invertible matrices are dense in set of nxn matrices

If $A$ is any $n\times n$ matrix with real or complex entries,Then there are invertible matrices arbitrarily close to $A$,under any norm for the $n\times n$ matrices.

This is easily proven as follows. Take any invertible matrix $B$(e.g. $B=I$), and consider the function(for $t\in ℝ$ or $ℂ$)

$$p(t)=det\left((1-t)A+tB\right).$$

Clearly, $p$ is a polynomial function. It is not identically zero, for $p(1)=detB\ne 0$.But a non-zero polynomial has only finitely many zeroes,So given any single point $t_0$, if $t$ is close enough but unequal to $t_0$,$p(t)$ must be non-zero. In particular, applying this for $t_0=0$,we see that the matrix $(1-t)A+tB$ is invertible for small $t\ne 0$.And the distance of this matrix from $A$ is $|t|\parallel B-A\parallel$,which becomes small as $t$ gets small.