Proof validation: Proving that the group of homeomorphisms of $[0,1]$ is infinite

Question

I'm trying to solve the following exercice:

Let $(X,\tau)$ be any topological space and $G$ the set of all homeomorphisms of $X$ into itself. Then the set $G$ is a group under the operation of composition of functions and it's called the "Group of homeomorphisms of $X$"

1 - If $X=[0,1]$, show that $G$ is infinite.

So, my initial thought was to assume that $G$ is finite and then show that we can build a new homeomorphism from the elements of $G$ that it's not in $G$. So what I did was the following:

My proof:

Let's suppose that $G$ is finite. Then we have that, as every function $f\in G$ has an inverse function $f^{-1} \in G$, that $\text{card } G=2n + 1,n\in \Bbb N$. The "$+1$" part comes from the fact that the identity function $I$ is in $G$, but $I^{-1}$ is simply $I$

So, we can list the elements of $G$ as following:

$$G = \{f_1,...,f_n,f^{-1}_1,...,f^{-1}_n,I\}$$

So now, lets define a new function as following:

$$f'= f_1 \circ \left( f_2 \circ (... \circ \ f_n )\right)$$

So we have that: $\forall f \in G, f' \neq f$. But we have that $G$ is a group so it's closed under the operation $\circ$. This is a contradiction, meaning that the set $G$ can't be finite.

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Is this proof correct? Because I assumed that $\forall f \in G, f' \neq f$, is this correct? Is this the most straight forward way of proving this? Because I sometimes find myself over-complicating proofs. Can you show me any other alternate ways of proving this?

Answer 1

For starters, you’ve overlooked the fact that a homeomorphism can be its own inverse; an example on $[0,1]$ is the function $f(x)=1-x$. And it’s really not clear that repeated composition is always going to give you a new function.

You would be better off actually constructing an explicit infinite family of autohomeomorphisms of $[0,1]$. Probably the simplest is the family of functions $f(x)=x^n$ for $n\in\Bbb Z^+$, though there are plenty of others. For instance, you might try to construct an uncountable family of piecewise linear autohomeomorphisms of $[0,1]$; this can be done in such a way that each function is defined on just two intervals, and the intervals are the same for all of the functions.

Answer 2

As you said, you haven’t prove that $f^\prime \neq f$, which is a flaw in your argumentation.

Also why to proceed by contradiction ? It is quite easy to show an infinite set of homeomorphisms. For example $x \mapsto x^n$ with $n \in \mathbb N$.