avewright/proofwiki-math huggingface dataset:
Theorem: Compact Sets in Countable Complement Space

Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then the compact sets of $T$ are exactly the finite subsets of $S$.

Proof:
From Finite Topological Space is Compact, any finite subset of $S$ is compact.
{{AimForCont}} $H \subseteq S$ is an infinite compact set.
Take a (countably) infinite sequence $\sequence {a_n}_{n \mathop \ge 0}$ of distinct elements of $H$.
Consider the open sets:
:$V_m := S \setminus \set {a_{m + n} }_{n \mathop \ge 0}$
for $m \ge 0$, which satisfy $V_{m_1} \subset V_{m_2}$ if $m_1 < m_2$.
Then:
:$\ds H \subseteq \bigcup_{m \mathop \ge 0} V_m$
is an open cover of $H$.
Since $H$ is compact, it has a finite open subcover, say:
:$\ds H \subseteq \bigcup_{i \mathop = 0}^N V_{m_i}$
with $m_0 < m_1 < \cdots < m_N$.
But then:
:$\ds \bigcup_{i \mathop = 0}^N V_{m_i} = V_{m_N}$ 
and so:
:$\ds a_{m_N + 1}, a_{m_N + 2}, \ldots \notin V_{m_N} = \bigcup_{i \mathop = 0}^N V_{m_i}$
which implies $a_{m_N + 1}, a_{m_N + 2}, \ldots \notin H$, a contradiction.
Hence the result by Proof by Contradiction.
{{qed}}
Theorem: Countable Complement Space is not Sigma-Compact

Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ is not a $\sigma$-compact space.

Proof:
From Compact Sets in Countable Complement Space, the only compact sets in $T$ are finite.
A countable union of finite sets can not be an uncountable set.
Hence the result by definition of $\sigma$-compact space.
{{qed}}
Theorem: Countable Complement Space is not Countably Compact

Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ is not a countably compact space.

Proof:
Consider $U \in \tau$.
By definition of countable complement topology, $\relcomp S U$ is countably infinite.
Then for each $x \in \relcomp S U$, $\relcomp S U \setminus \set x$ is countably infinite.
Then for each $x \in \relcomp S U$, $\relcomp S {\relcomp S U \setminus \set x} = U \cup \set x$ is open.
Hence $\set {U \cup \set x: x \in \relcomp S U}$ is an open cover.
This cover is countable because it is equivalent to $\relcomp S U$.
{{AimForCont}} $T$ is a countably compact space.
Then by definition there exists a finite subcover $\set {U \cup \set {x_1}, \ldots, U \cup \set {x_n} }$ of $T$.
Then:
:$\paren {U \cup \set {x_1} } \cup \dotsb \cup \paren {U \cup \set {x_n} } = U \cup \set {x_1, \ldots, x_n}$
Since $\relcomp S U$ is not finite, it follows that $U \cup \set {x_1, \ldots, x_n} \ne S$.
Thus there is no finite subcover of $T$.
It follows by Proof by Contradiction that $T$ cannot be a countably compact space.
{{qed}}
Theorem: Countable Complement Space is Lindelöf

Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ is a Lindelöf space.

Proof:
By definition, $T$ is a Lindelöf space {{iff}} every open cover of $S$ has a countable subcover.
Let $\CC$ be an open cover of $T$.
Let $U \in \CC$ be an arbitrary set in $\CC$.
$U$ covers all but a countable number of points of $T$.
So for each of those points we pick an element of $\CC$ which covers each of those points.
Hence we have a countable subcover of $T$.
So by definition $T$ is a Lindelöf space.
{{qed}}
Theorem: Countable Complement Space is not First-Countable

Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ is not a first-countable space.

Proof:
{{AimForCont}} some $x \in S$ has a countable local basis.
That means:
:there exists a countable set of sets $\BB_x \subseteq \tau$ 
such that:
:$\forall B \in \BB_x: x \in B$
and such that:
:every open neighborhood of $x$ contains some $B \in \BB_x$.
So:
{{begin-eqn}}
{{eqn | l = \bigcap \BB_x
      | r = \set x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = S \setminus \set x
      | r = S \setminus \bigcap \BB_x
      | c = 
}}
{{eqn | r = \bigcup_{B \mathop \in \BB_x} \paren {S \setminus B}
      | c = De Morgan's Laws: Difference with Intersection
}}
{{end-eqn}}
By definition, each of $S \setminus B$ is countable.
From Countable Union of Countable Sets is Countable it follows that $\ds \bigcup_{B \mathop \in \BB_x} \paren {S \setminus B}$ is also countable.
So $S \setminus \set x$ and therefore $S$ is also countable.
From this contradiction (as we have specified that $S$ is uncountable) it follows that our assumption that $x \in S$ has a countable local basis must be false.
Hence by definition $T$ can not be first-countable.
{{qed}}
Theorem: Closed Form for Number of Derangements on Finite Set

The number of derangements $D_n$ on a finite set $S$ of cardinality $n$ is:
{{begin-eqn}}
{{eqn | l = D_n
      | r = n! \paren {1 - \dfrac 1 {1!} + \dfrac 1 {2!} - \dfrac 1 {3!} + \cdots + \dfrac {\paren {-1}^n} {n!} }
      | c = 
}}
{{eqn | r = n! \sum_{k \mathop = 0}^n \dfrac {\paren {-1}^k} {k!}
      | c = 
}}
{{eqn | r = !n
      | c = where $!n$ denotes the subfactorial of $n$
}}
{{end-eqn}}

Proof:
Let $s_i$ be the $i$th element of set $S$.
Begin by defining set $A_m$, which is all of the permutations of $S$ which fixes $S_m$.
Then the number of permutations, $W$, with ''at least'' one element fixed, $m$, is:
:$\ds W = \size {\bigcup_{m \mathop = 1}^n A_m}$
Applying the Inclusion-Exclusion Principle:
{{begin-eqn}}
{{eqn | l = W
      | r = \sum_{m_1 \mathop = 1}^n \size {A_{m_1} }
      | c = 
}}
{{eqn | o = 
      | ro= -
      | r = \sum_{m_1, m_2 : 1 \mathop \le m_1 \mathop < m_2 \mathop \le n} \size {A_{m_1} \cap A_{m_2} }
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = \sum_{m_1, m_2, m_3 : 1 \mathop \le m_1 \mathop < m_2 \mathop < m_3 \mathop \le n} \size {A_{m_1} \cap A_{m_2} \cap A_{m_3} }
      | c = 
}}
{{eqn | o = 
      | ro= -
      | r = \cdots
      | c = 
}}
{{end-eqn}}
Each value $A_{m_1} \cap \cdots \cap A_{m_p}$ represents the set of permutations which fix $p$ values $m_1, \ldots, m_p$.
Note that the number of permutations which fix $p$ values only depends on $p$, not on the particular values of $m$.
Thus from Cardinality of Set of Subsets there are $\dbinom n p$ terms in each summation. 
So:
{{begin-eqn}}
{{eqn | l = W
      | r = \binom n 1 \size {A_1}
      | c = 
}}
{{eqn | o = 
      | ro= -
      | r = \binom n 2 \size {A_1 \cap A_2}
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = \binom n 3 \size {A_1 \cap A_2 \cap A_3}
      | c = 
}}
{{eqn | o = 
      | ro= -
      | r = \cdots
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = \paren {-1}^{p - 1} \binom n p \size {A_1 \cap \cdots \cap A_p}
      | c = 
}}
{{eqn | o = 
      | r = \cdots
      | c = 
}}
{{end-eqn}}
$\size {A_1 \cap \cdots \cap A_p}$ is the number of permutations fixing $p$ elements in position.
This is equal to the number of permutations which rearrange the remaining $n - p$ elements, which is $\paren {n - p}!$.
Thus we finally get:  
:$W = \dbinom n 1 \paren {n - 1}! - \dbinom n 2 \paren {n - 2}! + \dbinom n 3 \paren {n - 3}! - \cdots + \paren {-1}^{p - 1} \dbinom n p \paren {n - p}! \cdots$
That is:
:$\ds W = \sum_{p \mathop = 1}^n \paren {-1}^{p - 1} \binom n p \paren {n - p}!$
Noting that $\dbinom n p = \dfrac {n!} {p! \paren {n - p}!}$, this reduces to:
:$\ds W = \sum_{p \mathop = 1}^n \paren {-1}^{p - 1} \dfrac {n!} {p!}$ 
{{qed}}
Theorem: Closed Form for Number of Derangements on Finite Set

The number of derangements $D_n$ on a finite set $S$ of cardinality $n$ is:
{{begin-eqn}}
{{eqn | l = D_n
      | r = n! \paren {1 - \dfrac 1 {1!} + \dfrac 1 {2!} - \dfrac 1 {3!} + \cdots + \dfrac {\paren {-1}^n} {n!} }
      | c = 
}}
{{eqn | r = n! \sum_{k \mathop = 0}^n \dfrac {\paren {-1}^k} {k!}
      | c = 
}}
{{eqn | r = !n
      | c = where $!n$ denotes the subfactorial of $n$
}}
{{end-eqn}}

Proof:
A derangement is a permutation that fixes no elements.
So, to compute the total number of derangements, we compute the number of permutations that fix elements, that is leave elements in the same place.
Let $\pi$ be a permutation of $S$.
Let $x, y \in S$.
Let $i$ be a position in a given permutation of set $S$, where $1 \le i \le n$.
Let $\map {P_i} \pi$ be the property that $\map \pi x = y$ at position $i$.
Let $\map N {P_i, \dots, P_j}$ be the number of permutations where $\map {P_i} \pi, \dots, \map {P_j} \pi$ hold.
If we consider all the permutations of a set where the property $P_i$ holds, we have fixed a single element of the set.
So there are $\paren {n - 1}!$ such possible permutations.
Furthermore, we could have chosen $i$ as any element in the set so there are $\dbinom n 1$ terms to choose from.
So $\map N {P_i} = \paren {n - 1}!$ and there are $\dbinom n 1$ terms.
Similarly, $\map N {P_i, P_j} = \paren {n - 2}!$ and there are $\dbinom n 2$ terms.
So now we can compute the total number of derangements:
{{begin-eqn}}
{{eqn | l = D_n
      | r = n! - \dbinom n 1 \paren {n - 1}! + \dbinom n 2 \paren {n - 2}! - \ldots + \paren {-1}^n \dbinom n n \paren {n - n}!
      | c = 
}}
{{eqn | r = n! - \frac {n!} {1! \paren {n - 1}!} \paren {n - 1}! + \frac {n!} {2! \paren {n - 2}!} \paren {n - 2}! - \cdots + \paren {-1}^n \frac {n!} {n! \paren {n - n}!} \paren {n - n}!
      | c = {{Defof|Binomial Coefficient}}
}}
{{eqn | r = n! \paren {1 - \frac 1 {1!} + \frac 1 {2!} - \frac 1 {3!} + \cdots + \paren {-1}^n \frac 1 {n!} }
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Limit Points of Countable Complement Space

Let $T = \struct {S, \tau}$ be a countable complement space.
Let $H \subseteq S$ be an uncountable subset of $S$.
Then every point of $S$ is a limit point of $H$.
Let $H \subseteq S$ be an countable or finite subset of $S$.
Then $H$ contains all its limit points.

Proof:
Let $U \in \tau$ be any open set of $T$.
Let $H \subseteq S$ be an uncountable.
From Uncountable Subset of Countable Complement Space Intersects Open Sets, we have that $U \cap H$ is uncountable {{iff}} $H$ is uncountable.
Let $x \in S$.
Then every open set $U$ in $T$ such that $x \in U$ also contains an uncountable number of points of $H$ other than $x$.
Thus, by definition, $x$ is a limit point of $H$.
Now let $H$ be a countable subset of $S$.
By definition $\relcomp S H$ is open in $T$ and so $H$ is closed.
So by Closed Set Equals its Closure we have by definition that if $x$ is a limit point of $H$, then $x \in H$.
{{qed}}
Theorem: Countable Complement Space is not Separable

Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ is not a separable space.

Proof:
Let $U$ be a countable subset of $S$.
By the definition of $T$, $U$ is closed.
As $U$ is countable but $S$ is uncountable:
:$U \subsetneq S$
and so:
:$U \ne S$
From Closed Set Equals its Closure:
:$U^- = U$
where $U^-$ is the closure of $U$.
Thus:
:$U^- \ne S$
So by definition, $U$ is not everywhere dense in $T$.
As $U$ is arbitrary, there is no countable set which is everywhere dense in $T$.
The result follows by definition of separable space.
{{qed}}
Theorem: Countable Complement Space Satisfies Countable Chain Condition

Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ satisfies the countable chain condition.

Proof:
We have that a Countable Complement Space is Irreducible.
Therefore there is no disjoint set of open sets of $T$.
So the result holds vacuously.
{{qed}}
Theorem: Countable Complement Space is Connected

Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ is a connected space.

Proof:
We have that a Countable Complement Space is Irreducible.
The result follows from Irreducible Space is Connected.
{{qed}}
Theorem: Countable Complement Space is Locally Connected

Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ is a locally connected space.

Proof:
We have that a Countable Complement Space is Irreducible.
The result follows from Irreducible Space is Locally Connected.
{{qed}}
Theorem: Countable Complement Space is Pseudocompact

Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ is pseudocompact.

Proof:
We have that a Countable Complement Space is Irreducible.
The result follows from Irreducible Space is Pseudocompact.
{{qed}}
Theorem: Countable Complement Space is not Countably Metacompact

Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ is not countably metacompact.

Proof:
From Uncountable Subset of Countable Complement Space Intersects Open Sets, the intersection of any open sets is uncountable.
So for any open cover $\CC$ of $T$, every point is in an infinite number of open sets of $T$.
So no refinement of any open cover of $T$ can be point finite.
Hence the result by definition of countably metacompact.
{{qed}}
Theorem: Countable Complement Space is not Weakly Countably Compact

Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ is not weakly countably compact.

Proof:
By definition, $T$ is weakly countably compact {{iff}} every infinite subset of $S$ has a limit point in $S$.
Let $H \subseteq S$ be a countable set.
From Limit Points of Countable Complement Space, it contains all its limit points.
Let $x \in H$.
$\relcomp S {\relcomp S H \cup \set x} = H \setminus \set x$ is countable.
Then $\relcomp S H \cup \set x$ is an open neighborhood of $x$. 
However $H \cap \paren {\paren {\relcomp S H \cup \set x} \setminus \set x} = H \cap \relcomp S H = \O$.
So $x$ is not a limit point of $H$.
Since $x$ is arbitrary, there are no limit points in $H$. 
This shows that $T$ is not weakly countably compact.
{{qed}}
Theorem: Saturated Implies Universal

Let $\kappa$ be an infinite cardinal.
Let $\MM$ be a model of the $\LL$-theory $T$.
If $\MM$ is $\kappa$-saturated, then it is $\kappa^+$-universal, where $\kappa^+$ is the successor cardinal of $\kappa$.

Proof:
The idea of the proof is that $\MM$ being saturated means that when we want to define an elementary map $\NN \to \MM$, we can find an image $y \in \MM$ for an element $x \in \NN$ by realizing the type made up of the formulas that such a $y$ would need to satisfy.
Let $\NN$ be a model of $T$ with universe of cardinality strictly less than $\kappa$.
We will construct an elementary embedding of $\NN$ into $\MM$ by transfinite recursion.
Since $\card \NN < \kappa$, we can write its elements as $n_\alpha$ for ordinals $\alpha < \kappa$.
For each ordinal $\alpha < \kappa$, let $A_\alpha$ be the subset $\set {n_\beta: \beta < \alpha}$ of the universe of $\NN$.
Note for clarity that $n_\alpha \in A_{\alpha + 1}$ but $n_\alpha \notin A_\alpha$.
*Base case $\alpha = 0$:
Define $f_0 = \O$.
Note that $f_0$ is trivially an elementary embedding from $A_0 = \O$ into $\MM$.
*Limit ordinals $\alpha$, assuming $f_\beta$ is defined and elementary $A_\beta \to \MM$ for all $\beta < \alpha$:
Let $\ds f_\alpha = \bigcup_{\beta \mathop < \alpha} f_\beta$.
If $\phi$ is an $\LL$-sentence with parameters from $A_\alpha$, then since it involves only finitely many such parameters, they must all be contained in some $A_\beta$ for $\beta < \alpha$.  But $f_\alpha \restriction A_\beta = f_\beta$ is elementary, so $f_\alpha$ must be as well.
* Successor ordinals $\alpha = \beta + 1$, assuming $f_\beta$ is defined and elementary $A_\beta \to \MM$:
We need to extend $f_\beta$ to $A_\alpha = A_\beta \cup \set {n_\beta}$ so that truth of $\LL$-sentences with parameters from $A_\alpha$ is preserved.
Consider the subset $p = \set {\map \phi {v, \map {f_\beta} {\bar{a} } }: \bar{a} \text{ is a tuple from } A_\beta \text{ and } \NN \models \map \phi {n_\beta, \bar{a} } }$ of the set of $\LL$-formulas with one free variable and parameters from the image $\map {f_\beta} {A_\beta}$ of $A_\beta$ under $f_\beta$.
The set $p$ is a $1$-type over the image $\map {f_\beta} {A_\beta}$ in $\MM$.
Since $\card {A_\beta} < \kappa$ and by assumption $\MM$ is $\kappa$-saturated, this means that $p$ is realized in $\MM$ by some element $b$.
Thus, defining $f_\alpha$ to be $f_\beta \cup \set {(n_\beta, b)}$ makes it an elementary embedding $A_\alpha \to \MM$.
Now, define $\ds f = \bigcup_{\alpha \mathop < \kappa} f_\alpha$.
Then $f$ is an elementary embedding from $\NN$ to $\MM$ since $\ds \bigcup_{\alpha \mathop < \kappa} A_\alpha = \NN$, any finite set of parameters from $\NN$ must belong to one $A_\alpha$, and $f \restriction A_\alpha = f_\alpha$ is elementary.
{{qed}}
{{MissingLinks}}
Category:Model Theory for Predicate Logic
kib4rjmqd5qh81ibjue8ymfo5g54duc
Theorem: Double Pointed Countable Complement Topology fulfils no Separation Axioms

Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Let $T \times D$ be the double pointed topology on $T$.
Then $T \times D$ is not a $T_0$ space, $T_1$ space, $T_2$ space, $T_3$ space, $T_4$ space or $T_5$ space.

Proof:
From Separation Axioms on Double Pointed Topology, we have that:
{{begin-itemize}}
{{item||$T \times D$ is not a $T_0$ space}}
{{item||$T \times D$ is not a $T_1$ space}}
{{item||$T \times D$ is not a $T_2$ space.}}
{{end-itemize}}
Also from Separation Axioms on Double Pointed Topology, we have that:
{{begin-itemize}}
{{item||$T \times D$ is a $T_3$ space {{iff}} $T$ is a $T_3$ space}}
{{item||$T \times D$ is a $T_4$ space {{iff}} $T$ is a $T_4$ space}}
{{item||$T \times D$ is a $T_5$ space {{iff}} $T$ is a $T_5$ space.}}
{{end-itemize}}
From Countable Complement Space is not $T_3$:
:$T$ is not a $T_3$ space
From Countable Complement Space is not $T_4$:
:$T$ is not a $T_4$ space
From Countable Complement Space is not $T_5$:
:$T$ is not a $T_5$ space
Hence the result.
{{qed}}
Theorem: Double Pointed Countable Complement Topology is Weakly Countably Compact

Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Let $T \times D$ be the double pointed topology on $T$.
Then $T \times D$ is weakly countably compact.

Proof:
By definition, $T$ is weakly countably compact {{iff}} every infinite subset of $S$ has a limit point in $S$.
Let $D = \set {0, 1}$.
Let $\tuple {p, 0}$ belong to some infinite $A \subseteq S$.
Then its twin $\tuple {p, 1}$ is a limit point of $A$.
Hence the result by definition of weakly countably compact.
{{qed}}
Theorem: Open Extension of Double Pointed Countable Complement Topology is T4

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 $T_0$, it follows that $\paren {T \times D}^*_{\bar p}$ is not a $T_0$ space.
From Condition for Open Extension Space to be $T_5$, 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.
{{qed}}
Theorem: Saturated Models of same Cardinality are Isomorphic

Let $T$ be an $\LL$-theory.
Let $\kappa$ be an infinite cardinal.
If $\MM$ and $\NN$ are saturated models of $T$ and the cardinality of the universes of $\MM$ and $\NN$ are both $\kappa$, then $\MM$ and $\NN$ are isomorphic.

Proof:
=== Outline ===
The idea of the proof is that since the models are saturated, we can define an isomorphism $f: \NN \to \MM$ by picking the image $\map f x$ of an element $x \in \NN$ as something which realizes the type that $\map f x$ would need to satisfy in $\MM$.
This is done using a (transfinite) recursive construction which extends an elementary map at each step.
The proof is very similar to that of Saturated Implies Universal; in both proofs we attempt to construct a formula-preserving map.
In this case, however, the map must be bijective between the universes of the models.
Theorem: Compact Complement Topology is Topology

Let $T = \struct {\R, \tau}$ be the compact complement space.
Then $\tau$ is a topology on $T$.

Proof:
By definition, we have that $\O \in \tau$.
We also have that $\R \in \tau$ as $\relcomp \R \R = \O$ which is trivially compact.
Now suppose $A, B \in \tau$.
Let $H = A \cap B$.
Then:
{{begin-eqn}}
{{eqn | l = H
      | r = A \cap B
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \relcomp \R H
      | r = \relcomp \R {A \cap B}
      | c = 
}}
{{eqn | r = \relcomp \R A \cup \relcomp \R B
      | c = De Morgan's Laws: Complement of Intersection
}}
{{end-eqn}}
But as $A, B \in \tau$ it follows that $\relcomp \R A$ and $\relcomp \R B$ are both compact.
Hence from Union of Compact Sets is Compact their union is also compact and so $\relcomp \R H$ is compact.
So $H = A \cap B \in \tau$ as its complement is compact.
Now let $\UU \subseteq \tau$.
Then from De Morgan's Laws: Complement of Union:
:$\ds \relcomp \R {\bigcup \UU} = \bigcap_{U \mathop \in \UU} \relcomp \R U$
But as:
:$\forall U \in \UU: \relcomp \R U \in \tau$
each of the $\relcomp \R U$ is compact.
Hence from Intersection of Compact Sets is Compact, so is their intersection.
So $\ds \relcomp \R {\bigcup \UU}$ is compact which means $\ds \bigcup \UU \in \tau$.
So $\tau$ is a topology on $T$.
{{qed}}
Theorem: Infinite Ramsey's Theorem

Let $k, n \in \N$.
For any set $S$, let $S^{\paren n}$ denote the set $\set {\set {s_1, \ldots, s_n}: \text{each } s_i \in S}$ of cardinality $n$ subsets of $S$.
Let $X$ be an infinite set.
Then:
:for every partition $P$ of $X^{\paren n}$ into $k$ many components
:there is an infinite subset $Y \subseteq X$
such that:
:each member of $Y^{\paren n}$ is in the same component of $P$.

Proof:
We will prove the theorem for fixed $k$ by induction on $n$.
Theorem: Compact Complement Space is T1

Let $T = \struct {S, \tau}$ be a compact complement space.
Then $T$ is a $T_1$ space.

Proof:
We have that a Finite Topological Space is Compact.
So an arbitrary finite subspace of $T$ is compact.
Let $\tau^*$ be the set defined as:
:$\tau^* = \leftset {S \subseteq \R: S = \O \text { or } \relcomp \R S}$ is finite$\rightset {}$
where $\relcomp \R S$ denotes the complement of $S$ in $\R$.
Then $\tau^*$ is a subset of $\tau$ by definition of the compact complement topology.
But $\tau^*$ is the finite complement topology.
The result follows from Finite Complement Topology is Minimal $T_1$ Topology.
{{qed}}
Theorem: Compact Complement Topology is Irreducible

Let $T = \struct {\R, \tau}$ be the compact complement space on $\R$.
Then $T$ is an irreducible space.

Proof:
Let $U_1, U_2 \in \tau$ be open in $T$.
Let $\relcomp \R {U_1} = V_1$ and $\relcomp \R {U_2} = V_2$.
By definition of compact complement topology, $V_1, V_2 \subseteq \R$ such that $V_1$ and $V_2$ are both compact.
$V_1$ and $V_2$ are both bounded by definition of compact in $\R$.
From Union of Two Compact Sets is Compact, $V_1 \cup V_2$ is also compact.
Hence, by definition, $V_1 \cup V_2$ is likewise bounded:
:above by $\max \set {\sup V_1, \sup V_2}$
:below by $\min \set {\inf V_1, \inf V_2}$.
So $V_1 \cup V_2$ cannot equal $\R$ as $\R$ is not bounded.
So:
{{begin-eqn}}
{{eqn | l = \relcomp \R {V_1 \cup V_2}
      | o = \ne
      | r = \O
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \relcomp \R {V_1} \cap \relcomp \R {V_2}
      | o = \ne
      | r = \O
      | c = De Morgan's Laws: Complement of Union
}}
{{eqn | ll= \leadsto
      | l = U_1 \cap U_2
      | o = \ne
      | r = \O
      | c = 
}}
{{end-eqn}}
So two arbitrary open sets in $T$ are not disjoint.
Hence $T$ is an irreducible space by definition.
{{qed}}