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Show that $O(n,\mathbb{R})$ the set of all orthogonal matrices is closed in $M(n,\mathbb{R})$

Let $$f:M(n,\mathbb{R}) \mapsto \mathbb{R}$$

$$ f:A \mapsto det(A.A^t)$$

We know that $$ \text{det}(x_{i,j})= \sum_\sigma \text{sgn}(\sigma) \prod_{i=1}^{n} x_{\sigma(i),i} $$ The determinant function is continuous as it is a polynomial function in $x_1,\cdots x_n$. Then $det(A.A^t)$ is also a polynomial in $x_1,\cdots x_n$ . Any function from $M(n,\mathbb{R})$ to a polynomial in $x_1,\cdots , x_n$ is continuous. So the following map $f$ is continuous.

$O(n,\mathbb{R}) = f^{-1}(\{1\})$.

The singleton $\{1\}$ is a closed set in $\mathbb{R}$. Hence, $O(n,\mathbb{R})$ is a closed set in $M(n,\mathbb{R})$.

I have not seen any answer in stackexchange similar to this method. Is my procedure incorrect? Also can we extend the following method to prove that $SO(n,\mathbb{R})$ is continuous?

Here's my attempt for $SO(n,\mathbb{R})$ I was thinking of the following map $f:M(n,\mathbb{R}) \mapsto \mathbb{R}$ where $A \mapsto det(A).det(A^t)$ . I think that the following map will be continuous but I am not sure whether $f^{-1}(\{1\}) = SO(n,\mathbb{R})$?

ThirstForMaths
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    It's not true that $O(n)=f^{-1}({1})$. Take for example $n=2$ and $A=\begin{pmatrix}1/2&0\0&2\end{pmatrix}$. Then, $A$ is not orthogonal, however $f(A)=1$. – Surb Dec 02 '21 at 09:04
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    Go back to the definition of $O(n,\Bbb R)$, the function you will have to use is therein. – Didier Dec 02 '21 at 09:14
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    Hint : This function does not necessarily take values in $\mathbb{R}$ – G. Fougeron Dec 02 '21 at 09:31
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    BIG HINT: a matrix is orthogonal if and only if $AA^t=I$ (the identity matrix). Hence the function you need to use should be $$A \mapsto AA^t$$ – Crostul Dec 02 '21 at 09:57
  • The same question has already been asked multiple times. – PinkyWay Dec 02 '21 at 10:25

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