From 4bbda8f36b82215bc3db5abb05a81387938f475d Mon Sep 17 00:00:00 2001
From: mathmaster13 <74142852+mathmaster13@users.noreply.github.com>
Date: Fri, 27 Feb 2026 18:22:02 -0500
Subject: [PATCH 01/16] Create P000189.md

---
 spaces/S000140/properties/P000189.md | 9 +++++++++
 1 file changed, 9 insertions(+)
 create mode 100644 spaces/S000140/properties/P000189.md

diff --git a/spaces/S000140/properties/P000189.md b/spaces/S000140/properties/P000189.md
new file mode 100644
index 000000000..6e35968cb
--- /dev/null
+++ b/spaces/S000140/properties/P000189.md
@@ -0,0 +1,9 @@
+---
+space: S000140
+property: P000189
+value: true
+---
+
+Given a map $f$ from $X$ to {S15}, we know {S25|P189} and so $\mathbb{R}$
+is sent to a point. Thus $f$ factors through the quotient space
+$X/\mathbb{R}$, and $X/\mathbb{R}$ is the {S10}. {S10|P189} and so $f$ is constant.

From d39f84d8d926689071bd8fccb0d5ada29e49e97d Mon Sep 17 00:00:00 2001
From: mathmaster13 <74142852+mathmaster13@users.noreply.github.com>
Date: Fri, 27 Feb 2026 19:29:07 -0500
Subject: [PATCH 02/16] Create P000204.md

---
 spaces/S000140/properties/P000204.md | 13 +++++++++++++
 1 file changed, 13 insertions(+)
 create mode 100644 spaces/S000140/properties/P000204.md

diff --git a/spaces/S000140/properties/P000204.md b/spaces/S000140/properties/P000204.md
new file mode 100644
index 000000000..6315edc12
--- /dev/null
+++ b/spaces/S000140/properties/P000204.md
@@ -0,0 +1,13 @@
+---
+space: S000140
+property: P000204
+value: false
+---
+
+Clearly $\infty$ is not a cut point, since $X \setminus \{\infty\}$ is the
+space {S25}. So 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|P37}; thus $f$ is constant, which proves the assertion.

From 33001bb715dbc39cb75111d79c300b61d93dd630 Mon Sep 17 00:00:00 2001
From: mathmaster13 <74142852+mathmaster13@users.noreply.github.com>
Date: Fri, 27 Feb 2026 19:58:41 -0500
Subject: [PATCH 03/16] Create P000073.md

---
 spaces/S000140/properties/P000073.md | 15 +++++++++++++++
 1 file changed, 15 insertions(+)
 create mode 100644 spaces/S000140/properties/P000073.md

diff --git a/spaces/S000140/properties/P000073.md b/spaces/S000140/properties/P000073.md
new file mode 100644
index 000000000..17c259c5f
--- /dev/null
+++ b/spaces/S000140/properties/P000073.md
@@ -0,0 +1,15 @@
+---
+space: S000140
+property: P000073
+value: true
+---
+
+The closed subsets of $X$ have one of two forms. The first is $F \cup \{\infty\}$ where $F$ is a closed set in the {S25}. The second is countable closed subsets of {S25}.
+
+Suppose $F \cup \{\infty\}$ is {P39} and $F$ is a closed subset of the {S25}. Then $F$ is an open subset of $F \cup \{\infty\}$ and so is also {P39}.
+Since {S25|P73}, this means $F$ is a singleton or empty. But $F$ cannot be a singleton, since then $F \cup \{\infty\}$ is homeomorphic to {S1} and {S1|P39}.
+So $F \cup \{\infty\}$ is the singleton $\{\infty\}$.
+
+Meanwhile if $F$ is a nonempty countable closed subset of {S25} which is {P39}, then since {S25|P73}, $F$ is a singleton.
+
+Thus every nonempty irreducible closed subset of $X$ is a singleton.

From 060b3a4590c0787cf0cc35d8d349d8fb968c9fec Mon Sep 17 00:00:00 2001
From: mathmaster13 <74142852+mathmaster13@users.noreply.github.com>
Date: Fri, 27 Feb 2026 20:40:04 -0500
Subject: [PATCH 04/16] Create P000210.md

---
 spaces/S000140/properties/P000210.md | 16 ++++++++++++++++
 1 file changed, 16 insertions(+)
 create mode 100644 spaces/S000140/properties/P000210.md

diff --git a/spaces/S000140/properties/P000210.md b/spaces/S000140/properties/P000210.md
new file mode 100644
index 000000000..22d16df28
--- /dev/null
+++ b/spaces/S000140/properties/P000210.md
@@ -0,0 +1,16 @@
+---
+space: S000140
+property: P000210
+value: true
+---
+
+Fix $x \in \mathbb{R}$, and suppose there is a sequence of sets $S_n$ where for all $n$, $S_n$ converges to $x$. Then clearly for all $n$, $S_n \setminus \{\infty\}$ also converges to $x$.
+{S25|P210} and so there is a set $S \subset \left(\bigcup_n S_n\right) \setminus \{\infty\}$ such that $S$ converges to $x$ and $S \cap (S_n \setminus \{\infty\})$ is cofinite for all $n$. Then clearly $S \cap S_n$ is also cofinite for all $n$, and so the desired result is shown for $x$.
+
+Now suppose $A$ is a countably infinite set converging to $\infty$. In particular, $A$ can be enumerated as an injective sequence converging to $\infty$. Let $\{x_i\}_{i \in \mathbb{N}}$ be an injective sequence of points in $X$ converging to $\infty$. Without loss of generality we can assume $x_i$ is never $\infty$ and so $\{x_i\}_{i \in \mathbb{N}}$ is an injective sequence in $\mathbb{R}$.
+
+Suppose $\{x_i\}_{i \in \mathbb{N}}$ has a convergent subsequence $\{x_{i_j}\}_{j \in \mathbb{N}}$. Then the set $F = \{x_{i_j} : j \in \mathbb{N}\}$ is a closed countable subset of $\mathbb{R}$ and so $X \setminus F = \{\infty\} \cup (\mathbb{R} \setminus F)$ is a neighborhood of $\infty$ not containing any points in $F$. Thus $A$ does not converge to $\infty$, since there is a neighborhood of $\{\infty\}$ which misses infinitely many points in $A$.
+
+Now suppose $\{x_i\}_{i \in \mathbb{N}}$ has no convergent subsequence. Then the set $A$ has a discrete subset of {S25}; call this set $D$. $D$ is a countably infinite closed subset of {S25}, and so $X \setminus D = \{\infty\} \cup (\mathbb{R} \setminus D)$ is a neighborhood of $\{\infty\}$ which does not contain $D$ and thus misses infinitely many points in $A$.
+
+Thus there is no countably infinite set $A$ converging to $\infty$, and so the desired result is shown for $\infty$ vacuously.

From 0c3b3ee06acdae6fa1f15a153d157bc51b4e58bd Mon Sep 17 00:00:00 2001
From: mathmaster13 <74142852+mathmaster13@users.noreply.github.com>
Date: Fri, 27 Feb 2026 20:53:21 -0500
Subject: [PATCH 05/16] Create P000219.md

---
 spaces/S000140/properties/P000219.md | 7 +++++++
 1 file changed, 7 insertions(+)
 create mode 100644 spaces/S000140/properties/P000219.md

diff --git a/spaces/S000140/properties/P000219.md b/spaces/S000140/properties/P000219.md
new file mode 100644
index 000000000..44a0c0bf2
--- /dev/null
+++ b/spaces/S000140/properties/P000219.md
@@ -0,0 +1,7 @@
+---
+space: S000140
+property: P000219
+value: false
+---
+
+{S140|P65} and has a subspace homeomorphic to {S25}.

From 561b6633181e74726e127970209a704c512b8fdc Mon Sep 17 00:00:00 2001
From: mathmaster13 <74142852+mathmaster13@users.noreply.github.com>
Date: Fri, 27 Feb 2026 21:16:43 -0500
Subject: [PATCH 06/16] Update P000189.md

---
 spaces/S000140/properties/P000189.md | 4 +---
 1 file changed, 1 insertion(+), 3 deletions(-)

diff --git a/spaces/S000140/properties/P000189.md b/spaces/S000140/properties/P000189.md
index 6e35968cb..fb4cf3f7b 100644
--- a/spaces/S000140/properties/P000189.md
+++ b/spaces/S000140/properties/P000189.md
@@ -4,6 +4,4 @@ property: P000189
 value: true
 ---
 
-Given a map $f$ from $X$ to {S15}, we know {S25|P189} and so $\mathbb{R}$
-is sent to a point. Thus $f$ factors through the quotient space
-$X/\mathbb{R}$, and $X/\mathbb{R}$ is the {S10}. {S10|P189} and so $f$ is constant.
+{S25|P189} and {S25} is a dense subset of $X$.

From 8e91156b76873274b7f17954b979c9a0aaa9a8cd Mon Sep 17 00:00:00 2001
From: mathmaster13 <74142852+mathmaster13@users.noreply.github.com>
Date: Fri, 27 Feb 2026 23:51:20 -0500
Subject: [PATCH 07/16] Update P000210.md

---
 spaces/S000140/properties/P000210.md | 14 ++++----------
 1 file changed, 4 insertions(+), 10 deletions(-)

diff --git a/spaces/S000140/properties/P000210.md b/spaces/S000140/properties/P000210.md
index 22d16df28..2bd2921b0 100644
--- a/spaces/S000140/properties/P000210.md
+++ b/spaces/S000140/properties/P000210.md
@@ -2,15 +2,9 @@
 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?
 ---
 
-Fix $x \in \mathbb{R}$, and suppose there is a sequence of sets $S_n$ where for all $n$, $S_n$ converges to $x$. Then clearly for all $n$, $S_n \setminus \{\infty\}$ also converges to $x$.
-{S25|P210} and so there is a set $S \subset \left(\bigcup_n S_n\right) \setminus \{\infty\}$ such that $S$ converges to $x$ and $S \cap (S_n \setminus \{\infty\})$ is cofinite for all $n$. Then clearly $S \cap S_n$ is also cofinite for all $n$, and so the desired result is shown for $x$.
-
-Now suppose $A$ is a countably infinite set converging to $\infty$. In particular, $A$ can be enumerated as an injective sequence converging to $\infty$. Let $\{x_i\}_{i \in \mathbb{N}}$ be an injective sequence of points in $X$ converging to $\infty$. Without loss of generality we can assume $x_i$ is never $\infty$ and so $\{x_i\}_{i \in \mathbb{N}}$ is an injective sequence in $\mathbb{R}$.
-
-Suppose $\{x_i\}_{i \in \mathbb{N}}$ has a convergent subsequence $\{x_{i_j}\}_{j \in \mathbb{N}}$. Then the set $F = \{x_{i_j} : j \in \mathbb{N}\}$ is a closed countable subset of $\mathbb{R}$ and so $X \setminus F = \{\infty\} \cup (\mathbb{R} \setminus F)$ is a neighborhood of $\infty$ not containing any points in $F$. Thus $A$ does not converge to $\infty$, since there is a neighborhood of $\{\infty\}$ which misses infinitely many points in $A$.
-
-Now suppose $\{x_i\}_{i \in \mathbb{N}}$ has no convergent subsequence. Then the set $A$ has a discrete subset of {S25}; call this set $D$. $D$ is a countably infinite closed subset of {S25}, and so $X \setminus D = \{\infty\} \cup (\mathbb{R} \setminus D)$ is a neighborhood of $\{\infty\}$ which does not contain $D$ and thus misses infinitely many points in $A$.
-
-Thus there is no countably infinite set $A$ converging to $\infty$, and so the desired result is shown for $\infty$ vacuously.
+See {{mathse:5126685}}.

From 6f8e0e6a84f26532300ed5946f810a0a2ee10b9d Mon Sep 17 00:00:00 2001
From: mathmaster13 <74142852+mathmaster13@users.noreply.github.com>
Date: Fri, 27 Feb 2026 23:55:02 -0500
Subject: [PATCH 08/16] Update README.md

---
 spaces/S000140/README.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/spaces/S000140/README.md b/spaces/S000140/README.md
index 2a0b752e3..4b7617bf6 100644
--- a/spaces/S000140/README.md
+++ b/spaces/S000140/README.md
@@ -8,6 +8,6 @@ refs:
   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 $X=\mathbb R\cup \{\infty\}$, with $\mathbb R$ being open in $X$ and having the usual topology of {S25}, 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. 
 
 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}}.

From bc42514b8761331cadbb694bb5ff9c14d3544e4c Mon Sep 17 00:00:00 2001
From: mathmaster13 <74142852+mathmaster13@users.noreply.github.com>
Date: Sun, 1 Mar 2026 00:34:59 -0500
Subject: [PATCH 09/16] Update README.md

---
 spaces/S000140/README.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/spaces/S000140/README.md b/spaces/S000140/README.md
index 4b7617bf6..8b57c0f60 100644
--- a/spaces/S000140/README.md
+++ b/spaces/S000140/README.md
@@ -8,6 +8,6 @@ refs:
   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 $\mathbb R$ being open in $X$ and having the usual topology of {S25}, 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 usual 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}}.

From da59563e351ec595a3d766404760f805a0cdd934 Mon Sep 17 00:00:00 2001
From: mathmaster13 <74142852+mathmaster13@users.noreply.github.com>
Date: Thu, 5 Mar 2026 01:42:56 -0500
Subject: [PATCH 10/16] Update spaces/S000140/README.md

Co-authored-by: Patrick Rabau <70125716+prabau@users.noreply.github.com>
---
 spaces/S000140/README.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/spaces/S000140/README.md b/spaces/S000140/README.md
index 8b57c0f60..176723803 100644
--- a/spaces/S000140/README.md
+++ b/spaces/S000140/README.md
@@ -8,6 +8,6 @@ refs:
   name: What are the compactness properties of $\mathbb R$, extended by a point with co-countable open neighborhoods?
 ---
 
-Let $X=\mathbb R\cup \{\infty\}$. Let $\mathbb R$ be open in $X$ and have the usual 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. 
+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}}.

From 17c90802d7989e1aa4f96f7a5e2d77958bfaf487 Mon Sep 17 00:00:00 2001
From: mathmaster13 <74142852+mathmaster13@users.noreply.github.com>
Date: Thu, 5 Mar 2026 01:55:49 -0500
Subject: [PATCH 11/16] Update README.md

---
 spaces/S000140/README.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/spaces/S000140/README.md b/spaces/S000140/README.md
index 176723803..4f4f5eadd 100644
--- a/spaces/S000140/README.md
+++ b/spaces/S000140/README.md
@@ -1,6 +1,6 @@
 ---
 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'

From 64fc6c32587d2699d23d4e3d68c3a841813d9f45 Mon Sep 17 00:00:00 2001
From: mathmaster13 <74142852+mathmaster13@users.noreply.github.com>
Date: Thu, 5 Mar 2026 02:12:59 -0500
Subject: [PATCH 12/16] Update P000073.md

---
 spaces/S000140/properties/P000073.md | 6 +++---
 1 file changed, 3 insertions(+), 3 deletions(-)

diff --git a/spaces/S000140/properties/P000073.md b/spaces/S000140/properties/P000073.md
index 17c259c5f..95866aef1 100644
--- a/spaces/S000140/properties/P000073.md
+++ b/spaces/S000140/properties/P000073.md
@@ -4,12 +4,12 @@ property: P000073
 value: true
 ---
 
-The closed subsets of $X$ have one of two forms. The first is $F \cup \{\infty\}$ where $F$ is a closed set in the {S25}. The second is countable closed subsets of {S25}.
+The closed subsets of $X$ have one of two forms. The first is $F \cup \{\infty\}$ where $F$ is a closed set in $\mathbb R$. The second is countable closed subsets of $\mathbb R$.
 
-Suppose $F \cup \{\infty\}$ is {P39} and $F$ is a closed subset of the {S25}. Then $F$ is an open subset of $F \cup \{\infty\}$ and so is also {P39}.
+Suppose $F \cup \{\infty\}$ is {P39} and $F$ is a closed subset of $\mathbb R$. Then $F$ is an open subset of $F \cup \{\infty\}$ and so is also {P39}.
 Since {S25|P73}, this means $F$ is a singleton or empty. But $F$ cannot be a singleton, since then $F \cup \{\infty\}$ is homeomorphic to {S1} and {S1|P39}.
 So $F \cup \{\infty\}$ is the singleton $\{\infty\}$.
 
-Meanwhile if $F$ is a nonempty countable closed subset of {S25} which is {P39}, then since {S25|P73}, $F$ is a singleton.
+Meanwhile if $F$ is a nonempty countable closed subset of $\mathbb R$, and $F$ is {P39}, then since {S25|P73}, $F$ is a singleton.
 
 Thus every nonempty irreducible closed subset of $X$ is a singleton.

From cc2ac2152cf3d760f2e19d4851884535c4de745d Mon Sep 17 00:00:00 2001
From: mathmaster13 <74142852+mathmaster13@users.noreply.github.com>
Date: Thu, 5 Mar 2026 17:00:39 -0500
Subject: [PATCH 13/16] Update spaces/S000140/properties/P000204.md

Co-authored-by: Patrick Rabau <70125716+prabau@users.noreply.github.com>
---
 spaces/S000140/properties/P000204.md | 4 ++--
 1 file changed, 2 insertions(+), 2 deletions(-)

diff --git a/spaces/S000140/properties/P000204.md b/spaces/S000140/properties/P000204.md
index 6315edc12..ff3b1ceaf 100644
--- a/spaces/S000140/properties/P000204.md
+++ b/spaces/S000140/properties/P000204.md
@@ -4,8 +4,8 @@ property: P000204
 value: false
 ---
 
-Clearly $\infty$ is not a cut point, since $X \setminus \{\infty\}$ is the
-space {S25}. So let $x \in \mathbb{R}$ and consider a map $f$
+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 $X$

From d75e73506802645eb91cddebf7c5dfd44502e453 Mon Sep 17 00:00:00 2001
From: mathmaster13 <74142852+mathmaster13@users.noreply.github.com>
Date: Thu, 5 Mar 2026 17:01:28 -0500
Subject: [PATCH 14/16] Update spaces/S000140/properties/P000189.md

Co-authored-by: Patrick Rabau <70125716+prabau@users.noreply.github.com>
---
 spaces/S000140/properties/P000189.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/spaces/S000140/properties/P000189.md b/spaces/S000140/properties/P000189.md
index fb4cf3f7b..229173f74 100644
--- a/spaces/S000140/properties/P000189.md
+++ b/spaces/S000140/properties/P000189.md
@@ -4,4 +4,4 @@ property: P000189
 value: true
 ---
 
-{S25|P189} and {S25} is a dense subset of $X$.
+$\mathbb R$ is a dense subset of $X$ and {S25|P189}.

From c501f969aa178a07402463f7b7d8b9889ac129cc Mon Sep 17 00:00:00 2001
From: mathmaster13 <74142852+mathmaster13@users.noreply.github.com>
Date: Thu, 5 Mar 2026 18:38:30 -0500
Subject: [PATCH 15/16] Update P000204.md

---
 spaces/S000140/properties/P000204.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/spaces/S000140/properties/P000204.md b/spaces/S000140/properties/P000204.md
index ff3b1ceaf..3fa9b85e0 100644
--- a/spaces/S000140/properties/P000204.md
+++ b/spaces/S000140/properties/P000204.md
@@ -10,4 +10,4 @@ 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|P37}; thus $f$ is constant, which proves the assertion.
+{S11} and {S11|P36}; thus $f$ is constant, which proves the assertion.

From 3ae1d020e4a1626295372f58aa05154f6c6cd705 Mon Sep 17 00:00:00 2001
From: mathmaster13 <74142852+mathmaster13@users.noreply.github.com>
Date: Thu, 5 Mar 2026 18:49:57 -0500
Subject: [PATCH 16/16] Update P000073.md

---
 spaces/S000140/properties/P000073.md | 14 +++++---------
 1 file changed, 5 insertions(+), 9 deletions(-)

diff --git a/spaces/S000140/properties/P000073.md b/spaces/S000140/properties/P000073.md
index 95866aef1..32d0aa921 100644
--- a/spaces/S000140/properties/P000073.md
+++ b/spaces/S000140/properties/P000073.md
@@ -4,12 +4,8 @@ property: P000073
 value: true
 ---
 
-The closed subsets of $X$ have one of two forms. The first is $F \cup \{\infty\}$ where $F$ is a closed set in $\mathbb R$. The second is countable closed subsets of $\mathbb R$.
-
-Suppose $F \cup \{\infty\}$ is {P39} and $F$ is a closed subset of $\mathbb R$. Then $F$ is an open subset of $F \cup \{\infty\}$ and so is also {P39}.
-Since {S25|P73}, this means $F$ is a singleton or empty. But $F$ cannot be a singleton, since then $F \cup \{\infty\}$ is homeomorphic to {S1} and {S1|P39}.
-So $F \cup \{\infty\}$ is the singleton $\{\infty\}$.
-
-Meanwhile if $F$ is a nonempty countable closed subset of $\mathbb R$, and $F$ is {P39}, then since {S25|P73}, $F$ is a singleton.
-
-Thus every nonempty irreducible closed subset of $X$ is a singleton.
+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.