Continuous Function from Compact Hausdorff Space to Itself Fixes a Non-Empty Set/Lemma 1

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

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

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


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

Then, for each $i \in \N$:
 * $X_i$ is closed
 * $X_i$ is non-empty
 * $X_{i + 1} \subseteq X_i$.

Proof
For all $n \in \N$, let $\map P n$ be the proposition:


 * the set $X_n$ is non-empty and closed with $X_{n + 1} \subseteq X_n$.

Base Case
From Underlying Set of Topological Space is Clopen:


 * $X$ is closed.

We have that $X \ne \O$.

We also have that:


 * $X_2 = \map f {X_1} \subseteq X = X_1$

So $\map P 1$ is true.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.

So, suppose that:


 * $X_k$ is closed and non-empty,

and:


 * $X_{k + 1} \subseteq X_k$

This is our induction hypothesis.

Induction Step
Now we need to show that:


 * $X_{k + 1}$ is closed and non-empty,

and:


 * $X_{k + 2} \subseteq X_{k + 1}$

We have:


 * $X_{k + 1} = \map f {X_k}$

Note that:


 * $X_k \subseteq X$

Since $X_k$ is non-empty, we have:


 * $X_{k + 1} \ne \O$

From Closed Subspace of Compact Space is Compact, we have:


 * $X_k$ is compact.

From Continuous Image of Compact Space is Compact, we therefore have:


 * $X_{k + 1} = \map f {X_k}$ is compact.

From Compact Subspace of Hausdorff Space is Closed, we have:


 * $X_{k + 1}$ is closed.

So:


 * $X_{k + 1}$ is non-empty and closed.

From:


 * $X_{k + 1} \subseteq X_k$

We obtain:


 * $\map f {X_{k + 1} } \subseteq \map f {X_k}$

and so:


 * $X_{k + 2} \subseteq X_{k + 1}$

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.