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Show that set of all invertible $n\times n$ matrices with real entries (denoted by $GL_n(R)$) is open set in $M_n(R)$.

My attempt: by open set I think it means neighborhood of every point in set is contained in set, but this definition doesn't seem to help solve this question. Any other approach?

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  • $\begingroup$ Have you tried anything further than just writing down the definition of open sets? $\endgroup$
    – anomaly
    Sep 5, 2015 at 6:45
  • $\begingroup$ i think op needs some help to apply this definition. $\endgroup$
    – BigBang
    Sep 5, 2015 at 6:50

1 Answer 1

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If you can show that its complement is a closed set the also you prove that $$GL_{n}(\mathbb R)$$ is open.

Now see that the complement of $GL_{n}(\mathbb R)$ is the set of all matrices that have determinant $0$. And $$det : M_{n}(\mathbb R)\rightarrow \mathbb R$$ is a continuous function.

For continuous functions, pre-image of a closed set is closed and $\{0\}$ is a closed set in $\mathbb R$.

So the pre-image of this singleton closed set under the determinant map, i.e., the set of all non-invertible matrices, i.e., the complement of the set $GL_{n}(\mathbb R)$ is a closed set.

Hence $GL_{n}(\mathbb R)$ is open in $M_{n}(\mathbb R)$.

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    $\begingroup$ Why did you use the complement? The preimage of an open set by a continuous function is open as well, and $\mathbb{R} \setminus \{0\}$ is open in $\mathbb{R}$. Exactly the same proof without needing to consider the complement. $\endgroup$ Sep 5, 2015 at 8:22

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