I don´t know how to prove this. Obviously we use the usual topology on$ R^{n^2} $ and the obvious injection of $Gl_n(R)$ into $ R^{n^2} $ $Gl_n(R)$ is the set of invertibles matrices or equivalently, maybe be useful to use that $ GL_n \left( R \right) = \det ^{ - 1} \left( {R - \left\{ 0 \right\}} \right) $ and that the determinant it´s a continuous function.
$Gl_n(R)$ dense in $ R^{n^2} $
1 Answers
$GL$ is the complement of the zero set of a non-zero polynomial function defined on $\mathbb R^{n^2}$, namely the determinant. This suggest we simplify the problem by generalizing.
Let us show that
the complement of the zero set of a non-zero polynomial function in $\mathbb R^n$ is dense.
Suppose then $f$ is a polynomial function on $\mathbb R^n$ and that $U=\mathbb R^n\setminus f^{-1}(0)$ is not dense. There is then a point $p\in\mathbb R^n$ and an $\varepsilon>0$ such that $B_\varepsilon(p)\cap U=\emptyset$ or, in other words, $B_\varepsilon(p)\subseteq f^{-1}(0)$: this tells us that $f$ is zero on the whole of $B_\varepsilon(p)$. (Here $B_\varepsilon(p)$ is the ball of radius $\varepsilon$ centered at $p$)
Since $f$ is a polynomial, there is an $N\in\mathbb N$ such that the Taylor polynomial of $f$ at $p$ is actually equal to $f$ on all of $\mathbb R^n$. Now compute this Taylor polynomial, and see if you can end this.
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0I am sorry but i don't know what you don't understand. – 2014-01-04