Traits for S140: \mathbb{R} extended by a point with cocountable open neighborhoods

spaces/S000140/README.md

@@ -1,13 +1,13 @@
 ---
 uid: S000140
-name: Real numbers extended by a point with co-countable open neighborhoods
+name: $\mathbb{R}$ extended by a point with co-countable open neighborhoods
 refs:
 - mathse: 4850979
   name: Answer to 'Radial/pseudoradial implies Fréchet-Urysohn/sequential for locally countable spaces'
 - mathse: 4854178
   name: What are the compactness properties of $\mathbb R$, extended by a point with co-countable open neighborhoods?
 ---
 
-Let $X=\mathbb R\cup \{\infty\}$, with $\mathbb R$ having the Euclidean topology and open in $X$, and open neighborhoods of $\infty$ given by sets of the form $U\cup\{\infty\}$, where $U\subseteq \mathbb R$ is co-countable and open in the Euclidean topology. 
+Let $X=\mathbb R\cup \{\infty\}$. Let $\mathbb R$ be open in $X$ and have the topology of {S25}, and let open neighborhoods of $\infty$ be given by sets of the form $U\cup\{\infty\}$, where $U\subseteq \mathbb R$ is co-countable and open in the Euclidean topology. 
 
 Constructed in {{mathse:4850979}} as an example of a space that is {P173} and {P81}, yet fails to be {P79}, yielding a counterexample to a natural analogue of {T211}.  Elaborated on in {{mathse:4854178}}.

spaces/S000140/properties/P000073.md

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+---
+space: S000140
+property: P000073
+value: true
+---
+
+Let $S \subseteq X$ be a nonempty irreducible ({P39}) set. $\mathbb R$
+is open in $X$ and so $S \cap \mathbb R$ is also irreducible. Since {S25|P73} and {S25|P2}, we know that
+$S \cap \mathbb R$ is either empty or a singleton. Thus $S$ has at most two
+points. Now note that {S140|P2}, and so if $S$ had two points, then $S$
+would be homeomorphic to {S1}. But {S1|P39}, and so $S$ is a singleton.

spaces/S000140/properties/P000189.md

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+---
+space: S000140
+property: P000189
+value: true
+---
+
+$\mathbb R$ is a dense subset of $X$ and {S25|P189}.

spaces/S000140/properties/P000204.md

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+---
+space: S000140
+property: P000204
+value: false
+---
+
+The point $\infty$ is not a cut point, since $X \setminus \{\infty\}=\mathbb R$ is connected.
+Now, let $x \in \mathbb{R}$ and consider a map $f$
+from $X \setminus \{x\}$ to {S1}. Then the intervals $(-\infty, x)$ and
+$(x, \infty)$ in $\mathbb{R}$ are connected and thus get sent to a
+single point. Thus $f$ factors through the quotient space of $X$
+where $(-\infty, x)$ and $(x, \infty)$ become points. This space is
+{S11} and {S11|P36}; thus $f$ is constant, which proves the assertion.

spaces/S000140/properties/P000210.md

@@ -0,0 +1,10 @@
+---
+space: S000140
+property: P000210
+value: true
+refs:
+- mathse: 5126685
+  name: What properties hold for $\mathbb{R}$ extended by a point with co-countable open neighborhoods?
+---
+
+See {{mathse:5126685}}.

spaces/S000140/properties/P000219.md

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+---
+space: S000140
+property: P000219
+value: false
+---
+
+{S140|P65} and has a subspace homeomorphic to {S25}.