Can I show that intersection of finitely many compact subsets is non-zero

Question

Suppose that $X$ is Hausdorff as well as compact, and let $f : X → X$ be a continuous map. Let $X_0 = X, X_1 = f(X_0)$ and inductively define $X_{n+1} = f(X_n)$ for $n ≥ 1$. Show that $A =\bigcap_n X_n$ is non-empty.

Seeing as $X$ is Hausdorff and compact, any subset is closed. Hence, $X_i$ is compact and closed $\forall i\in\mathbb{N}$.

If I could show that the intersection of finitely many of these subsets is always non-zero, then it would follow that the intersection of all of them is non-zero (as $X$ is compact).

Any hints?

Answer 1

First: consider a function $f:X\rightarrow Y$, and $A\subset B \subset X$. I shall show that $f(A)\subset f(B)$. Consider an element $x\in f(A)$, and an element $y\in A$ such that $f(y)=x$. As $A\subseteq B$ and $y\in A$, we have $y\in B$. Therefore, $f(A)\subseteq f(B)$.

So if $X_i\subseteq X_{i-1}$, then $X_{i+1}=f(X_i)\subseteq f(X_{i-1})=X_i$. As $X=X_0$, the base case $X_1\subseteq X_0$ is verified and therefore, by induction, $X_i\subseteq X_{i-1},\forall i\in\mathbb{N}$. Hence, we have a nested sequence of subsets.

As $X$ is compact and Hausdorff, the $X_i$s are closed. The intersection of finitely many of these subsets is equal to the smallest subset, and hence isn't empty. So $A\neq \emptyset$.

Answer 2

HINT: Show that for every function $F\colon X\to X$, and every $A\subseteq X$: If $F(A)\subseteq A$, then $F(F(A))\subseteq F(A)$.