Deleted sequence of intervals topology

spaces/S000206/README.md

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+---
+uid: S000206
+name: Deleted Sequence of Intervals Topology
+refs:
+  - mathse: 4996886
+    name: Answer to "Are submetrizable semiregular spaces regular?"
+---
+
+David Gao's example of a {P112} {P10} space that is not {P11},
+as constructed in {{mathse:4996886}}.
+
+The underlying set is $\mathbb{R}$. Let $A = \bigcup_{n=1}^\infty [\frac{2}{3^n},\frac{1}{3^{n-1}}]$. The topology is generated by open sets in {S25} and $\mathbb{R} \setminus A$. The construction is similar to
+that of {S56}.

spaces/S000206/properties/P000010.md

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+---
+space: S000206
+property: P000010
+value: true
+refs:
+  - mathse: 4996886
+    name: Answer to "Are submetrizable semiregular spaces regular?"
+---
+
+See {{mathse:4996886}}.

spaces/S000206/properties/P000011.md

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+---
+space: S000206
+property: P000011
+value: false
+refs:
+  - mathse: 4996886
+    name: Answer to "Are submetrizable semiregular spaces regular?"
+---
+
+See {{mathse:4996886}}.

spaces/S000206/properties/P000017.md

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+---
+space: S000206
+property: P000017
+value: true
+refs:
+  - mathse: 4996886
+    name: Answer to "Are submetrizable semiregular spaces regular?"
+---
+
+The topology coincide with {S25} on $\mathbb{R} \setminus \{0\}$. It is easy to see that
+$\mathbb{R} \setminus \{0\}$ is {P17} under the subspace topology inherited from {S25}
+and thus also the subspace topology inherited from {S206}. $\{0\}$ is clearly
+compact.

spaces/S000206/properties/P000022.md

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+---
+space: S000206
+property: P000022
+value: false
+refs:
+  - mathse: 4996886
+    name: Answer to "Are submetrizable semiregular spaces regular?"
+---
+
+The formal identity map from $X$ to $\mathbb{R}$ is an unbounded continuous function from
+{S206} to {S25}.

spaces/S000206/properties/P000027.md

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+---
+space: S000206
+property: P000027
+value: true
+refs:
+  - mathse: 4996886
+    name: Answer to "Are submetrizable semiregular spaces regular?"
+---
+
+The topology is generated by a countable subbasis, namely, any countable subbasis for {S25}
+and $\mathbb{R}\setminus A$ for the set $A$ in the definition of the topology. Hence, it is also generated by a
+countable basis.

spaces/S000206/properties/P000037.md

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+---
+space: S000206
+property: P000037
+value: false
+refs:
+  - mathse: 4996886
+    name: Answer to "Are submetrizable semiregular spaces regular?"
+---
+
+If $X$ is {P37}, there will be a path $f:[0,1]\to X$ from $0$ to $1$. Any such path will also be a path from
+$0$ to $1$ in {S25}, so it must contain $[0,1]$ and thus also $A$ in the definition of the space.
+Hence, $A$ is a closed subset of the {P16} space $f([0,1])$, so $A$ is compact itself. However, it is easy to check
+that, for every $n\ge 1$, $I_n=[\frac{2}{3^n},\frac{1}{3^{n-1}}]$ is open in the subspace topology on $A$, and $A$ is the
+union of all such $I_n$ but not finitely many of them, contradicting compactness.

spaces/S000206/properties/P000041.md

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+---
+space: S000206
+property: P000041
+value: false
+refs:
+  - mathse: 4996886
+    name: Answer to "Are submetrizable semiregular spaces regular?"
+---
+
+Let $A$ be as in the definition of the space. Then $X \setminus A$ is an open neighborhood of $0$. Thus, if $X$ is
+{P41}, there must be an open set $U \subset X \setminus A$ containing $0$ that is connected. Note that the
+subspace topology on $X \setminus A$ coincides with the subspace topology inherited from {S25},
+so $U$ is a connected open neighborhood of $0$ in $X \setminus A$ under the Euclidean topology. It is easy to see that no such
+$U$ exists.

spaces/S000206/properties/P000044.md

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+---
+space: S000206
+property: P000044
+value: false
+refs:
+  - mathse: 4996886
+    name: Answer to "Are submetrizable semiregular spaces regular?"
+---
+
+$(1,2)$ and $(2,3)$ are two disjoint connected subsets of $X$, each with infinitely many points.

spaces/S000206/properties/P000046.md

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+---
+space: S000206
+property: P000046
+value: false
+refs:
+  - mathse: 4996886
+    name: Answer to "Are submetrizable semiregular spaces regular?"
+---
+
+$f:[0,1]\to X$, $f(t)=t+1$ is a non-constant path.

spaces/S000206/properties/P000056.md

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+---
+space: S000206
+property: P000056
+value: false
+refs:
+  - mathse: 4996886
+    name: Answer to "Are submetrizable semiregular spaces regular?"
+---
+
+Since {S25|P56}, it suffices to show nowhere dense sets in $X$ are also nowhere dense
+in the Euclidean topology. Let $A \subset X$ be nowhere dense, and let $\overline{A}$ be the closure of $A$ in $X$. Since the
+topology on $X$ coincides with the Euclidean topology away from $0$, the closure of $A$ under the Euclidean topology is either
+$\overline{A}$ or $\overline{A} \cup \{0\}$. Since the topology on $X$ is coarser than the Eucldiean topology, the
+Euclidean-interior of $\overline{A}$ is contained in the interior of $\overline{A}$ in $X$, which is empty as $A$ is nowhere
+dense in $X$. That is, $\overline{A}$ has empty interior in the Euclidean topology, from which it is immediate that
+$\overline{A} \cup \{0\}$ also has empty interior in the Euclidean topology. Thus, $A$ is nowhere dense in the Euclidean
+topology, as required.

spaces/S000206/properties/P000112.md

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+---
+space: S000206
+property: P000112
+value: true
+refs:
+  - mathse: 4996886
+    name: Answer to "Are submetrizable semiregular spaces regular?"
+---
+
+See {{mathse:4996886}}.

spaces/S000206/properties/P000189.md

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+---
+space: S000206
+property: P000189
+value: true
+refs:
+  - mathse: 4996886
+    name: Answer to "Are submetrizable semiregular spaces regular?"
+---
+
+Let $X = \bigsqcup_{i \in I} X_i$ where $I$ is a countable index set and $X_i \subset X$ are all closed. Note that the
+subspace $(-\infty,0)$ is homeomorphic to {S25}, which is {P189}, so exactly one
+$X_i$ intersects $(-\infty,0)$ and in fact contains $(-\infty,0)$. Denote it by $X_{i_0}$. Note that the closure of
+$(-\infty,0)$ is $(-\infty,0]$, so $0 \in X_{i_0}$. Similarly, exactly one $X_i$ intersects $(0,\infty)$ and in fact contains
+$(0,\infty)$, which we shall denote by $X_{i_1}$. Again, we have $0\in X_{i_1}$, so $X_{i_0}$ intersects $X_{i_1}$ and thus
+$X_{i_0}=X_{i_1}$ covers the entire space.