Vtyjkyuk

Let $(K,d)$ be a compact metric space and denote $ C(K,K):= f:K to K ; space f space textis continuous $. Endow the set $C(K,K)$ with the supremum metric $ rho (f,g):=mathrmsup d(f(x),g(x)) ; x in K $.

I'd like to show that the group of homeomorphisms of $(K,d)$ is a $G_delta$ subset of the metric space $( C(K,K),rho)$.

I managed to show that it is enough to prove that the set $ f in C(K,K) ; space f space textis injective $ is $G_delta$, but then I got stuck.

I'll appreciate any help.

Let $H$ be the set of homeomorphisms in $C(K,K)$. For $fin H$, let $U_n(f)$ be a countable nested collection of open [in $C(K,K)$] metric balls around $f$. We will shrink the radii later. For now, let's say $r_n(f)$ is the radius of $U_n(f)$.

If we set
$$ U_n =bigcap_fin H U_n(f)$$
then we want to show $H=cap U_n$. One direction is easy, so we show the other.

Let $gin cap U_n$. Then we have a sequence $f_n$ with $gin U_n(f_n)$. Let's insist that $r_n(f)<frac1n$, so that $f_n$ is a Cauchy sequence with $f_nrightarrow g$.

Since inversion is continuous on $H$, we can further shrink $r_n(f)$ [dependent on $f$] such that
$$ rho(f,alpha)<2r_n(f)implies rho(f^-1,alpha^-1<frac1n $$

This ensures $f_n^-1$ is also Cauchy, and thus $f_n^-1rightarrow hin C(K,K)$. But composition is continuous, so
beginalign*
f_ncirc f_n^-1 &rightarrow gcirc h\
f_n^-1circ f_n &rightarrow hcirc g
endalign*
Since the left-hand sides are constantly the identity map, we see $h=g^-1$, and so $gin H$, and we're done.