is $GL(n,\mathbb R)$ dense in $M(n,\mathbb R)$ Is $GL(n,\mathbb R)$ dense in $M(n,\mathbb R)$? I have proved it to be open,not closed,not connected but not sure about this property .How to do this?

The way I like writing this is as follows: Consider $A\in M_{n}(\mathbb R)$ and let $\varepsilon>0$. Let $\lambda _1, \ldots, \lambda_n$ be the eigenvalues of $A$. Take $\delta$ such that $0<\delta<\frac{\varepsilon}{n^{1/2}}$ and $\delta\neq \lambda_j$ for every  $j\in \{1, \ldots, n\}$. Define
$$A_\delta:=A-\delta I.$$
Then $A_\delta$ is invertible. In fact, on the contrary,
$$\det(A_\delta)=\det(A-\delta I)=0$$
and therefore $\delta$ would be an eigenvalue of $A$, so that $\delta =\lambda_j$ for some $j\in \{1, \ldots, n\}$. This contradicts the choice of $\delta$. Finally,
$$d_2(A, A_\delta)=\|A-A_\delta\|_2=\|A-(A-\delta I)\|_2=|\delta|\|I\|_2=\delta {n^{1/2}}<\frac{\varepsilon}{n^{1/2}} n^{1/2}=\varepsilon.$$
I assumed here that your metric is induced by norm:
$$\|A\|_2=\left(\sum_{i=1}^n \sum_{j=1}^n a_{ij}^2\right)^{1/2}.$$
