Neighborhood of the identity matrix Is it correct to say that there exists a neighborhood of the identity matrix which is a subset of the set of nonsingular matrices, because the determinant is continuous over the set of matrices?
 A: Yes, this is true. The determinant of a matrix $A$ can actually be computed as a linear combination of products of the entries, and so the determinant is a continuous function of the entries. The identity matrix has a determinant of $1$, so if we consider the the determinant as a function $D:\mathbb{R}^{n\times n}\to\mathbb{R}$, the inverse image of the interval $(0,2)$ is an open set in $\mathbb{R}^{n\times n}$ containing the identity matrix. Thus, this set is a neighborhood of $I$, and all matrices in this set are invertible.  
A: Yes! This is true for exactly the reason you describe! 
A: Yes!!!
Let
$\Bbb F = \Bbb R \; \text{or} \; \Bbb C. \tag 0$
Since 
$\det: M_n(\Bbb F) \to \Bbb F, \; A \to \det(A) \tag 1$
is continuous, which follows from the fact that it may be expressed as a polynomial in the entries of $A$, and since
$\det I = 1, \tag 2$
it follows that for any open disk about $1$ of the form
$D_\epsilon(1)  = \{ x \in \Bbb F \mid \vert x - 1 \vert < \epsilon, \; 0 < \epsilon < 1\}, \tag 3$
we have
$\det^{-1}(D_\epsilon(1)) \subset M_n(\Bbb F) \tag 4$
is open; furthermore,
$I \in \det^{-1}(D_\epsilon(1)), \tag 5$
by virtue of (2).  Now it is easy to see that
$A \in \det^{-1}(D_\epsilon(1)) \Longrightarrow \det A \ne 0 \notin D_\epsilon(1) \tag 6$
since $\epsilon < 1$.
We thus see that $\det^{-1}(D_\epsilon(1))$ is such a desired open set.
