Continuous mapping onto itself s.t. $f(A) = A$ where $A$ is closed

Question

Let $f$ be a continuous mapping of a Hausdorff non-separable space $(X,\tau)$ onto itself. Prove that there exists a proper non-empty closed subset $A$ of $X$ such that $f(A) = A$.

[ Hint: Let $x_0 \in X$ and define a set $S = \{x_n : n \in \mathbb{Z}\}$ such that $x_{n+1} = f(x_n)$ for every integer n ]

Is the above result true if $(X,\tau)$ is separable? (Justify your answer.)

I can find some examples of such maps and closed sets.

Let $f$ be a continuous function:

$$f: \mathbb{R} \to \mathbb{R}$$ $$f(x) = -x$$

and let set $A = \{-1, 1\}$. Then $A$ is closed, proper and non-empty and $f(A) = A$.

I need some help with general proof. If at some stage $f(x_n) = x_0$ then sequence $\{f(x_n)\}$ is finite and closed. Otherwise it's infinite countable. Can't guess how to use non-separability here ...

Answer 1

Let $A=\overline{\{x_n\,|\,n\in\mathbb{N}\}}$. Since $X$ is non-separable, $A\neq X$. On the other hand, $A$ is closed and $f(A)=A$.

On the other hand, if $S^1=\{z\in\mathbb{C}\,|\,|z|=1\}$ and $f\colon S^1\longrightarrow S^1$ is defined by $f(z)=e^{i\theta}z$, where $\theta\in\mathbb R$ is such that $\frac\theta\pi\notin\mathbb Q$, then there is no such $A$.

Answer 2

You can choose as $A=cl(S)$ and in this case your set is closed and proper (and oviously non-empty) because if $cl(S)=A=X$ than X is separable ($S$ is a countable set) and so $A\neq X$

Now we prove that $f(A)=A$:

$f(S)=S$ and so $ f(A)=f(cl(S))\subset cl(f(S))=cl(S)=A$ and for the other side you can use the property $T_2$