$GL_n(\mathbb{R})$ is open in $M_n(\mathbb{R})$ but not closed?

Question

It is well known that for any subgroup $F$ of a topological group $G$. If $F$ is open in $G$, then it must be closed $G$.

However, we know:

1. $GL_n(\mathbb{R})$ is open in $M_n(\mathbb{R})$
2. $GL_n(\mathbb{R})$ is a subgroup

But we also know $GL_n(\mathbb{R})$ is NOT closed, from the example in this link:

there exists a sequence $A_n$ in $GL(n,R)$ such that $A_n$ tends to $A$ as $n$ to infinity where $A$ is not invertible.For instance,considering $n=2$,let $A_n=\begin{pmatrix} 1 && 0\\ 0 && > \frac{1}{n}\end{pmatrix}$.Then $limA_n=A=\begin{pmatrix} 1 && 0\\ 0 && > 0\end{pmatrix} ,n\to{+\infty}$,where $det(A)=0$.The example tells us that $GL(n,R)$ is not closed in $M_n(R)$.

Isn't that contradictory?