Continuous Function from Compact Hausdorff Space to Itself Fixes a Non-Empty Set/Lemma 2

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

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

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


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

Define:


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

Then:
 * $A \subseteq \map f A$

Proof
Let $y \in A$.

Then, by the definition of intersection:


 * $y \in \map f {X_i}$ for each $i \in \N$.

That is:


 * for each $i \in \N$, there exists $x_i \in X_i$ such that $\map f {x_i} = y$.

So:


 * the intersection $\map {f^{-1} } {\set y} \cap X_i$ is non-empty for each $i \in \N$.

From Continuity Defined from Closed Sets:


 * $\map {f^{-1} } {\set y}$ is closed.

From Intersection of Closed Sets is Closed in Topological Space:


 * $\map {f^{-1} } {\set y} \cap X_i$ is closed.

Then, from Intersection of Nested Closed Subsets of Compact Space is Non-Empty, we have:


 * $\ds \bigcap_{i \mathop = 1}^\infty \paren{\map {f^{-1} } {\set y} \cap X_i} \ne \emptyset$

We then have:

So there exists $x \in A$ such that:


 * $\map f x = y$

That is:


 * $y \in \map f A$

So we obtain:


 * $A \subseteq \map f A$

as required.