Nonempty open subset of $M_n(\mathbb{R})$ spans $M_n(\mathbb{R})$ Every proper nonempty open subset in $M_n(\mathbb{R})$ spans $M_n(\mathbb{R})$
How do I proceed to prove this?
Edit:(As suggested by @dan_fulea)
Every open set contains an open ball so if I am able to show that the above statement is true for any open ball centered at some point then we are done.
If I consider the set $B=\{A \in M_n(\mathbb{R}): ||A-0||<1\}$
$B$ contains $\mathcal{E}_{ij} \in M_n(\mathbb{R})$ such that $e_{ij}=\frac12$ ($ij^{th}$ entry). Now the set $\{\mathcal{E}_{ij}:1\le i,j\le n\}$ is linearly independent and generates $M_n(\mathbb{R})$.
Since $B$ contains all the matrices $\mathcal{E}_{ij}$, we get $B$ generates $M_n{(\mathbb{R})}$. Hence any open set containing $B$ generates $M_n{(\mathbb{R})}$.

So the question now is reduced to finding an answer for open balls which are not centered at $0$.

Corollary:
Every proper subspace of $M_n(\mathbb{R})$ is nowhere dense in $M_n(\mathbb{R})$
As every proper subspace is closed, it cannot contain any open subset of $M_n(\mathbb{R})$. As it would generate $M_n(\mathbb{R})$, contradicting the subspace being closed.
 A: There are certainly many approaches. Here's one. The idea is to start at some point of the open set and travel towards some arbitrary destination matrix by linear interpolation. You stay inside the open set for at least a little bit. But spans that contain line segments contain the full line, so you can get to the destination that way.
More formally, let $U \subset M_n(\mathbb{R})$ be open and non-empty. Pick $U_0 \in U$ and $A \in M_n(\mathbb{R})$. Consider $tU_0 + (1-t)A$. Since $U$ is open, for sufficiently small $t_0>0$ we have $B := t_0U_0 + (1-t_0)A \in U$. But then
$$A = \frac{1}{1-t_0}(t_0 U_0 + (1-t_0)A) - \frac{t_0}{1-t_0} U_0 = \frac{1}{1-t_0} B - \frac{t_0}{1-t_0} U_0 \in \mathrm{span}(U).$$
A: Since $M_n(\mathbb{R})$ is homeomorphic to $\mathbb R^{n^2},$ it suffices to prove this for $\mathbb R^{m}=\mathbb R^{n^2}.$ As pointed out in the comments, we may assume that our open set is a ball containing the origin, $B_{2r}(0).$ (To prove this, note that every open set contains a ball, and that translation is a homeomorphism). The claim now follows as soon as we note that $\{r\vec e_i\}_{i=1}^m\subseteq B_{2r}(0)$ is a linearly independent set of $m$ vectors in $\mathbb R^m$ and so basic linear algebra tells us that every vector $\vec v\in \mathbb R^m$ can be expressed as a linear combination of the $\{r\vec e_i\}_{i=1}^m.$
