Continuous Function from Compact Hausdorff Space to Itself Fixes a Non-Empty Set

Theorem
Let $\struct {X, \tau}$ be a compact Hausdorff space.

Let $f : X \to X$ be a continuous function.

Then there exists a non-empty subset $A \subseteq X$ such that:


 * $f \sqbrk A = A$

Proof
Define a sequence of sets $\sequence {X_i}_{i \mathop \in \N}$ by:


 * $X_i = \begin{cases} X & : i = 1 \\ f \sqbrk {X_{i - 1} } & : i \ge 2 \end{cases}$

Since:


 * $f \sqbrk X \subseteq X$

we have:


 * $X_i \subseteq X$

for each $i \in \N$.

Define:


 * $\ds A = \bigcap_{n \mathop = 1}^\infty X_n$

We have:

We aim to show that:


 * $f \sqbrk A = A$

Lemma 1
Since $X_{i + 1} \subseteq X_i$ for all $i \in \N$, we can apply Intersection of Nested Closed Subsets of Compact Space is Non-Empty to obtain that:


 * $A$ is non-empty.

We then have:

so:


 * $f \sqbrk A \subseteq A$

It remains to show that:


 * $A \subseteq f \sqbrk A$

Lemma 2
We then have:


 * $A = f \sqbrk A$

as required.