# Are there holes in every neighbourhood in GL(n,R)?

$$GL(n,\mathbb{R})$$ can be seen as $$\mathbb{R}^{n^2}$$ with points of non invertible matrix removed which form "holes" in $$\mathbb{R}^{n^2}$$. These holes seems to be dense in $$\mathbb{R}^{n^2}$$ which will mean any neighbourhood of $$GL(n,\mathbb{R})$$ will contain holes, even for the neighbourhood that is homeomorphic to some neighbourhood of the oringin of $$\mathbb{R}^{n^2}$$. It is strange that a region with holes is homeomorphic to a region without holes.

Is this an example of a region with holes is homeomorphic to a region without holes, or is non invertible matrix not dense in $$\mathbb{R}^{n^2}$$?

• Also a region with 'holes' cannot be homeomorphic to a region without 'holes'. Think about the fact that when $n=1$, removing a point disconnects the region. Oct 29, 2021 at 0:57
• It should be easy to see that there aren't any noninvertible matrices close to, say, $\pmatrix{10&0\cr0&10\cr}$. Oct 29, 2021 at 0:59

$$GL_{n}(\mathbb{R})$$ is open in $$\mathbb{R}^{n^2}$$. This is a consequence of the determinant map being continuous. Also tells you that the set of non invertible matrices cannot be dense. The invertible matrices on the other hand are.
• The openness implies something further, that for every invertible matrix $A$ there is a $\delta$ such that all matrices $B$ satisfying $\|A-B\|<\delta$ are invertible. You can find a constructive proof of this from the fact that for a matrix $B$, $A-B=A(I-A^{-1}B)$. Can you see how taking B with norm sufficiently small makes $(I-A^{-1}B)$ invertible? Oct 29, 2021 at 1:04
• For a non invertible matrix $M$, for sufficiently small $\epsilon>0$, $M+\epsilon I$ is invertible. Oct 29, 2021 at 1:06