Is the set of invertible functions $f:A\rightarrow A$ compact? I would like to know if the space of invertible, or 1-to-1, functions $f:A\rightarrow A$ is a compact function space, or if restrictions on $A$ are required? 
Recommended resources would also be welcome.

$\renewcommand{\Re}{\mathbb{R}}$As a simple counterexample, consider the space of linear functions $f:\Re\to \Re$ with the operator norm $\|g\|=\alpha$ whenever $g(x) = \alpha x$. This defines the space of functions $(X,\|\cdot\|)$. Let $Y$ be the subset of $X$ of all invertible functions; equipped with the same norm. Consider the sequence:
$$f^\nu(x) = \frac{1}{2^\nu}x.$$
All $f^\nu$ are invertible, but $\|f^\nu\|=\frac{1}{2^\nu}\to 0$, so $Y$ is not norm-closed, thus not compact by the Heine-Borel theorem. 
Taking (linear) functions $f:A\to A$ where $A$ is a compact set will not remedy the problem. I don't think that the space of invertible functions is, in general, compact in the w$\ast$ topology (in more interesting infinite dimensional spaces). 
