Determining whether these sets are open, closed or neither Is the set of $n \times n$ invertible matrices open, closed, or neither?
My guess was that it was open. 
Is the set of $n \times n$ matrices with determinant $1$ open, closed, or neither?
My guess was that it was closed.
Not sure of the reasoning though. Keep in mind both of these are $\subset$ $M_{n \times n}(\mathbb{R})$.
 A: Invertible matrices = $GL_n ( \mathbb{R})$ = $det^{-1} ((- \infty , 0) \cup (0, \infty))$ Which is the inverse image of an open set and therefore open
Determinant 1 = $SL_n(\mathbb{R})$ = $det^{-1}(1)$ which is the inverse image of a closed set and therefore closed.
A: Think of open sets as those for which any element in the set has a neighbourhood contained in the set, and remember that determinant is a continuous function so that perturbing a matrix by a small amount won't change the determinant by too much.
If your matrix $A$ is invertible, it has a non-zero determinant, take a small neighbourhood $N \subseteq \mathbb{R}$ around $\det A \in \mathbb{R}$ which doesn't contain $0$.  Because $\det$ is continuous, the preimage of $N$, $\det^{-1}(N)$ is open.
To see it isn't closed, show that the complement isn't open in a similar way to the next part.
For those with determinant $1$, consider a perturbation of the identity matrix to see that no open ball centred on $I$ is contained in the set of matrices with determinant $1$.  To see it is closed, show the complement is open in a similar way to the first part.
