Continuous Function from Compact Hausdorff Space to Itself Fixes a Non-Empty Set
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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:
| \(\ds A\) | \(=\) | \(\ds \bigcap_{n \mathop = 1}^\infty X_n\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds X_1 \cap \paren {\bigcap_{n \mathop = 2}^\infty X_n}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \bigcap_{n \mathop = 2}^\infty X_n\) |
We aim to show that:
- $f \sqbrk A = A$
Lemma 1
$\Box$
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:
| \(\ds f \sqbrk A\) | \(=\) | \(\ds f \sqbrk {\bigcap_{n \mathop = 1}^\infty X_n}\) | ||||||||||||
| \(\ds \) | \(\subseteq\) | \(\ds \bigcap_{n \mathop = 1}^\infty f \sqbrk {X_n}\) | Image of Intersection under Mapping: General Result | |||||||||||
| \(\ds \) | \(=\) | \(\ds \bigcap_{n \mathop = 1}^\infty X_{n + 1}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \bigcap_{n \mathop = 2}^\infty X_n\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds A\) |
so:
- $f \sqbrk A \subseteq A$
It remains to show that:
- $A \subseteq f \sqbrk A$
Lemma 2
- $A \subseteq f \sqbrk A$
$\Box$
We then have:
- $A = f \sqbrk A$
as required.
$\blacksquare$