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$GL(n,\mathbb{R})$ 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 $\mathbb{R}^{n^2}$?

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  • $\begingroup$ 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. $\endgroup$ Oct 29, 2021 at 0:57
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    $\begingroup$ It should be easy to see that there aren't any noninvertible matrices close to, say, $\pmatrix{10&0\cr0&10\cr}$. $\endgroup$ Oct 29, 2021 at 0:59

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$GL_{n}(\mathbb{R})$ 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.

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  • $\begingroup$ It seems that for any invertible matrix, you can find another one that is near enough to the given one, is that right? Does this not imply dense? $\endgroup$
    – jw_
    Oct 29, 2021 at 0:58
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    $\begingroup$ 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? $\endgroup$ Oct 29, 2021 at 1:04
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    $\begingroup$ For a non invertible matrix $M$, for sufficiently small $\epsilon>0$, $M+\epsilon I$ is invertible. $\endgroup$ Oct 29, 2021 at 1:06

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