The Statement: Let $X$ be a compact metrizable space. Let $H(X)$ be the set of homeomorphisms of $X$. Then $H(X)$ is a $G_{\delta}$ set in $C(X, X)$, the space of all continuous functions from $X$ to $X$ with the topology induced by the uniform/sup metric. Here $G_{\delta}$ is a set of the form $\underset{n\in\mathbb{N}}{\cap}U_n$, where $U_n$ are open.
Source: Classical Descriptive Set Theory by Alexander S. Kechris.
My Attempt: It suffices to show that $H(X)$ is closed in $C(X, X)$, i.e., $H_c(X):=C(X, X)-H(X)$ is open. To that end, let $f\in H_c(X)$ be arbitrary. Then either $f$ is not bijective or $f^{-1}$ is not continuous.
Case I: $f$ is not bijective. Then $f(x)=f(y)$ for some $x\neq y$. Screeching halt.
Any help would be greatly appreciated.
Best Answer
So first of all $H(X)$ is not necessarily closed in $C(X,X)$. A concrete example is $X=[0,1]$ and then you define $f_n:[0,1]\to[0,1]$ via
$$f_n(0)=0$$ $$f_n(1)=1$$ $$f_n(1/2)=1/n$$
and extend this piecewise linearly to whole $[0,1]$. So each $f_n$ is glueing of two straight lines.
Then each $f_n$ is a homeomorphism but the limit is constant zero at $[0,1/2]$, while a straight line from $0$ to $1$ at $[1/2,1]$.
And thus we have a sequenece of homeomorphisms convergent to a continuous non-homeomorphism.
Btw, it is not hard to see that $H([0,1])$ is also not open, by slightly modifying a homeomorphism $[0,1]\to[0,1]$ inside small enough ball in $[0,1]$.
But indeed, the $H(X)$ is $G_\delta$ in $C(X,X)$. And the hint for that can be found in the book itself. Given $d_u$ to be the uniform/sup metric the book says:
This probably needs verification. But if it is the case, then $H(X)$ is a completely metrizable subspace of $C(X,X)$. And that is equivalent to the fact that $H(X)$ is $G_\delta$, assuming you already know that $C(X,X)$ itself is complete (which is true at least for compact $X$).