Cleanup traits of S145 (Free ultrafilter topology on ω) (#1186)

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}, which is not possible.
+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 $f(x) = \begin{cases} q & \text{if }x = p, \\ p & \text{if }x = q, \\ x & \text{otherwise}. \end{cases}$
+
+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:
+  - mathse: 5021484
+    name: Does a free ultrafilter topology have the fixed point property?
+---
+
+See {{mathse:5021484}}.

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 $f(x) = \begin{cases} q & \text{if }x = p, \\ x & \text{otherwise}. \end{cases}$
+
+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.
+
+Therefore, $f$ is a retract 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 $A$ be an infinite closed proper subset of $X$, and take an arbitrary $p \in X \setminus A$.
+For each $S \subseteq A$, the function $f_S:X\to X$ defined by
+$$f_S(x) = \begin{cases} p & \text{if }x\in S, \\ x & \text{otherwise}, \end{cases}$$
+is continuous since for every nonempty open set $U$, its inverse image $f^{-1}(U)$ contains $U\setminus A$, which is open and nonempty; hence $f^{-1}(U)$ is open.
+
+And there are continuum many such functions $f_S$.