Prove $\exists$ closed subset A st $f(A)=A$

Question

Let $(X,d)$ be a compact metric space and $f$ be continuous on $X$, show that there exists non-empty closed subset $A$ of $X$ such that $f(A)=A$

So, $f$ will be uniform continuous, but how does that help?

Answer 1

Let $C_0=X$ and $C_{n+1}=f(C_n)$ for all $n$. By induction, note that $C_n\supseteq C_{n+1}$ for all $n$. Let $C=\bigcap_n C_n$, and note that $C$ is nonempty, being the intersection of a decreasing sequence of nonempty compact sets.

To check that $f(C)=C$, consider a point $x\in C$. Since $x\in C_{n+1}$ for all $n$, then there are points $x_n\in C_n$ with $x=f(x_n)$ for all $n$. By compactness of $X$, there is a subsequence $(x_{n_k})_{k\in\mathbb N}$ of $(x_n)_{n\in\mathbb N}$ that converges to a point in $X$, call it $y$. For each $m$, the tail subsequence $(x_{n_k})_{k\ge m}$ is contained $C_m$ (since the $C_n$ are decreasing), and therefore so is its limit $y$. This means that $y\in \bigcap_n C_n=C$. Finally, since $f$ is continuous and $f(x_{n_k})=x$ for all $k$, then also $f(y)=x$. This proves that $x\in f(C)$, and therefore that $C\subseteq f(C)$.

On the other hand, if $t\in f(C)$, then there is a point $z\in C$ with $t=f(z)$. Since $z\in C_n$ for each $n$, then $t\in C_{n+1}$ for all $n$. Clearly, $t\in C_0=X$ as well, so $t\in C$. This proves that $f(C)\subseteq C$, and therefore that indeed $C=f(C)$, as desired.

Let me conclude with a general observation. A complete lattice is a partially ordered set such that each subset has a supremum and an infimum. If $(P,\le)$ is a partial order, a function $g:P\to P$ is order preserving iff whenever $x\le y$ in $P$, then also $g(x)\le g(y)$. The Knaster-Tarski theorem asserts that if $(L,\le)$ is a complete lattice, and $\pi:L\to L$ is order preserving, then $\pi$ admits a fixed point and, in fact, the set of fixed points of $\pi$ also constitutes a complete lattice.

It is easy to verify that if $X$ is a compact metric space, then the collection $K(X)$ of closed subsets of $X$, partially ordered by inclusion, forms a complete lattice: The infimum of a collection of closed sets is their intersection, and their supremum is the closure of their union. if $f:X\to X$ is continuous, then we can see $f$ as a function mapping closed subsets $D$ of $X$ to their pointwise image $f(D)$, and in this way we can consider $f$ as a map from $K(X)$ to itself, that it is obviously order-preserving. It follows that the collection of closed sets $A$ such that $f(A)=A$ forms a complete lattice (this particular case, of course, can be verified directly without having to go through the proof of the general theorem). Its minimum is the empty set, and its maximum is the set $C$ we built in the first paragraph.

Answer 2

This sketch of a proof is based on the "special" definition of compactedness on metric spaces : compact metric spaces. Take any point in $x\in X$ and consider the sequence $f(x), f(f(x)), \ldots$ then there is a subsequence that converges to a point $x_0$ then $A= \{x_0\}$.