Group of homeomorphism of [0,1]

Question

I'm considering the group of homeomorphisms from $[0,1]$ to $[0,1]$. Is this group with functions composition and supremum metric compact or locally compact?

I think that this group is not compact, because if we consider sequence of functions $x^{n}$ it converges to $0$ for $x=[0,1)$ and to $1$ for $x=1$, so we have a sequence of homeomorphisms which converges to function which is not bijection. Does it make sense?

Is this group locally compact?

Answer 1

The usual supremum metric $d_\infty$ on $\mathscr{H}([0,1])$ is not complete, one can replace it by $D(f,g)=d_\infty(f,g)+d_\infty(f^{-1},g^{-1})$ and have an equivalent complete metric. Composition and inverse are indeed continuous so we have a topological group.

It's not compact because a function like $f$ that is $2x$ on $[0,\frac14]$, $\frac12$ on $[\frac14,\frac34]$ and $2x-1$ on $[\frac34,1]$ is in the closure of $\mathscr{H}([0,1])$ as a subspace of $C(I,I)$ (in the $d_\infty$ metric) and so $\mathscr{H}([0,1])$ is not absolutely closed.

My hunch is that local compactness will also fail, but I have no proof of that.

Answer 2

Note that your space is homeomorphic to $l^2$ (the space of all square summable real sequences defined by $l^2 = \{(x_n)_{n\in\mathbb{N}}$, $x_n \in \mathbb{R}$, such that $\sum_{n\in\mathbb{N}} \left|x_n\right|^2 < \infty \}$).

$l^2$ is not compact or even locally compact. So since compactness is preserved by homeomorphism, your space is not compact.

As far as I know, local compactness is a topological property too, so your space is not even locally compact.

Answer 3

Yes, your basic idea is mostly correct, but a bit incomplete.

You should say that if a sequence of functions converges uniformly, then it converges pointwise to the same limit. Therefore, if a sequence $(f_n)_n$ converges pointwise to some $f$, then any subsequence $(f_{n_k})_k$ converging uniformly has to also converge uniformly to the same $f$, so if $f$ is not a homeomorphism, $(f_n)_n$ cannot have a uniformly convergent subsequence, thus violating (sequential) compactness.

The space is not locally compact, either. To show this, you can pick an arbitrary basic open subset and construct a sequence within whose pointwise limit has a jump discontinuity at $1$ (the idea is almost the same).