Cleanup traits of S145 (Free ultrafilter topology on ω)

spaces/S000145/properties/P000044.md

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+---
+space: S000145
+property: P000044
+value: true
+---
+
+Let $Y$ be a {P36} subset of $X$ with at least two points.
+If $Y$ is closed in $X$, by the property of ultrafilters all subsets of $Y$ are closed, thus $Y$ is {P52}.
+This implies that $Y$ is open, that is, every {P36} subsets of $X$ must be open.
+
+Now, because $X$ is {P36}, $X$ cannot be partitioned into two nonempty open subsets.
+We conclude that $X$ cannot be partitioned into two {P36} subsets, each with at least two points.

spaces/S000145/properties/P000086.md

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+---
+space: S000145
+property: P000086
+value: true
+---
+
+Let $p, q \in X$ be arbitrary distinct points.
+Define $f(x) = \begin{cases} q & x = p, \\ p & x = q, \\ x & \text{otherwise}, \end{cases}$ then $f(x)$ is involutary and continuous since for every open subset $U$ and every point $x \in X$, both $U \setminus \{x\} = U \cap \left( X \setminus \{x\} \right)$ and $U \cup \{x\}$ are open.

spaces/S000145/properties/P000089.md

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+---
+space: S000145
+property: P000089
+value: true
+refs:
+  - mo: 124144
+    name: Periodic point-free maps and free ultrafilters.
+---
+
+See {{mo:124144}}.

spaces/S000145/properties/P000101.md

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+---
+space: S000145
+property: P000101
+value: false
+---
+
+Let $p, q \in X$ be arbitrary distinct points.
+Define $f(x) = \begin{cases} q & x = p, \\ x & \text{otherwise}, \end{cases}$ then $f(x)$ is continuous since for every open subset $U$ and every point $x \in X$, both $U \setminus \{x\} = U \cap \left( X \setminus \{x\} \right)$ and $U \cup \{x\}$ are open.
+
+Then $f$ is a continuous function from $X$ to $X \setminus \{p\}$, which is not closed.

spaces/S000145/properties/P000138.md

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+---
+space: S000145
+property: P000138
+value: false
+---
+
+Let $A = \{a_0, a_1, \dots\}$ be any infinite closed subset of $X$, and take arbitrary $p \in X \setminus A$.
+Then for each $S \subseteq \mathbb N$, the function \[ f_S(x) = \begin{cases} p & x = a_i, i \in S, \\ x & \text{otherwise}, \end{cases} \] is continuous since the preimage of any open subset $U$ is of the form $U \setminus Q$ or $\left( U \cup \{p\} \right) \setminus Q$ where $Q \subseteq A$ is closed.
+
+These $f_S$ induced continuum many continuous function from $X$ to itself.