avewright/proofwiki-math huggingface dataset:
Theorem: Weierstrass-Casorati Theorem

Let $f$ be a holomorphic function defined on the open ball $\map B {a, r} \setminus \set a$.
Let $f$ have an essential singularity at $a$.
Then:
:$\forall s < r: f \sqbrk {\map B {a, s} \setminus \set a}$ is a dense subset of $\C$.

Proof:
{{WLOG}}, suppose $a = 0$ and $r = 1$.
{{AimForCont}} $\exists s < 1$ such that:
:$f \sqbrk {\map B {0, s} \setminus \set 0}$ is not a dense subset of $\C$.
Then, by definition of dense subset:
:$\exists z_0 \in \C: \exists r_0 > 0: \map B {z_0, r_0} \cap f \sqbrk {\map B {0, s} \setminus \set 0} = \O$
{{explain|This definition of dense still needs to be added to {{ProofWiki}}. Whichever definition of denseness (either everywhere dense or dense-in-itself) will need to be expanded for the Complex case so as to make it clear that the above definition follows.}}
Hence, the function $\varphi$ defined on $\map B {z_0, r_0}$ by:
:$\map \varphi z = \dfrac 1 {\map f z - z_0}$
is analytic on $\map B {0, s} \setminus \set 0$ and bounded near to $0$, because:
:$\forall z \in \map B {0, s} \setminus \set 0: \cmod {\map f z - z_0} > r_0 \implies \cmod {\map \varphi z} < \dfrac 1 {r_0}$
Therefore, we can extend the domain of $\varphi$ (using the Analytic Continuation Principle).
Let $\map \varphi 0 \ne 0$.
Then:
:$\map f 0 = z_0 + \dfrac 1 {\map \varphi 0}$
and the singular point of $f$ was removable.
Otherwise, let the power series of $\varphi$ be written:
:$\ds \map \varphi z = \sum_{n \mathop = 1}^{+\infty} a_n z^n$
Then as $\varphi \ne 0$:
:$E = \set {k \in \N: a_k \ne 0} \ne \O$
Let $p = \min E$.
Then $0$ is a pole of order $p$ of $f$.
In each case, the assumption that:
:$\exists s < 1: f \sqbrk {\map B {0, s} \setminus \set 0}$ is not a dense subset of $\C$ contradicts the fact that $0$ is an essential singularity of $f$.
Hence the result, by Proof by Contradiction.
{{qed}}
Theorem: Compact Complement Topology is Connected

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

Proof:
Follows from:
:Compact Complement Topology is Irreducible
:Irreducible Space is Connected
{{qed}}
Theorem: Compact Complement Topology is Locally Connected

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

Proof:
Follows from:
:Compact Complement Topology is Irreducible
:Irreducible Space is Locally Connected
{{qed}}
Theorem: Compact Complement Topology is not Ultraconnected

Let $T = \struct {\R, \tau}$ be the compact complement topology on $\R$.
Then $T$ is not an ultraconnected space.

Proof:
;Proof by Counterexample
{{Recall|Ultraconnected Space|ultraconnected space}}
{{:Definition:Ultraconnected Space/Definition 1}}
By definition, closed sets of $T$ are compact sets of $T$.
So, for example, $\closedint 0 1$ and $\closedint 2 3$ are disjoint compact sets and therefore closed sets of $T$.
Hence the result by definition of ultraconnected.
{{qed}}
Theorem: Compact Complement Topology is not Ultraconnected

Let $T = \struct {\R, \tau}$ be the compact complement topology on $\R$.
Then $T$ is not an ultraconnected space.

Proof:
We have:
: Compact Complement Space is $T_1$
: Non-Trivial Ultraconnected Space is not $T_1$
Hence the result.
{{qed}}
Theorem: Compact Complement Topology is Compact

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

Proof:
{{Recall|Compact Topological Space|compact space}}
{{:Definition:Compact Topological Space}}
Let $\CC$ be an arbitrary open cover of $\R$.
Let $U \in \CC$ be arbitrary.
By definition of the compact complement topology, $\R \setminus U$ is compact in the usual (Euclidean) topology.
We note that $\CC$ is an open cover of $\R \setminus U$.
As $\R \setminus U$ is compact in $\R$, there exists a finite subset $\VV \subseteq \CC$ which covers $\R \setminus U$.
Then $\VV \cup \set U$ is a finite subcover of $\CC$.
{{qed}}
Theorem: Countable Local Basis in Compact Complement Topology

Let $T = \struct {\R, \tau}$ be the compact complement topology on $\R$.
Let $p \in \R$.
Then sets of the form:
:$\openint \gets {-n} \cup \openint {p - \dfrac 1 n} {p + \dfrac 1 n} \cup \openint n \to$
form a countable local basis for $p$.

Proof:
Let $p \in \R$.
Let:
:$\BB_p = \set {\openint \gets {-n} \cup \openint {p - \dfrac 1 n} {p + \dfrac 1 n} \cup \openint n \to: n \in \N}$
Let $n \in \N$ and $P_n \in \BB$, so that:
:$P_n = \openint \gets {-n} \cup \openint {p - \dfrac 1 n} {p + \dfrac 1 n} \cup \openint n \to$
Then:
:$\R \setminus P_n = \closedint {-n} {p - \dfrac 1 n} \cup \closedint {p + \dfrac 1 n} n$
which is the union of two compact sets in $\R$ and therefore itself compact.
Clearly $p \in P_n$.
So $\BB$ is a set of open neighborhoods of $p$ in $T$.
Now let $U \in \tau$ such that $p \in U$.
Then $V := \R \setminus U$ is compact and is therefore bounded.
Suppose $\sup V = M, \inf V = m$.
Then $\exists n_B \in \N: n_T > M, -n_B > m$.
But as $p \notin V$ we have:
:$\exists \epsilon \in \R: p + \epsilon \notin V, p - \epsilon \notin V$
So $\exists n \in \N: \epsilon > \dfrac 1 n$
and so by making $n$ large enough you can fix it so that:
:$\openint \gets {-n} \cup \openint {p - \dfrac 1 n} {p + \dfrac 1 n} \cup \openint n \to \subseteq U$
so fulfilling the conditions for $\BB$ to be a local basis for $p$ which is countable.
{{qed}}
Theorem: Compact Complement Topology is First-Countable

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

Proof:
Let $\CC$ be an open cover of $\R$.
Let $p \in \R$.
Consider the set:
:$\BB_p = \set {\openint {-\infty} {-n} \cup \openint {p - \dfrac 1 n} {p + \dfrac 1 n} \cup \openint n \infty: n \in \N}$
From Countable Local Basis in Compact Complement Topology, $\BB_p$ is a countable local basis for $T$.
Hence the result, by definition of first-countable space.
{{qed}}
Theorem: Compact Complement Topology is Sequentially Compact

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

Proof:
We have:
:Compact Complement Topology is Compact
:Compact Space is Countably Compact
:Compact Complement Topology is First-Countable
:First-Countable Space is Sequentially Compact iff Countably Compact
{{qed}}
Theorem: Compact Complement Topology is Coarser than Euclidean Topology

Let $T = \struct {\R, \tau}$ be the compact complement topology on $\R$.
Then $\tau$ is coarser than the usual (Euclidean) topology in $\R$.

Proof:
Let $U \in \tau$.
Then $V = \R \setminus U$ is compact.
So by definition $V$ is closed in the Euclidean topology.
That is, $U = \R \setminus V$ is open in the Euclidean topology.
That is, every open set in the compact complement topology is also open in the Euclidean topology.
Hence the result, by definition of coarser topology.
{{qed}}
Theorem: Existence of Uncomputable Mappings

There exists a mapping that is not computable.

Proof:
Let $\CC$ be the set of all computer programs.
Every computer program in $\CC$ is a finite string of symbols from some alphabet.
Thus, from Set of Finite Strings is Countably Infinite and Subset of Countable Set is Countable, $\CC$ is countable.
Let $\BB \subset \CC$ be the set of all computer programs that implement mappings from the set of natural numbers $\mathbb N$ to the Boolean domain $\set {0, 1}$.
From Subset of Countable Set is Countable, $\BB$ is countable.
Therefore, there exists a injection $p : \BB \to \mathbb N$.
Let $p_n \in \BB$ be the $n^{th}$ computer program according to $p$.
Let $f : \mathbb N \to \set {0, 1}$ be the mapping:
:$\forall n \in \N: \map f n = \begin {cases} 0 & : \map {p_n} n = 1 \\ 1 & : \text {otherwise} \end {cases}$
In other words, $\map f n$ negates the output of the $n^{th}$ computer program in $\BB$ when its argument is $n$.
If this mapping is computable, then there is some $m \in \mathbb N$ such that $p_m \in \BB$ and:
:$\forall n \in \mathbb N : \map {p_m} n = \map f n$.
But, if we attempt to compute $\map {p_m} m$, we obtain:
:$\map {p_m} m = \begin {cases} 0 & : \map {p_m} m = 1 \\ 1 & : \text {otherwise} \end {cases}$
Therefore, we have:
:$\map {p_m} m = 1 \implies \map {p_m} m = 0$
:$\map {p_m} m = 0 \implies \map {p_m} m = 1$.
or
:$\map {p_m} m \neq \map {p_m} m$
Such a contradictory program cannot exist.
Therefore, by Modus Tollendo Tollens, the mapping $f$ is not computable.
{{qed}}
Theorem: Fort Topology is Topology

Let $T = \struct {S, \tau_p}$ be a Fort space.
Then $\tau_p$ is a topology on $T$.

Proof:
We have that $p \in \relcomp S \O = S$, so $\O \in \tau_p$.
We have that $\relcomp S S = \O$ which is finite, so $S \in \tau_p$.
Now consider $A, B \in \tau_p$, and let $H = A \cap B$.
If $p \notin A$ or $p \notin B$ then $p \notin A \cap B$ and so $H \in \tau_p$.
Now suppose $p \in A$ and $p \in B$.
Then:
{{begin-eqn}}
{{eqn | l = H
      | r = A \cap B
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \relcomp S H
      | r = \relcomp S {A \cap B}
      | c = 
}}
{{eqn | r = \relcomp S A \cup \relcomp S B
      | c = De Morgan's Laws: Complement of Union
}}
{{end-eqn}}
In order for $A$ and $B$ to be open sets it follows that $\relcomp S A$ and $\relcomp S B$ are both finite.
Hence their union is also finite and so $\relcomp S H$ is finite.
So $H = A \cap B \in \tau_p$ as its complement is finite.
Now let $\UU \subseteq \tau_p$.
Then from De Morgan's Laws: Complement of Union:
:$\ds \relcomp S {\bigcup \UU} = \bigcap_{U \mathop \in \UU} \relcomp S U$
We have either of two options:
{{begin-itemize}}
{{item|(1):|$\forall U \in \UU: p \in \relcomp S U$
in which case:
:$\ds p \in \bigcap_{U \mathop \in \UU} \relcomp S U$}}
{{item|(2):|$\exists U \in \UU: \relcomp S U$ is finite
in which case:
:$\ds \bigcap_{U \mathop \in \UU} \relcomp S U$ is finite, from Intersection is Subset}}
{{end-itemize}}
So in either case $\ds \bigcup \UU \in \tau_p$.
{{qed}}
Theorem: Fort Space is Excluded Point Space with Finite Complement Space

Let $T = \struct {S, \tau_p}$ be a Fort space on an infinite set $S$.
Then $\tau_p$ is the minimal topology that is generated by the excluded point topology and the finite complement topology.

Proof:
Let $T_1 = \struct {S, \tau_1}$ be the excluded point space on $S$ from $p$.
Let $T_2 = \struct {S, \tau_2}$ be the finite complement space on $S$.
By definition:
:$\tau_1 = \set {H \subseteq S: p \in \relcomp S H} \cup \set S$ 
:$\tau_2 = \leftset {H \subseteq S: \relcomp S H}$ is finite$\rightset {} \cup \set \O$
By definition of Fort space, we have:
:$U \in \tau_1 \implies U \in \tau_p$
:$U \in \tau_2 \implies U \in \tau_p$
So $\tau_1 \cup \tau_2 \subseteq \tau_p$.
Similarly:
:$U \in \tau_p \implies U \in \tau_1 \lor U \in \tau_2$
and so $\tau_p \subseteq \tau_1 \cup \tau_2$.
So $\tau_p = \tau_1 \cup \tau_2$ and the result follows from Union is Smallest Superset.
{{qed}}
Theorem: Fort Space is T1

Let $T = \struct {S, \tau_p}$ be a Fort space on an infinite set $S$.
Then $T$ is a $T_1$ space.

Proof:
From Fort Space is Excluded Point Space with Finite Complement Space, $T$ is an expansion of a finite complement space.
Then we have that a Finite Complement Space is $T_1$.
Then from $T_1$ Property is Preserved under Expansion, $T$ is a $T_1$ space.
{{qed}}
Theorem: Fort Space is T5

Let $T = \struct {S, \tau_p}$ be a Fort space on an infinite set $S$.
Then $T$ is a $T_5$ space.

Proof:
Let $A, B \in \tau_p$ such that $A$ and $B$ are separated.
If $p \notin A$ and $p \notin B$ then $A$ and $B$ are both open.
Otherwise $p$ must be in exactly one of them, because if $p$ were in both they could not be separated.
{{WLOG}}, suppose that $p \in A$.
Then $p \notin B$ so $B$ is open.
{{AimForCont}} $B$ is not closed.
Then $B$ is infinite by definition of Fort space.
But every open set containing $p$ is cofinite in $S$.
So $p$ would be in $B^-$, where $B^-$ is the closure of $B$.
But then {{hypothesis}}:
:$p \in A$
So $A \cap B^- \ne \O$ and so $A$ and $B$ are not separated.
This contradicts our assertion that $A$ and $B$ are separated.
So $B$ must be closed.
Therefore $\relcomp S B$ is an open set such that $A \subseteq \relcomp S B$, and the $T_5$ separation axiom is satisfied.
{{qed}}
Theorem: Fort Space is Completely Normal

Let $T = \struct {S, \tau_p}$ be a Fort space on an infinite set $S$.
Then $T$ is a completely normal space.
Consequently, $T$ satisfies all weaker separation axioms.

Proof:
We have:
:Fort Space is $T_1$
:Fort Space is $T_5$
and so by definition $T$ is completely normal.
{{qed}}
See Sequence of Implications of Separation Axioms for confirmation of the statement about weaker separation axioms.
Theorem: Uncountable Fort Space is not Perfectly Normal

Let $T = \struct {S, \tau_p}$ be a Fort space on an uncountable set $S$.
Then $T$ is not a perfectly normal space.

Proof:
From Clopen Points in Fort Space, $\set p$ is closed in $T$.
Consider a countable intersection of open sets of $T$ which contain $p$.
By definition, all these are cofinite in $S$ and so uncountable.
So this intersection must itself contain all but a countable number of points of $S$.
So $\set p$ is not a $G_\delta$ set.
Hence $T$ is not a perfectly normal space as not each one of its closed sets is a $G_\delta$ set.
{{qed}}
Theorem: Countable Fort Space is Perfectly Normal

Let $T = \struct {S, \tau_p}$ be a Fort space on a countably infinite set $S$.
Then $T$ is a perfectly normal space.

Proof:
We have from Closed Set of Countable Fort Space is $G_\delta$ that every closed set in $T$ is a $G_\delta$ set.
From Fort Space is $T_5$ and $T_5$ Space is $T_4$, we have that a Fort space is a $T_4$ space.
From Fort Space is $T_1$ it follows by definition that $T$ is a perfectly normal space.
{{qed}}
Theorem: Fort Space is Compact

Let $T = \struct {S, \tau_p}$ be a Fort space on an infinite set $S$.
Then $T$ is a compact space.

Proof:
Let $\CC$ be an open cover of $T$.
Then $\exists U \in \CC$ such that $p \in U$ and so $\relcomp S U$ is finite.
For each $x \in \relcomp S U$ there exists some $C_x \in \CC$ such that $x \in C$.
So $U$, together with each of those $C_x \in \CC$, is a finite subcover of $\CC$.
Hence the result by definition of compact space.
{{qed}}
Theorem: Fort Space is Sequentially Compact

Let $T = \struct {S, \tau_p}$ be a Fort space on an infinite set $S$.
Then $T$ is a sequentially compact space.

Proof:
Let $\sequence {x_n}_{n \mathop \in \N}$ be an infinite sequence in $T$.
Suppose $\sequence {x_n}$ takes an infinite number of distinct values in $S$.
Then there is an infinite subsequence $\sequence {x_{n_r} }_{r \mathop \in \N}$ with distinct terms.
Let $U$ be a neighborhood of $p$.
Then $S \setminus U$ is a finite set by definition.
Thus there exists $N \in \N$ such that $\forall r > N: x_{n_r} \in U$.
Thus $\sequence {x_{n_r} }$ converges to $p$.
Otherwise $\sequence {x_n}$ only takes a finite number of distinct values.
Then, since $\sequence {x_n}$ is infinite, there exists $x \in S$ such that:
:$\forall N \in \N: \exists n > N: x = x_n$
This implies that we can take a subsequence of $\sequence {x_n}$ which is constant, and which converges to that constant.
We can conclude then that, by definition, $T$ is a sequentially compact space.
{{qed}}
Theorem: Uncountable Fort Space is not First-Countable

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

Proof:
Let $\UU$ be a countable set of open neighborhoods of $p$.
Let $U \in \UU$.
Then as $p \in U$:
:$p \notin \relcomp S U$
So for $U$ to be open in $T$, it must follow that $\relcomp S U$ is finite.
From De Morgan's Laws: Complement of Intersection we have:
:$\ds H := \bigcup_{U \mathop \in \UU} \relcomp S U = \relcomp S {\bigcap \UU}$
From Countable Union of Countable Sets is Countable it follows that $H$ can be no more than countable.
So $\ds \bigcap \UU = \relcomp S H$ must be uncountable.
So (trivially):
:$H \ne \set p$
So:
:$\exists q \ne p: q \in \bigcap \UU$
So $\relcomp S {\set q}$ is an open neighborhood of $P$ which does not contain any of the elements of $\UU$.
So $\UU$ is not a countable local basis of $p$.
Hence by definition $T$ is not a first-countable space.
{{qed}}
Theorem: Generating Function for Boubaker Polynomials

The Boubaker polynomials, defined as:   
:$\map {B_n} x = \begin{cases}
1 & : n = 0 \\
x & : n = 1 \\
x^2 + 2 & : n = 2 \\
x \map {B_{n - 1} } x - \map {B_{n - 2} } x & : n > 2 \end{cases}$
have as an ordinary generating function:
:$\ds \map {f_{B_n, \operatorname {ORD} } } {x, t} = \sum_{n \mathop = 0}^{\infty} \map {B_n} x t^n = \frac {1 + 3 t^2} {1 + t \paren {t - x} }$

Proof:
{{begin-eqn}}
{{eqn | l = \map f {x, t}
      | r = 1 + x t + \paren {x^2 + 2} t^2 + \sum_{n \mathop = 3}^\infty \paren {x \map {B_{n - 1} } x t^n - \map {B_{n - 2} } x t^n}
      | c = Recursive definition of $\map {B_n} x$ for $n > 2$
}}
{{eqn | r = 1 + x t + \paren {x^2 + 2} t^2 + x t \sum_{n \mathop = 3}^\infty \map {B_{n - 1} } x t^{n - 1} - t^2 \sum_{n \mathop = 3}^\infty \map {B_{n - 2} } x t^{n - 2}
}}
{{eqn | r = 1 + x t + \paren {x^2 + 2} t^2 + x t \paren {\map f {x, t} - 1 - x t} - t^2 \paren {\map f {x, t} - 1}
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | ll= \leadsto
      | l = \paren {1 - x t + t^2} \map f {x, t}
      | r = 1 + x t + x^2 t^2 + 2 t^2 - x t - x^2 t^2 + t^2
}}
{{eqn | r = 1 + 3 t^2
}}
{{eqn | ll=\leadsto
      | l = \map f {x, t}
      | r = \frac {1 + 3 t^2} {1 + t \paren {t - x} }
}}
{{end-eqn}}
{{qed}}
Category:Boubaker Polynomials
Category:Examples of Generating Functions
63hgjmuav4215avpgz7t5zckuspzjs0
Theorem: Exponential Generating Function for Boubaker Polynomials

The Boubaker polynomials, defined as:           
:<nowiki>$\map {B_n} x = \begin {cases}
1 & : n = 0 \\
x & : n = 1 \\
x^2 + 2 & : n = 2 \\
x \map {B_{n - 1} } x - \map {B_{n - 2} } x & : n > 2 \end {cases}$</nowiki>
have as an exponential generating function:
:$\ds \map {f_{B_n, \operatorname {EXP} } } {x, t} = \sum_{n \mathop = 0}^\infty \map {B_n} x \frac {t^n} {n!} = 4 x e^{t \frac x 2} \frac {\map \sin {t \sqrt {1 - \paren {\frac x 2}^2} } } {\sqrt {1 - \paren {\frac x 2}^2} } - 2 e^{t \frac x 2} \map \cos {t \sqrt {1 - \paren {\frac x 2}^2} } - 3$
{{Proofread}}

Proof:
From the definition of the Boubaker polynomials, we have:    
:$\begin {cases} \map {B_0} x = 1 \\ \map {B_1} x = x \end {cases}$
We have also, for $T_n$, the Chebyshev polynomials of the first kind and $U_n$, the Chebyshev polynomials of the second kind:        
:$\begin {cases} \map {T_0} x = 1 \\ \map {T_1} x = x \end {cases}$
and:
:$\begin {cases} \map {U_{-1} } x = 0 \\ \map {U_0} x = 1 \\ \map {U_1} x = 2 x \end {cases}$
Then, from the definition of the exponential generating functions of the Chebyshev polynomials of the first kind:
:$\ds \map {f_{T_n, \operatorname {EXP} } } {x, t} = e^{t \frac x 2} \map \cos {t \sqrt {1 - \paren {\frac x 2}^2} }$
and the Chebyshev polynomials of the second kind: 
:$\ds \map {f_{U_n, \operatorname {EXP} } } {x, t} = e^{t \frac x 2} \frac {\map \sin {t \sqrt {1 - \paren {\frac x 2}^2} } } {\sqrt {1 - \paren {\frac x 2}^2} }$
and from the relation:
:$\map {B_n} x = 4 x \map {U_{n - 1} } {\dfrac x 2} - 2 \map {T_n} {\dfrac x 2}$
which is valid only for $n > 1$, we have:
:$\ds \map {B_0} x \dfrac {t^0} {0!} + \map {B_1} x \dfrac {t^1} {1!} + \sum_{n \mathop = 2}^\infty \map {B_n} x \dfrac {t^n} {n!} = 4 x \sum_{n \mathop = 2}^\infty \map {U_{n - 1} } x \dfrac {t^n} {n!} - 2 \sum_{n \mathop = 2}^\infty \map {T_n} x \dfrac {t^n} {n!} + \map {B_0} x \dfrac {t^0} {0!} + \map {B_1} x \dfrac {t^1} {1!} - 4 x \sum_{n \mathop = 0}^1 \map {U_{n - 1} } x \dfrac {t^n} {n!} - 2 \sum_{n \mathop = 0}^1 \map {T_n} x \dfrac {t^n} {n!}$     
Finally, by simplifying the two sides and considering the exponential generating functions and the values of the first terms of $T_n$ and $U_n$, we obtain:      
:$\ds \sum_{n \mathop = 0}^\infty \map {B_n} x \frac{t^n} {n!} = 4 x e^{t \frac x 2} \dfrac {\map \sin {t \sqrt {1 - \paren {\frac x 2}^2} } } {\sqrt {1 - \paren {\frac x 2}^2} } - 2 e^{t \frac x 2} \map \cos {t \sqrt {1 - \paren {\dfrac x 2}^2} } - 3$
Hence the result.     
{{qed}}
{{MissingLinks}}
Category:Boubaker Polynomials
Category:Examples of Exponential Generating Functions
jsfonnruvtow6h6qbb9ymglm01kw1dd
Theorem: Clopen Points in Fort Space

Let $T = \struct {S, \tau_p}$ be a Fort space on an infinite set $S$.
Let $q \in S: q \ne p$.
Then $\set q$ is both open and closed in $T$.
$\set p$ itself, on the other hand, is closed but not open.

Proof:
We have that $\set q$ is finite so $\relcomp S {\set q}$ is cofinite.
So $\relcomp S {\set q}$ is open and so $\set q$ is closed.
Then we have that $p \notin \set q$ so $\set q$ is open.
However, $p \notin \relcomp S {\set p}$ and $\relcomp S {\set p}$ is infinite.
So $\set p$ is not open in $S$.
But $\set p$ is finite, so $\relcomp S {\set p}$ is cofinite.
Hence, as for $\set q$, we have that $\set p$ is closed.
{{qed}}
Theorem: Fort Space is Totally Separated

Let $T = \struct {S, \tau_p}$ be a Fort space on an infinite set $S$.
Then $T$ is a totally separated space.

Proof:
Let $a, b \in S$ such that $a \ne b$.
From Clopen Points in Fort Space, all points in $S$ apart from $p$ are both open and closed in $T$.
{{WLOG}}, suppose that $a \ne p$.
Then as $\set a$ is closed, $\relcomp S {\set a}$ is open in $T$.
So we have a partition $\set a \mid \relcomp S {\set a}$ of $S$ such that $a \in \set a, b \in \relcomp S {\set a}$.
So, by definition  $T$ is a totally separated space.
{{qed}}
Theorem: Fort Space is not Extremally Disconnected

Let $T = \struct {S, \tau_p}$ be a Fort space on an infinite set $S$.
Then $T$ is not an extremally disconnected space.

Proof:
{{Recall|Extremally Disconnected Space|extremally disconnected space}}
{{:Definition:Extremally Disconnected Space/Definition 1}}
Let $H \subseteq S$ be an infinite subset of $S$ such that $p \in \relcomp S H$ and $\relcomp S H$ is infinite.
Then $H$ is open in $T$, as $p \in \relcomp S H$.
From Limit Points in Fort Space, as $H$ is infinite, $p$ is the only limit point of $H$.
So $H^- = H \cup \set p$ where $H^-$ is the closure of $H$.
But as $\relcomp S {H^-}$ is infinite and does not contain $p$, it follows by definition of Fort space that $H^-$ is not open in $T$.
So by definition $T$ is not an extremally disconnected space.
{{qed}}
Theorem: Fort Space is Zero Dimensional

Let $T = \struct {S, \tau_p}$ be a Fort space on an infinite set $S$.
Then $T$ is a zero dimensional space.

Proof:
Let $q \in S$ such that $q \ne p$.
Then from Clopen Points in Fort Space, $\set q$ is clopen.
So $\forall q \in S, q \ne p: \set {\set q}$ is a local basis for $q$.
If we take the open neighborhoods of $p$ we get a local basis $\UU_p$ with the following property:
:Since $p \in U \in \UU_p$, its complement does not contain $p$ and so it is open.
This implies that $U$ is also closed.
The union of the local basis forms a basis for the topology.
This basis is formed with clopen sets.
So, by definition, $T$ is a zero dimensional space.
{{qed}}
Theorem: Formula and its Negation Cannot Both Cause Forking

Let $T$ be a complete $\LL$-theory.
Let $\mathfrak C$ be a monster model for $T$.
Let $A \subseteq B$ be subsets of the universe of $\mathfrak C$.
Let $\map \pi {\bar x}$ be an $n$-type over $B$.
If $\pi$ does not fork over $A$, then for any formula $\map \phi {\bar x, \bar b}$, either $\pi \cup \set \phi$ or $\pi \cup \set {\neg \phi}$ does not fork over $A$.

Proof:
We prove the contrapositive.
Suppose both $\pi \cup \set \phi$ and $\pi \cup \set {\neg \phi}$ fork over $A$.
By definition of forking:
:$\pi \cup \set \phi$ implies $\map {\phi_1} {\bar x, \bar c_1} \vee \cdots \vee \map {\phi_k} {\bar x, \bar c_k}$ and
:$\pi \cup \set {\neg \phi}$ implies $\map {\psi_1} {\bar x, \bar d_1} \vee \cdots \vee \map {\psi_k} {\bar x, \bar d_h}$,
where each $\phi_i$ and $\psi_j$ divide over $A$.
Then $\pi$ implies the disjunction
:$\ds \bigvee_i \map \phi {\bar x, \bar c_i} \vee \bigvee_j \map \psi {\bar x, \bar d_i}$
with each component formula dividing over $A$.
By definition, this means that $\pi$ forks over $A$.
{{qed}}
Category:Model Theory for Predicate Logic
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