Is $SL(n,\mathbb R)$ nowhere dense in $M(n,\mathbb R)$? 
Is $SL(n,\mathbb R)$ nowhere dense in $M(n,\mathbb R)$?

I know the definition and some properties of nowhere dense set. I know the property that if $W$ be a linear space with dimension at most $n-1$ then it is nowhere dense in a space with dim $ n.$ So, it is nowhere dense. Can it be proved in this way? Please help me. Thanks in advance.
 A: Let $A\in Sl(n,\mathbb{R})$, $det(e^tA)=e^{nt}det(A)=e^{nt}$. This implies that $e^tA\in Sl(n,\mathbb{R})$ if and only if $e^{nt}=1$ which is equivalent to $nt=t=0$. $lim_{t\rightarrow 0}e^tA=A$ (since $lim_{t\rightarrow 0}e^t=1$) implies that for every open subset $U$ containing $A$, there exists $c>0$ such that $|t|<c$ implies that $e^tA\in U$, we can choose $0<t<c$, then $e^tA\in U$ and $e^tA$ is not in $Sl(n,\mathbb{R})$. This is equivalent to saying that $Sl(n,\mathbb{R})$ is nowhere dense in $M(n,\mathbb{R})$.
A: This is not the complete answer but I hope it helps:

A set $E$ in a metric space $X$ is no where dense $\iff$ $X \setminus\overline{E}$ is dense in $X$.

Here $SL(n,\Bbb R)$ is closed.  Now consider $\mathcal{B}:=M(n,\Bbb R) \setminus \overline{SL(n,\Bbb R)}=M(n,\Bbb R) \setminus SL(n,\Bbb R)$ 
Now the question is: Is $\mathcal{B}$ dense in $M(n,\Bbb R)$ ?
In order to prove $\mathcal{B}$ is dense, we prove every point in $M(n,\Bbb R)$ is a limit point of $\mathcal{B}$ or a point of $\mathcal{B}$.
So if your arbitrary point is already in $\mathcal{B}$, then  we are done
Otherewise that point belongs to $SL(n,\Bbb R)$. In this case, Can you produce the sequence of matrices in $M(n,\Bbb R)$ such that it converges to that point ?
If so, we are done!
A: While your observation that a linear subspace with codimension at least $1$ is nowhere dense is true, it doesn't quite help here, since $SL(n,\mathbb{R})$ is not a linear subspace of $M(n,\mathbb{R})$. However, you can strengthen your observation to work for manifolds. The version for manifolds is the following.

If $M$ is an embedded submanifold of $N$ with codimension at least $1$, then $M$ is nowhere dense in $N$.

Proving this isn't very hard, and just relies on picking a nice coordinate chart for $N$, such that on $M$, the last coordinate is $0$, which can always be done for embedded submanifolds, but not necessarily immersed submanifolds. Once you have this result, your question is answered by observing that $SL(n,\mathbb{R})$ is an embedded submanifold of $M(n, \mathbb{R})$ with codimension $1$.
