Given a singular matrix, I am tring to find an invertible matrix... (Finite Dimensional Space)

Question

In coordinates and in a finite-dimensional space, how would I prove that given any singular $n$x$n$ matrix $A$, any $\epsilon\gt0$ and any matrix norm $||.||$, there is an invertible $n$x$n$ matrix $B$ such that $||A-B|| \lt \epsilon$ ?

Is it worthwhile to consider

$spectrum(A)=\{\lambda | A-\lambda I$ is singular$\}$ ?

Answer 1

Hint: The determinant is a polynomial in the matrix entries. If it was identically zero in any open subset of $\mathbb R^{n\times n}$, it would need to be the zero polynomial, but it isn't.