nyuuzyou/wikis huggingface dataset:
Open Continuous Image of Paracompact Space is not always Countably Metacompact

== Theorem ==
Let $T_A = \struct {X_A, \tau_A}$ be a topological space which is paracompact.
Let $T_B = \struct {X_B, \tau_B}$ be another topological space.
Let $\phi: T_A \to T_B$ be a mapping which is both continuous and open.
Then it is not necessarily even the case that $T_B$ is countably metacompact, let alone paracompact.
== Proof ==
Let $T_X = \struct {X, \tau}$ be a countable discrete space.
Let $T_Y = \struct {\set {0, 1}, \tau_0}$ be a Sierpiński space.
Let $T_A = T_X \times T_Y$ be the product space of $T_X$ and $T_Y$ whose product topology is $\tau_A$
Then $T_A$ is paracompact from Product of Countable Discrete Space with Sierpiński Space is Paracompact.
Let $T_B = \struct {S, \tau_p}$ be a countable particular point space.
Then $T_B$ is not countably metacompact from Infinite Particular Point Space is not Countably Metacompact
Let $\phi: T_A \to T_B$ be a mapping defined as:
:$\forall \tuple {x, y} \in T_A: \map \phi {x, y} = \begin {cases}
p & : y = 0 \\
x & : y = 1 \end {cases}$
Then $\phi$ is open and continuous as follows:
Let $U \in \tau_A$ such that $U \ne \O$.
By construction of $T_A$, $U$ is of the form $\tuple {U_A, U_B}$ where $U_A \in \tau$ and $U_B \in \tau_0$.
By definition of $\tau_0$, $U_B \in \set {\set 0, \set {0, 1} }$.
It follows that $U$ contains elements of the form $\tuple {x, 0}$.
Hence by definition of $\phi$ we have that:
:$\exists \tuple {x, y} \in U: \map \phi {x, y} = p$
and so:
:$p \in \phi \sqbrk U$ 
By definition of $T_B$ it follows that $\phi \sqbrk U$ is an open set of $T_B$.
Thus $\phi$ maps from open sets to open sets and so $\phi$ is an open mapping.
Let $V \in T_B$ be an open set such that $V \ne \O$.
Then by definition:
:$p \in V$
Let $F = \phi^{-1} \sqbrk V$.
As $p \in V$ we have that $F$ is of the form:
:$F = A \times B$
where:
:$A \subseteq X$
and:
:$B \in \set {\set 0, \set {0, 1} }$
Because $T_X$ is a countable discrete space, $A$ is open in $T_X$.
By definition of $T_Y$, $B$ is open in $T_Y$.
Hence by definition of product space, $A \times B = F$ is open in $T_A$.
Thus the preimage of $V$ under $\phi$ is an open set.
Hence $\phi$ is a continuous mapping.
So we have constructed a mapping $\phi$ which is both open and continuous from a paracompact space to a space which is not countably metacompact.
== Sources ==
* : Part $\text {II}$: Counterexamples: $9 \text { - } 10$. Infinite Particular Point Topology: $19$
Category:Paracompact Spaces
Category:Countably Metacompact Spaces
Category:Continuous Mappings
Category:Open Mappings
Open Continuous Injection is Embedding

== Theorem ==
Let $T_1 = \struct{S_1, \tau_1}$ and $T_2 = \struct{S_2, \tau_2}$ be topological spaces.
Let $f: S_1 \to S_2$ be an open and continuous injection.
Then $f$ is an embedding of $T_1$ into $T_2$.
== Proof ==
Let $g: S_1 \to f \sqbrk {S_1}$ be the restriction of $f$ to $S_1 \times f \sqbrk {S_1}$.
It must be shown that $g$ is a homeomorphism.
=== $g$ is a bijection ===
From Restriction of Mapping to Image is Surjection, $g$ is a surjection.
From Restriction of Injection is Injection, $g$ is a injection.
Thus, $g$ is a bijection by definition.
=== $g$ is continuous ===
Follows from Restriction of Continuous Mapping is Continuous: Topological Spaces
=== $g$ is open ===
Let $A \subseteq S_1$ be open in $T_1$.
Since $f$ is an open mapping, $f \sqbrk {A}$ is open in $T_2$.
Because $f$ is an open mapping, $f \sqbrk {S_1}$ is open in $T_2$, as $S_1$ is open in $T_1$ by Definition of a topology.
Thus Open Set in Open Subspace applies: $f \sqbrk {A}$ is open in $T_2$ it is open in the subspace topology of $f \sqbrk {S_1}$.
Therefore, $f \sqbrk {A}$ is open in the subspace topology of $f \sqbrk {S_1}$.
Since $f$ and $g$ agree on $S_1$, $f \sqbrk {A} = g \sqbrk {A}$.
Therefore, $g$ is an open mapping.
== Also See ==
* Closed Continuous Injection is Embedding
== Sources ==
* : $\S 3$: New Spaces From Old: Subspaces. Topological Embeddings
Category:Embeddings (Topology)
Open Convex Set in Hausdorff Locally Convex Space is Separated from Points outside Set by Continuous Linear Functional

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \PP}$ be a Hausdorff locally convex space over $\GF$ equipped with its standard topology.
Let $X^\ast$ be the topological dual of $X$.
Let $C$ be an open convex set such that ${\mathbf 0}_X \in C$. 
Let $x_0 \in X \setminus C$. 
Then there exists $f \in X^\ast$ such that: 
:$\map \Re {\map f x} < \map \Re {\map f {x_0} }$
for each $x \in X$. 
== Proof ==
=== Case 1: $\GF = \R$ ===
Let $\mu_C$ be the Minkowski functional of $C$. 
From Minkowski Functional of Convex Absorbing Set is Sublinear Functional, $\mu_A$ is a sublinear functional.
Let $Y = \span \set {x_0}$. 
Define $g : Y \to \R$ by: 
:$\map g {\lambda x_0} = \lambda \map {\mu_C} {x_0}$
for each $\lambda \in \R$.
For $\lambda \in \R_{> 0}$, we have: 
:$\lambda \map {\mu_C} {x_0} = \map {\mu_C} {\lambda x_0}$
by .
For $\lambda \in \R_{\le 0}$, we have: 
:$\lambda \map {\mu_C} {x_0} \le 0 \le \map {\mu_C} {\lambda x_0}$
So we have: 
:$\map g y \le \map {\mu_C} y$
for all $y \in Y$. 
From Hahn-Banach Theorem: Real Vector Space, there exists a linear functional $f : X \to \R$ extending $g$ and satisfying: 
:$\map f x \le \map {\mu_C} x$
for each $x \in X$. 
We show that $f$ is continuous.
Since $C$ is an open neighborhood of ${\mathbf 0}_X$, $-C$ is is an open neighborhood of ${\mathbf 0}_X$ by Dilation of Open Set in Topological Vector Space is Open.
Then $U = C \cap \paren {-C}$ is an open neighborhood of ${\mathbf 0}_X$.
From Characterization of Continuous Linear Functionals on Topological Vector Space, it is enough to show that $f$ is bounded on $U$. 
Let $x \in U$.
Then $x \in C$.
From Minkowski Functional of Open Convex Set containing Zero Vector in Topological Vector Space recovers Set, we have: 
:$C = \set {y \in X : \map {\mu_C} x < 1}$
Hence we have that:
:$\map f x \le \map {\mu_C} x < 1$
Since $x \in U$, we also have $x \in -C$.
So:
:$\map f x = -\map f {-x} \ge -\map {\mu_C} x > -1$ 
Hence we deduce that: 
:$\cmod {\map f x} < 1$
for all $x \in U$.
So $f$ is bounded on $U$, and we deduce that $f$ is continuous by Characterization of Continuous Linear Functionals on Topological Vector Space.
=== Case 2: $\GF = \C$ ===
Let $X_\R$ be the realification of $X$. 
By Case 1, there exists $f \in X_\R^\ast$ such that: 
:$\map f x < \map f {x_0}$ 
From Continuous Real Linear Functional on Complex Topological Vector Space is Real Part of Continuous Complex Linear Functional, there exists $g \in X^\ast$ such that: 
:$\map f x = \map \Re {\map f x}$
for each $x \in X$. 
Then: 
:$\map \Re {\map f x} < \map \Re {\map f {x_0} }$
Category:Locally Convex Spaces
Category:Topological Dual Spaces
Open Cover may not have Lebesgue Number

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $\CC$ be an open cover of $M$.
Then it may not necessarily be the case that $\CC$ has a Lebesgue number.
== Proof ==
Let $M := \openint 0 1 \subseteq \R$ under the Euclidean metric.
Let $\CC := \set {\openint {\dfrac 1 n} 1: n \ge 2}$.
For any $\epsilon \in \R_{>0}$, take $n > \dfrac 1 \epsilon$ and $x = \dfrac 1 n$.
There is no $\openint {\dfrac 1 m} 1$ such that $\map {B_\epsilon} x \subseteq \openint {\dfrac 1 m} 1$, since $\map {B_\epsilon} x = \map {B_\epsilon} {\dfrac 1 n} = \openint 0 {\dfrac 1 n + \epsilon}$.
== Sources ==
* : $7.2$: Sequential compactness
Category:Metric Spaces
Open Cover with Closed Locally Finite Refinement is Even Cover

== Theorem ==
Let $T = \struct {X, \tau}$ be a topological Space.
Let $\UU$ be an open cover of $T$ with a closed locally finite refinement.
Then:
:$\UU$ is an even cover
== Proof ==
Let $\AA$ be a closed locally finite refinement of $\UU$.
By definition of refinement:
:$\forall A \in \AA : \exists U \in \UU : A \subseteq U$
For each $A \in \AA$, let $U_A \in \UU$ such that $A \subseteq U_A$.
Let $T \times T = \struct {X \times X, \tau_{X \times X} }$ denote the product space of $T$ with itself.
For each $A \in \AA$, let:
:$V_A = \paren {U_A \times U_A} \cup \paren {\paren {X \setminus A} \times \paren {X \setminus A} }$
Let:
:$V = \ds \bigcap_{A \mathop \in \AA} V_A$
For each $x \in X$, let:
:$\map V x = \set {y \in X : \tuple {x, y} \in V}$
where:
:$V$ is seen as a relation on $X \times X$
:$\map V x$ denotes the image of $x$ under $V$.
=== Lemma 1 ===
=== Lemma 2 ===
It follows that $\UU$ is an even cover by definition.
Category:Covers
Category:Open Cover with Closed Locally Finite Refinement is Even Cover
Open Cover with Closed Locally Finite Refinement is Even Cover/Lemma 1

== Theorem ==
Let $T = \struct {X, \tau}$ be a topological Space.
Let $\UU$ be an open cover of $T$.
Let $\AA$ be a closed locally finite refinement of $\UU$.
For each $A \in \AA$, let $U_A \in \UU$ such that $A \subseteq U_A$.
For each $A \in \AA$, let:
:$V_A = \paren {U_A \times U_A} \cup \paren {\paren {X \setminus A} \times \paren {X \setminus A} }$
For each $x \in X, A \in \AA$, let:
:$\map {V_A} x = \set {y \in X : \tuple {x, y} \in V_A}$
where:
:$V_A$ is seen as a relation on $X \times X$
:$\map {V_A} x$ denotes the image of $x$ under $V_A$.
Let:
:$V = \ds \bigcap_{A \mathop \in \AA} V_A$
For each $x \in X$, let:
:$\map V x = \set {y \in X : \tuple {x, y} \in V}$
where:
:$V$ is seen as a relation on $X \times X$
:$\map V x$ denotes the image of $x$ under $V$.
Then:
:$\set {\map V x : x \in X}$ is a refinement of $\UU$.
== Proof ==
Let $x \in X$.
By definition of refinement:
:$\AA$ is a cover of $X$
By definition of cover:
:$\exists A \in \AA : x \in A$
=== Lemma 3 ===
From lemma 3
:$\map V x \subseteq \map {V_A} x = U_A \in \UU$
Since $x$ was arbitrary, then:
:$\forall x \in X : \exists U \in \UU : \map V x \subseteq U$
Also, we have:
:$\forall x \in X : x \in \map V x$
It follows that $\set {\map V x : x \in X}$ is a refinement of $\UU$ by definition.
Category:Open Cover with Closed Locally Finite Refinement is Even Cover
Open Cover with Closed Locally Finite Refinement is Even Cover/Lemma 2

== Theorem ==
Let $T = \struct {X, \tau}$ be a topological Space.
Let $\UU$ be an open cover of $T$.
Let $\AA$ be a closed locally finite refinement of $\UU$.
For each $A \in \AA$, let $U_A \in \UU$ such that $A \subseteq U_A$.
Let $T \times T = \struct {X \times X, \tau_{X \times X} }$ denote the product space of $T$ with itself.
For each $A \in \AA$, let:
:$V_A = \paren {U_A \times U_A} \cup \paren {\paren {X \setminus A} \times \paren {X \setminus A} }$
Let:
:$V = \ds \bigcap_{A \mathop \in \AA} V_A$
Let $\Delta_X$ denote the diagonal on $X$.
Then:
:$V$ is a neighborhood of the diagonal $\Delta_X$ in $T \times T$.
== Proof ==
Let $x \in X$.
By definition of locally finite:
:$\exists W \in \tau : x \in W : \set {A \in \AA : W \cap A \ne \O}$ is finite.
Let:
:$A \in \AA : W \cap A = \O$
From Subset of Set Difference iff Disjoint Set:
:$W \subseteq X \setminus A$
From Cartesian Product of Subsets:
:$W \times W \subseteq \paren {X \setminus A} \times \paren {X \setminus A} \subseteq V_A$
Since $A$ was arbitrary, we have established:
:$\forall A \in \AA : W \cap A = \O \leadsto W \times W \subseteq V_A$
From Set is Subset of Intersection of Supersets:
:$W \times W \subseteq \bigcap \set {V_A : A \in \AA : W \cap A = \O}$
From Intersection with Subset is Subset:
:$(1): \quad W \times W = \paren {W \times W} \cap \bigcap \set {V_A : A \in \AA : W \cap A = \O}$
We have:
By definition of product topology:
:$W \times W$ is open in $T \times T$
=== Lemma 4 ===
Recall:
:$\set{A \in \AA : W \cap A \ne \O}$ is finite
By :
:$\paren{W \times W} \cap V$ is open in $T \times T$
By definition of Cartesian product:
:$\tuple{x, x} \in W \times W$
By definition of diagonal $\Delta_X$:
:$\tuple{x, x} \in \Delta_X$
By definition of open neighborhood:
:$\forall A \in \AA : \tuple{x, x} \in \Delta_X \subseteq V_A$
By definition of set intersection:
:$\tuple{x, x} \in \paren{W \times W} \cap V$
Hence:
:$V$ is a neighborhood of $\tuple{x, x}$ in $T \times T$ by definition.
Since $x \in X$ was arbitrary, then:
:$\forall x \in X : V$ is a neighborhood of $\tuple {x, x}$ in $T \times T$
From Set is Neighborhood of Subset iff Neighborhood of all Points of Subset:
:$V$ is a neighborhood of the diagonal $\Delta_X$ in $T \times T$.
Category:Open Cover with Closed Locally Finite Refinement is Even Cover
Open Cover with Closed Locally Finite Refinement is Even Cover/Lemma 3

== Theorem ==
Let $T = \struct {X, \tau}$ be a topological Space.
Let $\UU$ be an open cover of $T$.
Let $\AA$ be a closed locally finite refinement of $\UU$.
For each $A \in \AA$, let:
:$U_A \in \UU$ 
such that:
:$A \subseteq U_A$
For each $A \in \AA$, let:
:$V_A = \paren {U_A \times U_A} \cup \paren {\paren {X \setminus A} \times \paren {X \setminus A} }$
For each $x \in X, A \in \AA$, let:
:$\map {V_A} x = \set {y \in X : \tuple {x, y} \in V_A}$
where:
:$V_A$ is a relation on $X \times X$
:$\map {V_A} x$ denotes the image of $x$ under $V_A$.
Then:
:$\forall A \in \AA, x \in A : \map {V_A} x = U_A$
== Proof ==
We have:
Category:Open Cover with Closed Locally Finite Refinement is Even Cover
Open Cover with Closed Locally Finite Refinement is Even Cover/Lemma 4

== Theorem ==
Let $T = \struct {X, \tau}$ be a topological Space.
Let $\UU$ be an open cover of $T$.
Let $\AA$ be a closed locally finite refinement of $\UU$.
For each $A \in \AA$, let $U_A \in \UU$ such that $A \subseteq U_A$.
For each $A \in \AA$, let:
:$V_A = \paren {U_A \times U_A} \cup \paren {\paren {X \setminus A} \times \paren {X \setminus A} }$
Let $T \times T = \struct {X \times X, \tau_{X \times X} }$ denote the product space of $T$ with itself.
Then:
:$\forall A \in \AA : V_A$ is an open neighborhood of the diagonal $\Delta_X$ in $T \times T$
== Proof ==
Let $A \in \AA$.
By definition of closed set:
:$X \setminus A$ is open in $T$
By definition of product topology:
:$U_A \times U_A, \paren {X \setminus A} \times \paren {X \setminus A}$ are open in $T \times T$
By :
:$V_A = \paren {U_A \times U_A} \cup \paren {\paren {X \setminus A} \times \paren {X \setminus A}}$ is open in $T \times T$
Let $x \in X$.
By definition of set difference:
:either $x \in A$ or $x \in X \setminus A$.
By definition of subset:
:either $x \in U_A$ or $x \in X \setminus A$.
By definition of cartesian product:
:either $\tuple {x, x} \in U_A \times U_A$ or $\tuple {x, x} \in \paren {X \setminus A} \times \paren {X \setminus A}$
By definition of set union:
:$\tuple {x, x} \in V_A$
Since $x$ was arbitrary:
:$\forall x \in X : \tuple {x, x} \in V_A$
By definition of diagonal:
:$\Delta_X \subseteq V_A$
Since $A$ was arbitrary:
:$\forall A \in \AA : V_A$ is an open neighborhood of the diagonal $\Delta_X$ in $T \times T$
Category:Open Cover with Closed Locally Finite Refinement is Even Cover
Open Domain is Connected iff it is Path-Connected

== Theorem ==
Let $D \subseteq \C$ be a open subset of the set of complex numbers.
Then $D$ is connected $D$ is path-connected.
== Proof ==
=== Necessary Condition ===
Complex Plane is Metric Space shows that $\C$ is topologically equivalent to the Euclidean space $\R^2$.
Continuous Image of Connected Space is Connected/Corollary 1 shows that connectedness is a topological property.
The result follows from Connected Open Subset of Euclidean Space is Path-Connected.
=== Sufficient Condition ===
The result follows from Path-Connected Space is Connected.
Category:Complex Analysis
Open Extension Space is Compact

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $T^*_{\bar p} = \struct {S^*_p, \tau^*_{\bar p} }$ be the open extension space of $T$.
Then $T^*_{\bar p}$ is a compact space.
== Proof ==
By definition of open extension space, the only open set of $T^*_{\bar p}$ containing $p$ is $S^*_p$.
So any open cover $\CC$ of $T^*_{\bar p}$ must have $S^*_p$ in it.
So $\set {S^*_p}$ will be a subcover of $\CC$, whatever $\CC$ may be.
And $\set {S^*_p}$ (having only one set in it) is trivially a finite cover of $T^*_{\bar p}$.
Hence the result, by definition of compact space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $16$. Open Extension Topology: $9$
Category:Compact Spaces
Category:Open Extension Topology
Open Extension Space is Connected

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $T^*_{\bar p} = \struct {S^*_p, \tau^*_{\bar p} }$ be the open extension space of $T$.
Then $T^*_{\bar p}$ is a connected space.
== Proof ==
The only open set of $T$ which contains $p$ is $S^*_p$.
Therefore it is impossible to set up a separation of $T$, as $S^*_p$ will always need to be an element of such a separation.
== Sources ==
* : Part $\text {II}$: Counterexamples: $16$. Open Extension Topology: $9$
Category:Connected Spaces
Category:Open Extension Topology
Open Extension Space is Path-Connected

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $T^*_{\bar p} = \struct {S^*_p, \tau^*_{\bar p} }$ be the open extension space of $T$.
Then $T^*_{\bar p}$ is path-connected.
== Proof ==
:Open Extension Space is Ultraconnected
:Ultraconnected Space is Path-Connected
== Sources ==
* : Part $\text {II}$: Counterexamples: $16$. Open Extension Topology: $9$
Category:Path-Connected Spaces
Category:Open Extension Topology
Open Extension Space is Ultraconnected

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $T^*_{\bar p} = \struct {S^*_p, \tau^*_{\bar p} }$ be the open extension space of $T$.
Then $T^*_{\bar p}$ is ultraconnected.
== Proof ==
Apart from $S^*_p$, every open set of $T^*_{\bar p}$ does not contain $p$, by definition of open extension space.
So, apart from $\O$, every closed set of $T$ does contain $p$, by definition of closed set.
So every pair of closed sets of $T$ has an intersection which contains at least $p$.
So there are no non-empty disjoint closed sets of $T$.
Hence the result, by definition of ultraconnected.
== Sources ==
* : Part $\text {II}$: Counterexamples: $16$. Open Extension Topology: $9$
Category:Ultraconnected Spaces
Category:Open Extension Topology
Open Extension Topology is T4

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $T^*_{\bar p} = \struct {S^*_p, \tau^*_{\bar p} }$ be the open extension space of $T$.
Then $T^*_{\bar p}$ is a $T_4$ space.
== Proof ==
We have that an Open Extension Space is Ultraconnected.
That means none of its closed sets are disjoint.
Hence, vacuously, any two of its disjoint closed subsets are separated by neighborhoods.
The result follows by definition of $T_4$ space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $16$. Open Extension Topology: $9$
Category:T4 Spaces
Category:Open Extension Topology
Open Extension Topology is Topology

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $\tau^*_{\bar p}$ be the open extension topology of $\tau$.
Then $\tau^*_{\bar p}$ is a topology on $S^*_p = S \cup \set p$.
== Proof ==
By definition:
:$\tau^*_{\bar p} = \set {U: U \in \tau} \cup \set {S^*_p}$
We have that $S^*_p \in \tau^*_{\bar p}$ by definition.
We also have that $\O \in \tau$ so $\O \in \tau^*_{\bar p}$.
Now let $U_1, U_2 \in \tau^*_{\bar p}$.
Then:
Finally consider $\UU \subseteq \tau^*_{\bar p}$.
Assuming $S^*_p \notin \UU$ we have that $\UU \subseteq \tau$.
So:
:$\ds\bigcup \UU \in \tau$
and so:
:$\ds \bigcup \UU \in \tau^*_p$
If $S^*_p \in \UU$ then:
:$\ds \bigcup \UU = S^*_p \in \tau^*_p$
So $\tau^*_p$ is a topology on $S \cup \set p$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $16$. Open Extension Topology: $8$
Category:Open Extension Topology
Open Extension Topology is not Perfectly T4

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $T^*_{\bar p} = \struct {S^*_p, \tau^*_{\bar p} }$ be the open extension space of $T$.
Then $T^*_{\bar p}$ is not a perfectly $T_4$ space.
== Proof ==
By definition:
:$\tau^*_{\bar p} = \set {U: U \in \tau} \cup \set {S^*_p}$
We have that $S$ is an open set in $T$ and so open set in $T^*_{\bar p}$.
So $\set p = S^*_p \setminus S$ is closed in $T^*_{\bar p}$.
The only open set in $T^*_{\bar p}$ which contains $p$ is $S^*_p$.
So $\set p$ can not be the intersection of open sets of $T^*_{\bar p}$, whether that intersection be countable or not.
So by definition $T^*_{\bar p}$ is not a perfectly $T_4$ space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $16$. Open Extension Topology: $9$
Category:Perfectly T4 Spaces
Category:Open Extension Topology
Open Extension Topology is not T1

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $T^*_{\bar p} = \struct {S^*_p, \tau^*_{\bar p} }$ be the open extension space of $T$.
Then $T^*_{\bar p}$ is not a $T_1$ (Fréchet) space.
== Proof ==
By definition:
:$\tau^*_{\bar p} = \set {U: U \in \tau} \cup \set {S^*_p}$
Let $x \in S^*_p, x \ne p$.
Let $U = \set x$.
Then $U \in \tau^*_p$ such that $x \in U, p \notin U$.
But the only $v \in \tau^*_p$ such that $p \in V$ is the set $S^*_p$, and we have that $x \in S^*_p$.
So there is no $V \in \tau^*_p$ such that $x \notin V, p \in V$, by definition of the open extension topology.
Hence $T^*_{\bar p}$ can not be a $T_1$ (Fréchet) space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $16$. Open Extension Topology: $9$
Category:T1 Spaces
Category:Open Extension Topology
Open Extension Topology is not T3

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $T^*_{\bar p} = \struct {S^*_p, \tau^*_{\bar p} }$ be the open extension space of $T$.
Then $T^*_{\bar p}$ is not a $T_3$ space.
== Proof ==
As $S$ is (trivially) open in $T$, it is also open in $T^*_{\bar p}$.
As $S^*_p = S \cup \set p$, it follows that $\set p$ is closed in $T^*_{\bar p}$.
But by definition, the only open set of $T^*_{\bar p}$ that contains $\set p$ is $S^*_p$ itself.
So there can be no $x \in S^*_p: x \notin \set p$ contained in an open set of $T^*_{\bar p}$ which is disjoint from $S^*_p$.
Hence the result.
== Sources ==
* : Part $\text {II}$: Counterexamples: $16$. Open Extension Topology: $9$
Category:T3 Spaces
Category:Open Extension Topology
Open Extension of Double Pointed Countable Complement Topology is T4 Space

== Theorem ==
Let $T = \struct {S, \tau_S}$ be a countable complement topology on an uncountable set $S$.
Let $D = \struct {\set {0, 1}, \tau_D}$ be the indiscrete topology on two points.
Let $T \times D$ be the double pointed topology on $T$.
Let $\paren {T \times D}^*_{\bar p}$ be the open extension topology on $S \times \set {0, 1} \cup \set p$ where $p \notin S \times \set {0, 1}$.
Then $\paren {T \times D}^*_{\bar p}$ is a $T_4$ space, and no other separation axioms are fulfilled.
That is, $\paren {T \times D}^*_{\bar p}$ is not a $T_0$ space, $T_1$ space, $T_2$ space, $T_3$ space or $T_5$ space.
== Proof ==
From Open Extension Topology is T4 we have that $\paren {T \times D}^*_{\bar p}$ is a $T_4$ space.
From Double Pointed Countable Complement Topology fulfils no Separation Axioms, we have that $T \times D$ is not a $T_0$ space or a $T_5$ space.
From Condition for Open Extension Space to be $T_0$ Space, it follows that $\paren {T \times D}^*_{\bar p}$ is not a $T_0$ space.
From Condition for Open Extension Space to be $T_5$ Space, it follows that $\paren {T \times D}^*_{\bar p}$ is not a $T_5$ space.
From Open Extension Topology is not $T_1$, we have that $\paren {T \times D}^*_{\bar p}$ is not a $T_1$ space.
From Open Extension Topology is not $T_3$ we have that $\paren {T \times D}^*_{\bar p}$ is not a $T_3$ space.
Finally, from $T_2$ Space is $T_1$ Space, as $\paren {T \times D}^*_{\bar p}$ is not a $T_1$ space it is not a $T_2$ space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $21$. Double Pointed Countable Complement Topology: $8$
Category:T4 Spaces
Category:Open Extension Topology
Category:Double Pointed Topologies
Category:Countable Complement Topology