Show that $GL_n(\mathbb{R})$ is open in $\mathcal{M}_n(\mathbb{R})$ by showing that the complement is closed? I would like to  that $GL_n(\mathbb{R})$ is open in $\mathcal{M}_n(\mathbb{R})$, I know there are posts on it but I would like to know what can be said about my proof especially if it is false and what can be made to improve the argument I used if this argument is valid of course..
My attempt :
Consider the complement of $GL_n(\mathbb{R})$ in $\mathcal{M}_n(\mathbb{R})$ in order to show that it is a closed set.
Consider a sequence of matrices $(A_n)_{n\in\mathbb{N}}\in (\mathcal{M}_n(\mathbb{R})\setminus GL_n(\mathbb{R}))^{\mathbb{N}}$ which converges to $A$. First, we notice that
$$
\forall n\in\mathbb{N}, \forall B\in GL_n(\mathbb{R}), \exists\delta>0 : \lVert A_n - B\rVert > \delta
$$
since this sequence does not intersect $GL_n(\mathbb{R})$. And we know that :
$$
\forall\epsilon>0, \exists N\in\mathbb{N} : n\geq N\implies \lVert A_n - A\rVert < \epsilon
$$
The idea is to choose $\epsilon < \delta$. This gives us an $N\in\mathbb{N}$ such that
$$
0 \leq\lVert A_n - A\rVert < \epsilon <\delta < \lVert A_n - B\rVert
$$
Then we have $\forall B\in GL_n(\mathbb{R}), \exists N\in\mathbb{N}, \forall n\geq N$:
$$
\lVert A - B\rVert = \lVert (A_n - B) - (A_n - A)\rVert\geq \big\lvert \lVert A_n - B\rVert - \lVert A_n - A\rVert \big\rvert > 0
$$
and we conclude by the fact that since it holds for all $B\in GL_n(\mathbb{R})$, $A$ must not belong to $GL_n(\mathbb{R})$ so the complement of $GL_n(\mathbb{R})$ in $\mathcal{M}_n(\mathbb{R})$ is closed.
Is this proof correct and if not please could you explain why ?
Thak you a lot !
 A: The complement is the set $C:=\{A:\det A=0\}$, and the function $\det:M_n(\mathbb R)\to\mathbb R$ is continuous. Since $\{0\}$ is closed in  $\mathbb R$, continuity implies that $C$ is closed.
A: In the very first inequality, $\delta$ depends on $B$ and on $n$, but you treat it as being independent of $n$ later on in the proof when you say

The idea is to choose $\epsilon < \delta$. This gives us an $N\in\mathbb{N}$ such that
$$
0 \leq\lVert A_n - A\rVert < \epsilon <\delta < \lVert A_n - B\rVert
$$

You didn't write it, but implicitly you want this to hold for all $n\ge N$. You can choose $\epsilon$ smaller than any particular $\delta(n,B)$, but you cannot necessarily guarantee that $\epsilon$ will be smaller than $\delta(n,B)$ for infinitely many distinct values of $n$. Similarly, when you say

Then we have $\forall B\in GL_n(\mathbb{R}), \exists N\in\mathbb{N}, \forall n\geq N$:
$$
\lVert A - B\rVert = \lVert (A_n - B) - (A_n - A)\rVert\geq \big\lvert \lVert A_n - B\rVert - \lVert A_n - A\rVert \big\rvert > 0
$$

Again, presumably the very last inequality cannot be guaranteed to hold because $\delta$ depends on both $n$ and $B$, but you ask for it to hold independently of either of these parameters.

Here is a similar "argument" that illustrates the problem clearly. Suppose we want to prove that $G \stackrel{\Delta}{=} \mathbb R\setminus\big(\{0\}\cup\{1/k:k=1,2,\dots\}\big)$ is open by showing the complement is closed. Consider a sequence $\{a_n\}$ in $\mathbb R\setminus G$ that converges to a number $a$, and imagine for the sake of illustration that our sequence $a_n = 1/n$. (It converges to $0$, but we don't need to know that to illustrate what's wrong with the argument you gave, just that it's a possible sequence.)
For every $x\in G$, and every $n$, there is a $\delta(n,x)$ so that $|a_n-x|>\delta$. If $x$ lies in the interval $(\frac1{n+1},\frac1n)$, then $\delta(n,x) < 1/n$. If you keep running your argument, later on when you want to choose $\epsilon$, you want to choose it to be smaller  than $\delta(n,x)$ for all $n\ge N$, for some large $n$. This is impossible to do if $\epsilon > 0$.
