avewright/proofwiki-math huggingface dataset:
Theorem: Closed Sets of Either-Or Topology

Let $T = \struct {S, \tau}$ be the either-or space.
Then the closed sets of $T$ are:
:$\O$
:$S$
:$\set {-1}$
:$\set 1$
:$\set {-1, 1}$
:Every subset $H$ of $\closedint {-1} 1$ such that $\set 0 \subseteq H$.

Proof:
From Open and Closed Sets in Topological Space we have that $\O$ and $S$ are closed sets trivially.
From the definition of closed set, we have:
:$U$ is open in $T$ {{iff}} $S \setminus U$ is closed in $T$
:$U$ is closed in $T$ {{iff}} $S \setminus U$ is open in $T$
where $S \setminus U$ denotes the relative complement of $U$ in $S$.
Now we have that:
:$\openint {-1} 1$ is open in $T$ so $S \setminus \openint {-1} 1 = \set {-1, 1}$ is closed in $T$.
:Both $\hointr {-1} 1 \supseteq \openint {-1} 1$ and $\hointr {-1} 1 \supseteq \openint {-1} 1$ therefore are open in $T$, so $\set {-1}$ and $\set 1$ are closed in $T$.
Apart from $S$ itself, there are no other subsets of $S = \closedint {-1} 1$ which have $\openint {-1} 1$ as a subset.
Finally, let $U \in \tau: \set 0 \nsubseteq U$.
Then by the definition of relative complement $\set 0 \subseteq S \setminus U$.
Now suppose $\set 0 \subseteq V$.
Then again by the definition of relative complement $\set 0 \nsubseteq S \setminus V$.
So $S \setminus V$ is open in $T$ and so $V$ is closed in $T$.
{{qed}}
Theorem: Type Space is Compact

Let $\MM$ be an $\LL$-structure.
Let $A$ be a subset of the universe of $\MM$.
The type space $\map {S_n^\MM} A$ of $n$-types over $A$ is compact.

Proof:
It will suffice to show that every open cover of $\map {S_n^\MM} A$ by the basic open sets $[\phi]$ of the topology has a finite subcover.
Let $\UU = \set { [\phi_i] : i \in I}$ be a cover of $\map {S_n^\MM} A$ by basic open sets.
This means that every complete $n$-type over $A$ contains some $\phi_i$.
We will find a finite subcover of $\UU$.
Let $\Gamma = \set {\neg \phi_i: i \in I}$.
Then $\map {\operatorname{Th}_\AA} M \cup \Gamma$ cannot be satisfied, since if $\NN \models \map {\operatorname{Th}_\AA} M \cup \map \Gamma {\bar b}$, then the type $\map {\operatorname{tp}_\NN} {\bar b / A}$ is a complete $n$-type in $\map {S_n^\MM} A$ which does not contain any $\phi_i$.
By the Compactness Theorem, $\map {\operatorname{Th}_\AA} M \cup \Gamma$ must have a finite subset $\Delta$ which is not satisfiable.
Since $\Delta$ is not satisfiable but $\map {\operatorname{Th}_\AA} M$ is, $\Delta$ must contain some of the $\neg \phi_i$ from $\Gamma$.
Furthermore, we must have that every model of $\map {\operatorname{Th}_\AA} M$ fails to satisfy at least one of these finitely many $\neg \phi_i$ in $\Delta$.
We claim that the finite set $\FF = \set {[\phi_i] : \neg \phi_i \in \Delta}$ is a finite subcover of $\UU$.
Since $\FF$ is clearly a subset of $\UU$, we need only show that every $p \in \map {S_n^\MM} A$ is contained in some $[\phi_i] \in \FF$.
That is, we must show that each $p \in \map {S_n^\MM} A$ contains one of these $\phi_i$.
Let $p \in \map {S_n^\MM} A$.
By definition, this means that $p \cup \map {\operatorname{Th}_\AA} M$ is satisfiable by some $\NN$.
But, as mentioned above, since $\NN \models \map {\operatorname{Th}_\AA} M$, we have that $\NN \not\models \neg \phi_i$ for some $\neg \phi_i \in \Delta$.
Since $p$ is complete, it must contain either $\phi_i$ or $\neg \phi_i$.
However, if $p$ contained $\neg \phi_i$, then $\NN \models \neg \phi_i$, which contradicts $\NN \not\models \neg \phi_i$.
Thus, $p$ contains $\phi_i$.
This demonstrates that $\FF$ is a finite subcover for $\UU$.
Thus, $\map {S_n^\MM} A$ is compact.
{{qed}}
Category:Model Theory for Predicate Logic
1f55mrigabfszcn1a0yddk1t148y4k3
Theorem: Either-Or Topology is T0

Let $T = \struct {S, \tau}$ be the either-or space.
Then $T$ is a $T_0$ space.

Proof:
Let $x, y \in S$ such that $x \ne y$.
{{WLOG}}, let $x \ne 0$.
Then $U = \set x$ is open in $T$ from the definition of the either-or topology.
We have that $x$ and $y$ are arbitrary.
So:
:$\forall x, y \in S: \exists U \in \tau: x \in U, y \notin U$
and the result follows by definition of $T_0$ space.
{{qed}}
Theorem: Either-Or Topology is not T1

Let $T = \struct {S, \tau}$ be the either-or space.
Then $T$ is not a $T_1$ space.

Proof:
Let $x = \dfrac 1 2$.
We have that $V = \set x$ such that $x \in V, 0 \notin V$.
However, by definition of the either-or topology, the only open sets of $T$ containing $0$ also contain $\openint {-1} 1$, and so must also contain $x$.
So we have that $\nexists U, V \in \tau: 0 \in U, x \notin U, x \in V, 0 \notin V$.
Hence the result by definition of $T_1$ space.
{{qed}}
Theorem: Omitting Types Theorem

Let $\LL$ be the language of predicate logic with a countable signature.
Let $T$ be an $\LL$-theory.
Let $\set {p_i: i \in \N}$ be a countable set of non-isolated $n$-types of $T$.
There is a countable $\LL$-structure $\MM$ such that $\MM \models T$ and $\MM$ omits each $p_i$.

Proof:
This proof is done by constructing a model using a method known as a Henkin construction.
This results in a model all of whose elements are the interpretations of constant symbols from some language.
The construction for this proof in particular is done so that the theory this model satisfies asserts that each tuple of constants fails to satisfy at least one $\phi$ from each type $p_i$.
As a result, the model we construct will not realize any of the types.
We will prove the theorem for the special case where we only omit one type.
That is, assume $\set {p_i: i \in \N}$ is $\set p$ for some $p$.
Comments on the general case will be made after the proof.
Our goal is to eventually be able to apply Maximal Finitely Satisfiable Theory with Witness Property is Satisfiable to a suitable $\LL^*$-theory $T^*$, where $\LL^*$ contains $\LL$ and $T^*$ contains $T$.
Let $\LL^*$ be $\LL \cup \set {c_i : i \in \N}$ where each $c_i$ is a new constant symbol.
Note that $\LL^*$ is still countable.
We will recursively construct $T^*$ by adding a sentence $\theta_k$ at each stage.
In order to ensure that it has the necessary properties, we will construct it in stages, each of which has three steps.
The goal of the first step of each stage is ensure that $T^*$ will be maximal upon completion of the definition.
The goal of the second step of each stage is to ensure that $T^*$ will satisfy the witness property.
The goal of the third step of each stage is to ensure that $p$ is not realized by any tuple of constants.
Since $\LL^*$ is countable, the set of $\LL$-formulas is countable, and hence in particular, the $\LL^*$-sentences can be listed as $\phi_0, \phi_1, \ldots$
Similarly, since $\LL^*$ is countable, the set of $n$-tuples of its constant symbols can be listed as $\bar d_0, \bar d_1, \dots$
Theorem: Vinogradov's Theorem

Let $\Lambda$ be the von Mangoldt function.
For $N \in \Z$, let:
:$\ds \map R N = \sum_{n_1 + n_2 + n_3 \mathop = N} \map \Lambda {n_1} \, \map \Lambda {n_2} \, \map \Lambda {n_3}$
be a weighted count of the number of representations of $N$ as a sum of three prime powers.
Let $\SS$ be the arithmetic function:
:$\ds \map \SS N = \prod_{p \mathop \nmid N} \paren {1 + \frac 1 {\paren {p - 1}^3} } \prod_{p \mathop \divides N} \paren {1 - \frac 1 {\paren {p - 1}^2} }$
where:
:$p$ ranges over the primes
:$p \nmid N$ denotes that $p$ is not a divisor of $N$
:$p \divides N$ denotes that $p$ is a divisor of $N$.
Then for any $A > 0$ and sufficiently large odd integers $N$:
:$\map R N = \dfrac 1 2 \map \SS N N^2 + \map \OO {\dfrac {N^2} {\paren {\log N}^A} }$
where $\OO$ denotes big-$\OO$ notation.

Proof:
Throughout the proof, for $\alpha \in \R$, let the following notation be understood:
:$\map e \alpha := \map \exp {2 \pi i \alpha}$
Let $B > 0$, and set $Q = \paren {\log N}^B$.
For $1 \le q \le Q, 0 \le a \le q$ such that $\gcd \set {a, q} = 1$, let:
:$\map \MM {q, a} := \set {\alpha \in \closedint 0 1: \size {\alpha - \dfrac a q} \le \dfrac  Q N}$
Let:
:$\ds \MM := \bigcup {1 \mathop \le q \mathop \le Q} \bigcup_{\substack {0 \mathop \le a \mathop \le q} {\gcd \set {a, q} \mathop = 1} } \map \MM {q, a}$
be referred to as the ''major arcs''.
Let:
:$m := \closedint 0 1 \setminus \MM$
be referred to as the ''minor arcs''.
Theorem: Vinogradov's Theorem/Minor Arcs

For any $B > 0$:
:$\ds \int_\MM \map F \alpha^3 \map e {-\alpha N} \rd \alpha \ll \frac {N^2} {\paren {\ln N}^{B/2 - 5} }$

Proof:
=== Lemma 1 ===
{{:Vinogradov's Theorem/Minor Arcs/Lemma 1}}
{{qed|lemma}}
{{finish|The below need proofs}}
Theorem: Vinogradov's Theorem/Major Arcs

Let $B \in \R_{>0}$.
Then:
:$\ds \int_\MM \map F \alpha^3 \map e {-N \alpha} \rd \alpha = \frac {N^2} 2 \map \SS N + \map \OO {\frac {N^2} {\paren {\ln N}^{B/2} } }$
where the implied constant depends only on $B$.

Proof:
=== Lemma 1 ===
{{:Vinogradov's Theorem/Major Arcs/Lemma 1}}
{{qed|lemma}}
{{finish|These need proof.}}
Theorem: Exponential Dominates Polynomial

Let $\exp$ denote the real exponential function.
Let $k \in \N$.
Let $\alpha \in \R_{>0}$.
Then: 
:$\exists N \in \N: \forall x \in \R_{>N}: \map \exp {\alpha x} > x^k$

Proof:
Choose any $N > \dfrac {\paren {k + 1}!} {\alpha^{k + 1} }$, where $!$ denotes the factorial.
By Taylor Series Expansion for Exponential Function we have for $x \in \R_{\ge 0}$:
:$\ds \map \exp {\alpha x} = \sum_{m \mathop \ge 0} \frac {\paren {\alpha x}^m}{m!} > \frac {\paren {\alpha x}^{k + 1} } {\paren {k + 1}!}$
Therefore, for $x > N$:
{{begin-eqn}}
{{eqn | l = \map \exp {\alpha x}
      | o = >
      | r = \frac {\paren {\alpha x}^{k + 1} } {\paren {k + 1}!}
}}
{{eqn | r = \frac {x \alpha^{k + 1} } {\paren {k + 1}!} x^k
      | c = factoring out an $x$
}}
{{eqn | o = >
      | r = \frac {N \alpha^{k + 1} } {\paren {k + 1}!} x^k
      | c = because $x > N$
}}
{{eqn | o = >
      | r = \frac {\paren {\dfrac {\paren {k + 1}!} {\alpha^{k + 1} } } \alpha^{k + 1} } {\paren {k + 1}!} x^k
      | c = Definition of $N$
}}
{{eqn | r = x^k
      | c = 
}}
{{end-eqn}}
{{qed}}
Category:Exponential Function
Category:Polynomial Theory
14tw3moz4gg8dnv18pz28xipegy0znt
Theorem: Power Dominates Logarithm

Let $\epsilon \in \R_{>0}$.
Let $B \in \N$ be arbitrary.
Then there exists $N \in \N$ such that:
:$\forall n > N: \paren {\ln n}^B < n^\epsilon$

Proof:
Let $\epsilon \in \R_{>0}$ be a (strictly) positive real number.
First we show that there exists $N$ such that $\ln n < n^\epsilon$ for all $n > N$,.
By Exponential is Strictly Increasing, the real exponential function is strictly increasing.
Therefore:
:$\ln n < n^\epsilon \iff n < \map \exp {n^\epsilon}$
Choose $k \in \N$ such that $k \epsilon > 2$ and $N = k!$.
By Taylor Series Expansion for Exponential Function:
:$\ds \map \exp {n^\epsilon} = \sum_{m \mathop \ge 0} \frac {n^{m \epsilon} } {m!} > \frac {n^{k \epsilon} } {k!}$
Then for all $n > N$:
{{begin-eqn}}
{{eqn | l = \map \exp {n^\epsilon}
      | r = \frac {n^{k \epsilon} } {k!}
      | o = >
      | c = 
}}
{{eqn | r = \frac N {k!} n^{k \epsilon - 1}
      | o = >
      | c =
}}
{{eqn | r = n
      | o = >
      | c = because $k \epsilon > 2$ and $N = k!$
}}
{{end-eqn}}
This completes the proof when $B = 1$.
Now let $B \in \N$ be arbitrary.
By the above we can find $N \in \N$ such that:
:$\forall n > N: \ln n < n^{\epsilon / B}$
Then:
:$\forall n > N: \paren {\ln n}^B < \paren {n^{\epsilon / B} }^B = n^\epsilon$
Hence the result.
{{qed}}
Category:Logarithms
Category:Powers
mr8r280xdm1fx5nub776kxgjqpqekmz
Theorem: Cover of Interval By Closed Intervals is not Pairwise Disjoint

Let $\closedint a b$ be a closed interval in $\R$.
{{explain|Title mentions only "interval"; this does not affect truth of statement so may "closed" above line be removed as superfluous?}}
Let $\JJ$ be a set of two or more closed intervals contained in $\closedint a b$ such that $\ds \bigcup \JJ = \closedint a b$.
Then the intervals in $\JJ$ are not pairwise disjoint.

Proof:
{{AimForCont}} that the intervals of $\JJ$ are pairwise disjoint.
Let $I = \closedint p q$ be the unique interval of $\JJ$ such that $a \in I$.
Let $J = \closedint r s$ be the unique interval of $\JJ$ containing the least real number not in $I$.
These choices are possible since $\JJ$ has at least two elements, and they are supposed disjoint.
If $q = r$ then $I \cap J \ne \O$, a contradiction.
If $r < q$ then $I \cap J \ne \O$, a contradiction.
If $r > q$ then there is some real number with $q < \alpha < r$.
Therefore $\alpha \notin I$.
Since $J$ contains the least real number not in $I$ it follows that there is $\beta \in J$ with $\beta < \alpha$.
But we also have that $\alpha < r \le \beta$ for all $\beta \in J$, a contradiction.
This exhausts all the possibilities, and we conclude that the intervals of $\JJ$ are not pairwise disjoint.
{{qed}}
Category:Measure Theory
rr3xq87leqsmvhekpt9vtghciohtyze
Theorem: Open Extension Topology is T4

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.
{{qed}}
Theorem: Excluded Point Topology is T4

Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.
Then $T$ is a $T_4$ space.

Proof:
We have that an Excluded Point Space is Ultraconnected.
That means none of its closed sets are disjount.
Hence, vacuously, any two of its disjoint closed subsets of $S$ are separated by neighborhoods.
The result follows by definition of $T_4$ space.
{{qed}}
Theorem: Excluded Point Topology is T4

Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.
Then $T$ is a $T_4$ space.

Proof:
We have:
: Excluded Point Topology is Open Extension Topology of Discrete Topology
: Open Extension Topology is $T_4$
Hence the result.
{{qed}}
Theorem: Excluded Point Topology is T4

Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.
Then $T$ is a $T_4$ space.

Proof:
We have:
:Excluded Point Topology is $T_5$
:$T_5$ Space is $T_4$
{{qed}}
Theorem: Open Extension Topology is not T3

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.
{{qed}}
Theorem: Excluded Point Topology is not T3

Let $T = \struct {S, \tau_{\bar p} }$ be a excluded point space.
Then $T$ is not a $T_3$ space.

Proof:
:Excluded Point Topology is Open Extension Topology of Discrete Topology
:Open Extension Topology is not T3
{{qed}}
Theorem: Condition for Open Extension Space to be Separable

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 separable space {{iff}} $T$ is.

Proof:
Let $T = \struct {S, \tau}$ be a separable space.
Then there exists a countable subset $H \subseteq S$ which is everywhere dense in $T$.
That is, $H^- = S$ where $H^-$ is the closure of $H$ in $S$.
So in $T_{\bar p}^*$, $S\subseteq H^-=S^-\subseteq S^*_p$.
Hence $H^- = S$ or $H^- = S \cup \set p = S^*_p$.
From Topological Closure is Closed we deduce that $H^-$ is closed in $T_{\bar p}^*$.
From the definition of open extension topology, the closed sets of $T_{\bar p}^*$ are $S^*_p \setminus U$ where $U \in \tau$.
Because $\set p \notin \tau$, $S$ is not closed.
Thus it must be that $H^- = S^*_p$.
Hence $H$ is a countable subset $H \subseteq S^*_p$ which is everywhere dense in $T^*_{\bar p}$.
{{qed|lemma}}
{{AimForCont}} $T^*_{\bar p} = \struct {S^*_p, \tau}$ is separable, but that $T = \struct {S, \tau}$ is not separable.
Let $H \subseteq S^*_p$ be a countable subset which is dense in $T^*_{\bar p}$.
If $H \subseteq S$, then $H^- = S$ in $T$, but this cannot be since $T$ is not separable.
Then $p \in H$.
Consider the subset $V = H \setminus \set p$.
From Topological Closure is Closed we have that $V^-$ is closed in $T_{\bar p}^*$.
From the definition of open extension space, all non-empty closed sets contain the point $p$.
So $V \cup \set p = H \subseteq V^-$. 
From Set Closure is Smallest Closed Set in Topological Space, $S^*_p = H^- \subseteq V^- \implies V$ is dense in $T_{\bar p}^*$.
But because $V \subseteq S$, $V$ is dense in $S$ also.
From Subset of Countably Infinite Set is Countable, $V$ is countable as it is a subset of a countable set.
Thus $T$ is separable, which is a contradiction.
Finally, our assumption is false: $T$ is indeed separable.
{{qed}}
Theorem: Finitely Satisfiable Theory has Maximal Finitely Satisfiable Extension

Let $T$ be a finitely satisfiable $\LL$-theory.
Then there exists a finitely satisfiable $\LL$-theory $T'$ which contains $T$ as a subset such that:
:for all $\LL$-sentences $\phi$, either $\phi \in T'$ or $\neg \phi \in T'$.

Proof:
The set of all finitely satisfiable $\LL$-theories containing $T$ forms an ordered set using subset inclusion as the ordering.
{{explain|Link to "subset is ordering"}}
Let $C$ be a non-empty chain in this ordered set.
Let $\ds T_C = \bigcup_{\Sigma \mathop \in C} \Sigma$.
Let $\Delta$ be a finite subset of $T_C$.
Then there exists a single $\Sigma$ in $C$ which contains $\Delta$.
Since this $\Sigma$ is finitely satisfiable by definition, this means that $\Delta$ is satisfiable.
Hence $T_C$ is finitely satisfiable.
Since each $\Sigma \in C$ is contained in $T_C$, this means that $T_C$ is an upper bound for $C$ in the ordered set.
Thus, by Zorn's Lemma, there is a finitely satisfiable $\LL$-theory $T'$ containing $T$ such that $T'$ contains all other such theories.
Let $\phi$ be an $\LL$-sentence.
Let $\phi \nsubseteq T'$.
{{AimForCont}} $T' \cup \set \phi$ were finitely satisfiable.
Then by definition of $T'$, $T'$ would contain $T' \cup \set \phi$ as a subset.
This would mean that $T'$ contains $\phi$, which contradicts the assumption.
Thus $T' \cup \set \phi$ is not finitely satisfiable.
By the lemma, $T' \cup \set {\neg \phi}$ is finitely satisfiable.
Thus, by definition of $T'$, $T'$ contains $T' \cup \set {\neg \phi}$ as a subset.
Hence $T'$ contains $\neg \phi$.
{{qed}}
Theorem: Finitely Satisfiable Theory has Maximal Finitely Satisfiable Extension

Let $T$ be a finitely satisfiable $\LL$-theory.
Then there exists a finitely satisfiable $\LL$-theory $T'$ which contains $T$ as a subset such that:
:for all $\LL$-sentences $\phi$, either $\phi \in T'$ or $\neg \phi \in T'$.

Proof:
Let $\AA$ be the set of finitely satisfiable extensions of $T$.
By the lemma, for each element $S$ of $\AA$ and each $\LL$-sentence $\phi$, either $S \cup \set \phi \in \AA$ or $S \cup \set {\neg \phi} \in \AA$.
$\AA$ has finite character, by the following argument:
Let $S \in \AA$.
Let $F$ be a finite subset of $S$.
Then $S$ is satisfiable and hence finitely satisfiable.
Thus in $\AA$.
Let $S$ be a theory on $\LL$.
Let every finite subset of $S$ be finitely satisfiable.
Then every finite subset of $S$ is satisfiable.
Therefore $S$ is finitely satisfiable.
Thus $\AA$ has finite character.
By the Restricted Tukey's Theorem, $\AA$ has an element $T'$ such that:
:for each $\LL$-sentence $\phi$, either $\phi \in T'$ or $\neg \phi \in T'$.
{{qed}}
Theorem: Condition for Open Extension Space to be First-Countable

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 first-countable space {{iff}} $T$ is.

Proof:
Let $T = \struct {S, \tau}$ be a first-countable space.
Then every point in $S$ has a countable local basis.
Every open set of $T$ is an open set of $T^*_{\bar p}$ by definition.
So if $x$ has a countable local basis in $T$, then it has one in $T^*_{\bar p}$ as well.
Finally we note that $p$ is by definition in exactly one open set of $T^*_{\bar p}$, that is, $S^*_p$, and thus (trivially) has a countable local basis in $T^*_{\bar p}$.
Then every point in $S^*_p$ has a countable local basis in $T^*_{\bar p}$.
So if $T$ is a first-countable space, then $T^*_{\bar p}$ is also a first-countable space.
Now suppose $T = \struct {S, \tau}$ is not a first-countable space.
Then $\exists x \in S$ such that $x$ has no countable local basis in $T$.
As every open set of $T$ is an open set of $T^*_{\bar p}$ by definition, it follows that $x$ has no countable local basis in $T^*_{\bar p}$ either.
So if $T $ is not a first-countable space, then neither is $T^*_{\bar p}$ a first-countable space.
{{qed}}
Theorem: Condition for Open Extension Space to be Second-Countable

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 second-countable space {{iff}} $T$ is.

Proof:
Let $T = \struct {S, \tau}$ be a second-countable space.
Then $\tau$ has a countable basis.
Every open set of $T$ is an open set of $T^*_{\bar p}$ by definition.
So if $\tau$ has a countable basis in $T$, then $\tau^*_{\bar p} = \tau \cup \set {S^*_p}$ has one as well.
So if $T$ is a second-countable space, then $T^*_{\bar p}$ is also a second-countable space.
Now suppose $T = \struct {S, \tau}$ is not a second-countable space.
Then $\tau$ has no countable basis in $T$.
As every open set of $T$ is an open set of $T^*_{\bar p}$ by definition, it follows that $\tau^*_{\bar p} = \tau \cup \set {S^*_p}$ has no countable basis either.
So if $T$ is not a second-countable space, then neither is $T^*_{\bar p}$ a second-countable space.
{{qed}}
Theorem: Either-Or Topology is not T3

Let $T = \struct {S, \tau}$ be the either-or space.
Then $T$ is not a $T_3$ space.

Proof:
Consider the set $\set 0$ which is closed from Closed Sets of Either-Or Topology.
Let $x = \dfrac 1 2$.
Let $U \in \tau$ such that $\set 0 \subseteq U$.
Then by definition $\openint {-1} 1 \subseteq U$ and so $x \in U$.
So we have found a closed set of $T$ and a point in $S$ which do not satisfy the conditions for $T$ to be a $T_3$ space.
Hence the result.
{{qed}}
Theorem: Limit Points of Either-Or Topology

Let $T = \struct {S, \tau}$ be the either-or space.
Let $H \subseteq S$ be any subset of $S$.
Then no element of $S$ can be a limit point of $H$ except $0$.

Proof:
Let $x \in S$ such that $x \ne 0$.
Then, as $0 \notin \set x$, we have by definition of the either-or topology that $x$ is open in $T$.
So whatever $H$ is, $\set x$ never contains any points of $H$ which are different from $x$.
So $x$ can not be a limit point of $H$.
However, every open set of $T$ which contains $0$ also contains the interval $\openint {-1} 1$ and so can contain points of $H$.
It follows that the only non-empty sets of $T$ for which $0$ is ''not'' a limit point are $\set {-1}$, $\set 0$ and $\set 1$ and their unions.
{{qed}}
Theorem: Either-Or Topology is T5

Let $T = \struct {S, \tau}$ be the either-or space.
Then $T$ is a $T_5$ space.

Proof:
Let $A, B \subseteq S$ such that $A \cap B \ne \O$.
Suppose neither $A$ nor $B$ contain $0$.
Then $A$ and $B$ are both open in $T$.
From Limit Points of Either-Or Topology, the only limit point that either $A$ or $B$ may have is $0$.
So either $A^- = A \cup \set 0$ or $A^- = A$, where $A^-$ is the closure of $A$
Similarly for $B$.
So $A^- \cap B = \O$, and so are separated, and such that:
:$\exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$
where $U = A$ and $V = B$, fulfilling the condition for a $T_5$ space.
Suppose one of $A$ and $B$ contains $0$.
{{WLOG}}, suppose $0 \in A$.
Then for $A$ and $B$ to be separated, $0 \notin B^-$ and so $B^-$ has to be $\set {-1}$ or $\set 1$ or their union.
In any of these cases we have that $B$ and $S \setminus B$ are disjoint open sets containing $B$ and $A$ respectively.
So in all cases we see that if $A$ and $B$ are separated, then:
:$\exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$
where $U = A$ and $V = B$.
and so by definition $T$ is a $T_5$ space.
{{qed}}