_n(\Bbb R)\mid \Sigma(M)_{ij}=\delta_{ij},i,j\in\{0,\ldots,n\}\}.$$ From $(1),$ we have $M_k\to M\iff ({m_k}_{ij})\to (m_{ij}),\forall i,j\in\Bbb N.$

Now, using the following theorem:

Let $(x_k)_k$ and $(y_k)_k$ be convergent sequences in $\Bbb R$ . Then:

1. $\lim\limits_{k\to\infty}(x_k\pm y_k)=\lim\limits_{k\to\infty}x_k\pm\lim\limits_{k\to\infty}y_k.$
2. $\lim\limits_{k\to\infty}x_k\cdot y_k=\lim\limits_{k\to\infty}x_k\cdot\lim\limits_{k\to\infty}y_k.$

From $(1)$ and the theorem above, for an arbitrary sequence $(M_k)_k$ of orthogonal matrices converging to $M\in M_n(\Bbb R)$, it follows $$\delta_{ij}=\lim_{k\to\infty}\Sigma_{ij}(M_k)=\Sigma_{ij}(M)\implies M\in O_n.$$

Therefore, since the sequence $(M_k)_k$ in $O_n$ was arbitrary, we conclude that the limit of any convergent sequence of orthogonal matrices is again an orthogonal matrix, therefore, $O_n$ is closed.

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Can someone verify my proof? Is there anything wrong/any room for improvement?

Answer 1

You proof seems good to me.

Using matrix formalism can make it simpler: Let $\{M_p\}_{p\in\mathbb N}$ be the sequence of orthogonal matrices (so $M^T_pM_p=I$) and let $$M=\lim_{p\rightarrow+\infty}M_p$$ Then $$\begin{split} \|M^TM-I\| &= \|(M-M_p)^T M +M_p^T(M-M_p) +\underbrace{M_p^TM_p-I}_{=0}\|\\ &\leq \|(M-M_p)^T M\| + \|M_p^T(M-M_p)\| \\ &\leq \underbrace{\|(M-M_p)^T\|}_{\rightarrow 0}\| M\| + \|M_p^T\|\underbrace{\|(M-M_p)\|}_{\rightarrow 0} \\ \end{split}$$ Thus $M^TM=I$ and $M$ is orthogonal.

In general, when you can express a property (e.g. orthogonality) via a continuous function $f$, that is having the property is equivalent to $f(M_p)=0$, then limits preserve that property: $$0 = \lim_{p\rightarrow+\infty}f(M_p) = f(\lim_{p\rightarrow+\infty}M_p)$$

In your case, $f(M_p)=M_p^T M_p - I$.