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I want to show that $O(n)=\{M\in GL(n,\mathbb{R}): M^T M=id\}$ is compact

I would do it as follows.

Let us consider $$t:GL(n,\mathbb{R})\rightarrow GL(n,\mathbb{R}); \,\,\,M\mapsto M^TM$$We know that $t$ is continuous since it is multilinear. Now we remark that $t^{-1}(id)=O(n)$, but using the of the metric induced topology we know that $\{id\}$ is closed and thus $t^{-1}(id)=O(n)$ is also closed. Now I want to show that $O(n)$ is also bounded.

Let us remark that a subset $A$ of $\mathbb{R}^n$ is bounded if there exists $K\in \mathbb{N}$ such that $|x|<K$ for all $x\in A$.

Therefore let us consider the euclidean norm $||M||=\sqrt{\sum_{1\leq i,j\leq n}m_{i,j} ^2}$. Now take $M\in O(n)$ then $M^TM=id$. This implies that $$||M||=\sqrt{\langle M,M\rangle}=\sqrt{||M^T M||}=\sqrt{||id||}=\sqrt{n}$$ So we have shown that each element is bounded and thus by definition $O(n)$ is bounded. Now using Heine Borel we can conclude that $O(n)$ is compact.

Is this correct so?

user1294729
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  • $\operatorname{id}$ is a point in $GL(n,\mathbb{R})$ which is Hausdorff. So points are closed. You don't need the indiscrete topology. More over, the indiscrete topology is different than the one induced by the norm. – Daniel H. Hartman Jan 09 '22 at 18:52
  • So you mean when we take the indiscrete topology then ${M}$ is open for all $M\in O(n)$ and since $M\neq M'$ we find two disjoint open subsets s.t. one contains $M$ and the other $M'$ and thus it is Hausdorff? Or is there another reason – user1294729 Jan 09 '22 at 18:56
  • Sorry but why was this question closed? I only wanted to discuss my solution, I know the question is a duplicate but not the answer, my answer is not the same and the other post's do not answer my question! – user1294729 Jan 09 '22 at 19:17
  • Your attempt is exactly the one given in the duplicate. – Arctic Char Jan 09 '22 at 22:22
  • yes I'm sorry for trying to solve exercises which are asked long times ago. But what would you do if you see that someone has the same approach but hasn't proved anything rigorously. How can I know that I didn't made some small but grave mistakes although we had the same idea? (please notice that I'm new in this topic and thus not really familiar with the way of proving things) – user1294729 Jan 09 '22 at 22:37
  • It is fine that you post your attempt here and see if others can check the solution. But when the solution is correct and there is nothing more to comment, the best action is to close it as a duplicate. – Arctic Char Jan 09 '22 at 23:02
  • aha so I didn't know that if it is correct, it gets closed without a comment. Is this always so? – user1294729 Jan 10 '22 at 06:10

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