$T$ be a contraction on an infinite complete metric space , then is there a non singleton compact subset of $X$ which is invariant under $T$?

Question

Let $(X,d)$ be an infinite complete metric space , $T:X \to X$ be a contraction i.e. $\exists 0\le k<1 : d(T(x),T(y)) \le k d(x,y) , \forall x,y \in X$ ; then it is known that $T$ has a unique fixed point so there is a compact subset $C \subseteq X$ (namely the fixed point single-ton ) such that $T(C) \subseteq C$ , I want to ask , is there an infinite compact subset $A \subseteq X$ such that $T(A) \subseteq A$ ? Or at least , is there always a non single-ton compact set $B \subseteq X$ such that $T(B) \subseteq B$ ?

Answer 1

Of course the answer is no in general. For example, say $X$ is an infinite set with the discrete metric (and say $T(x)=x_0$ for all $x$). Then there are no infinite compact sets, invariant or not.

The answer is "typically" yes:

How do you prove that there is a fixed point? You start with $x_0$, define $x_{n+1}=T(x_n)$ and show that $x_n\to x$.

Since $x_n\to x$ the set $$A=\{x\}\cup\{x_n:n=0,1,\dots\}$$is compact, and it's certainly invariant under $T$.

The reason that's not a definite yes is that the set $A$ may be finite.