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 $\sequence {X_i}_{i \mathop \in \N}$ by:

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

Define:

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

Then:

$A \subseteq f \sqbrk A$

Proof

Let $y \in A$.

Then, by the definition of set intersection:

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

From Continuity Defined from Closed Sets:

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

From Intersection of Closed Sets is Closed in Topological Space:

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

From Continuous Function from Compact Hausdorff Space to Itself Fixes a Non-Empty Set: Lemma 1, we have:

$X_{i + 1} \subseteq X_i$ for each $i \in \N$.

So, from Set Intersection Preserves Subsets:

$f^{-1} \sqbrk {\set y} \cap X_{i + 1} \subseteq {f^{-1} } \sqbrk {\set y} \cap X_i$

We can therefore apply Intersection of Nested Closed Subsets of Compact Space is Non-Empty to obtain:

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

We then have:

\(\ds \bigcap_{i \mathop = 1}^\infty \paren {f^{-1} \sqbrk {\set y} \cap X_i}\)

\(=\)

\(\ds f^{-1} \sqbrk {\set y} \cap \bigcap_{i \mathop = 1}^\infty X_i\)

General Associativity of Set Intersection

\(\ds \)

\(=\)

\(\ds f^{-1} \sqbrk {\set y} \cap A\)

\(\ds \)

\(\ne\)

\(\ds \O\)

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

$\map f x = y$

That is:

$y \in f \sqbrk A$

So we obtain:

$A \subseteq f \sqbrk A$

as required.

$\blacksquare$