The limit of a singular matrix?

Question

Now to show the set of invertible $n \times n$ matrices are an open set in the set of all $n \times n$ matrices one can show the set of singular matrices are closed in the set of all $n \times n$ matrices. Therefore if $A_n$ is a sequence of singular matrices that converge to say $B$ then $B$ must be a singular matrix. I am having trouble trying to show this though. So assuming we have a normed vector space $$\lim_{n \to \infty} ||A_n-B||=0.$$ I guess I need to show $\det(A_n-B)=0$ for $n\ge N$.

Answer 1

Or you probably know that the pre-image of a closed set under a continuous map is closed. Consider the determinant as a function from the space $\mathbb R^{n^2}$ of all $n\times n$ matrices to the real numbers. I.e. $$ \det:\mathbb R^{n^2}\to \mathbb R.$$

This is a continuous function because the determinant is just a polynomial in the entries of the matrices. The singular matrices as you know are the pre-image of the set $\{0\}$ which you know is closed in $ \mathbb R$. So the singular matrices are closed in the space of all $n\times n$ matrices as well.

Edit: Note that this proof is not unique to matrices over $\mathbb R$ but any topological ring $R$, i.e a ring with a topology such that addition and multiplication are continuous functions with respect to this topology. Thanks to user1551 for setting me straight.