nyuuzyou/wikis huggingface dataset:
Formulation 1 Rank Axioms Implies Rank Function of Matroid/Lemma 1

== Theorem ==
Let $S$ be a finite set.
Let $\rho : \powerset S \to \Z$ be a mapping from the power set of $S$ to the integers.
Let $\rho$ satisfy formulation 1 of the rank axioms:
Then:
:$\forall A, B \subseteq S: A \subseteq B \implies \map \rho A \le \map \rho B$
== Proof ==
:
:$\exists A, B \subseteq S : A \subseteq B$ and $\map \rho A > \map \rho B$
Let $B \subseteq S$:
:$\exists A \subseteq B : \map rho A > \map \rho B$
Let $A_0 \subseteq B$:
:$\card {A_0} = \max \set{\card A : A \subseteq B \land \map \rho A > \map \rho B}$
As $\map \rho {A_0} > \map \rho B$:
:$A_0 \ne B$
From Set Difference with Proper Subset:
:$\exists y \in B \setminus A_0$
We have:
This is a contradiction.
Hence:
:$\forall A, B \subseteq S: A \subseteq B \implies \map \rho A \le \map \rho B$
Category:Formulation 1 Rank Axioms Implies Rank Function of Matroid
Formulation 1 Rank Axioms Implies Rank Function of Matroid/Lemma 2

== Theorem ==
Let $S$ be a finite set.
Let $\rho : \powerset S \to \Z$ be a mapping from the power set of $S$ to the integers.
Let $\rho$ satisfy formulation 1 of the rank axioms:
Then:
:$\forall A \subseteq S: \map \rho A \le \card A$
== Proof ==
:
:$\exists A \subseteq S : \map \rho A > \card A$
Let:
:$A_0 \subseteq S : \card{A_0} = \min \set{\card A : \map \rho A > \card A}$
We have:
Hence:
:$A_0 \ne \O$
Let $y \in A_0$.
From Cardinality of Set Difference:
:$\card {A_0 \setminus \set y} = \card{A_0} - 1 < \card {A_0}$
We have:
This contradicts the choice of $A_0$:
:$\card{A_0} = \min \set{\card A : \map \rho A > \card A}$
It follows that:
:$\forall A \subseteq S: \map \rho A \le \card A$
Category:Formulation 1 Rank Axioms Implies Rank Function of Matroid
Formulation 1 Rank Axioms Implies Rank Function of Matroid/Lemma 3

== Theorem ==
Let $S$ be a finite set.
Let $\rho : \powerset S \to \Z$ be a mapping from the power set of $S$ to the integers.
Let $\rho$ satisfy formulation 1 of the rank axioms:
Let:
:$A \subseteq S : \map \rho A = \card A$
Let:
:$B \subseteq S : \forall b \in B \setminus A : \map \rho {A \cup \set b} \ne \card{A \cup \set b}$
 
Then:
:$\map \rho {A \cup B} = \map \rho A$
==== Lemma 1 ====
==== Lemma 2 ====
== Proof ==
==== Case 1 : $B \setminus A = \O$ ====
Let $B \setminus A = \O$.
From Set Difference with Superset is Empty Set:
:$B \subseteq A$
From Union with Superset is Superset:
:$A \cup B = A$
It follows that:
:$\map \rho {A \cup B} = \map \rho A$
==== Case 2 : $B \setminus A \ne \O$ ====
Let:
:$B \setminus A = \set{b_1, b_2, \dots, b_k}$
For each $i \in \closedint 1 k$, let:
:$B_i = \begin{cases}
A & : i = 1 \\
A \cup \set{x_1, x_2, \dots, x_{i-1}} & : i \in \closedint 2 k
\end{cases}$
:$\mathscr B_i = \set{B_i \cup \set y : y \in B \setminus B_i}$
We note that:
We prove by induction:
:$\forall i \in \closedint 1 k : \forall Y \in \mathscr B_i : \map \rho Y = \map \rho A$
For all $i \in \closedint 1 k$, let $\map P i$ be the proposition:
:$\forall Y \in \mathscr B_i : \map \rho Y = \map \rho A$
===== Basis for the Induction =====
We have:
:$\mathscr B_1 = \set{A \cup \set y : y \in \set{b_1, b_2, \dots, b_k}}$
$\map P 1$ is the case:
:$\forall j \in \closedint 1 k : \map \rho {A \cup \set{b_j}} = \map \rho A$
Let $j \in \closedint 1 k$.
We have:
By rank axiom $(\text R 2)$:
:$\map \rho A = \map \rho {A \cup \set {b_j}}$
This establishes our base case.
===== Induction Hypothesis =====
Now we need to show that, if $\map P i$ is true, where $i \ge 1$, then it logically follows that $\map P {i + 1}$ is true.
So this is our induction hypothesis:
:$\forall Y \in \mathscr B_i : \map \rho Y = \map \rho A$
Then we need to show:
:$\forall Y \in \mathscr B_{i+1} : \map \rho Y = \map \rho A$
===== Induction Step =====
This is our induction step.
Let $Y \in \mathscr B_{i+1}$.
Then:
:$Y = A \cup \set{b_1, b_2, \dots, b_i, y}$
where $y \in {b_{i+1}, \dots, b_k}$
We note that:
:$A \cup \set{b_1, b_2, \dots, b_{i-1}, b_i} \in \mathscr B_i$
:$A \cup \set{b_1, b_2, \dots, b_{i-1}, y} \in \mathscr B_i$
We have:
Hence:
:$\map \rho {A \cup \set{b_1, b_2, \dots, b_{i-1}, b_i } } = \map \rho {A \cup \set{b_1, b_2, \dots, b_{i-1} } }$
and
:$\map \rho {A \cup \set{b_1, b_2, \dots, b_{i-1} } } = \map \rho A$
Similarly:
:$\map \rho {A \cup \set{b_1, b_2, \dots, b_{i-1}, y } } = \map \rho {A \cup \set{b_1, b_2, \dots, b_{i-1} } }$
By Rank axiom $(\text R 3)$:
:$\map \rho {A \cup \set{b_1, b_2, \dots, b_{i-1}, b_i, y } } = \map \rho {A \cup \set{b_1, b_2, \dots, b_{i-1} } } = \map \rho A$
It follows that:
:$\forall Y \in \mathscr B_{i+1} : \map \rho Y = \map \rho A$
This establishes our induction step.
Hence by induction:
:$\forall i \in \closedint 1 k : \forall Y \in \mathscr B_i : \map \rho Y = \map \rho A$
We have:
In either case:
:$\map \rho {A \cup B} = \map \rho A$
Category:Formulation 1 Rank Axioms Implies Rank Function of Matroid
Formulation 1 Rank Axioms Implies Rank Function of Matroid/Lemma 4

== Theorem ==
Let $S$ be a finite set.
Let $\rho : \powerset S \to \Z$ be a mapping from the power set of $S$ to the integers.
Let $\rho$ satisfy formulation 1 of the rank axioms:
Let $M = \struct{S, \mathscr I}$ where:
:$\mathscr I = \set{X \subseteq S : \map \rho X = \card X}$
Then $M$ satisfies matroid axiom $(\text I 1)$.
== Proof ==
We have:
So:
:$\O \in \mathscr I$
Hence:
:$M$ satisfies matroid axiom $(\text I 1)$.
Category:Formulation 1 Rank Axioms Implies Rank Function of Matroid
Formulation 1 Rank Axioms Implies Rank Function of Matroid/Lemma 5

== Theorem ==
Let $S$ be a finite set.
Let $\rho : \powerset S \to \Z$ be a mapping from the power set of $S$ to the integers.
Let $\rho$ satisfy formulation 1 of the rank axioms:
Let $M = \struct{S, \mathscr I}$ where:
:$\mathscr I = \set{X \subseteq S : \map \rho X = \card X}$
Then $M$ satisfies matroid axiom $(\text I 2)$.
=== Lemma 2 ===
== Proof 1 ==
== Proof 2 ==
Category:Formulation 1 Rank Axioms Implies Rank Function of Matroid
Formulation 1 Rank Axioms Implies Rank Function of Matroid/Lemma 5/Proof 1

== Theorem ==
=== Lemma 2 ===
 
== Proof ==
Let
:$X \in \mathscr I$
:$\exists Y \subseteq X : Y \notin \mathscr I$
Let:
:$Y_0 \subseteq X : \card {Y_0} = \max \set{\card Z : Z \subseteq X \land Z \notin \mathscr I}$
By definition of $\mathscr I$:
:$Y_0 \notin \mathscr I \leadsto \map \rho {Y_0} \ne \card {Y_0}$
From Lemma 2:
:$\map \rho {Y_0} < \card {Y_0}$
As $X \in \mathscr I$ then:
:$Y_0 \ne X$
Hence:
:$Y_0$ is a proper subset of $X$
From Set Difference with Proper Subset:
:$X \setminus Y_0 \ne \O$
Let $y \in X \setminus Y_0$.
We have:
This is a contradiction.
So:
:$\forall Y \subseteq X : Y \in \mathscr I$
Hence:
:$M$ satisfies matroid axiom $(\text I 2)$.
== Sources ==
* Chapter $13$ Matroids $\S 13.2$ Other Matroid Axioms, Theorem $13.10$
Category:Formulation 1 Rank Axioms Implies Rank Function of Matroid
Formulation 1 Rank Axioms Implies Rank Function of Matroid/Lemma 5/Proof 2

== Theorem ==
=== Lemma 2 ===
== Proof ==
We prove the contrapositive statement:
:$\forall X, Y \subseteq S: Y \notin \mathscr I \land Y \subseteq X \implies X \notin \mathscr I$
Let $X, Y \subseteq S : Y \notin \mathscr I$ and $Y \subseteq X$.
====== Case 1: $Y = X$ ======
Let $Y = X$.
Then $X \notin \mathscr I$.
====== Case 2: $Y \subset X$ ======
Let $Y \subset X$.
By definition of $\mathscr I$:
:$\map \rho Y \neq \size Y$
From Lemma 2:
:$\map \rho Y < \size Y$
Let:
:$X \setminus Y = \set{z_1, z_2, \ldots , z_k}$
We have:
It follows that $X \notin \mathscr I$.
In either case we have:
:$X \notin \mathscr I$
This proves the contrapositive statement:
:$\forall X, Y \subseteq S: Y \notin \mathscr I \land Y \subseteq X \implies X \notin \mathscr I$
It follows that $\mathscr I$ satisfies matroid Axiom $(\text I 2)$.
Category:Formulation 1 Rank Axioms Implies Rank Function of Matroid
Formulation 1 Rank Axioms Implies Rank Function of Matroid/Lemma 6

== Theorem ==
Let $S$ be a finite set.
Let $\rho : \powerset S \to \Z$ be a mapping from the power set of $S$ to the integers.
Let $\rho$ satisfy formulation 1 of the rank axioms:
Let $M = \struct{S, \mathscr I}$ where:
:$\mathscr I = \set{X \subseteq S : \map \rho X = \card X}$
Then $M$ satisfies matroid axiom $(\text I 3)$.
===== Lemma 1 =====
===== Lemma 3 =====
== Proof ==
Let
:$U \in \mathscr I$
:$V \subseteq S$
:$\card U < \card V$
We prove the contrapositive statement:
:$\paren{\forall x \in V \setminus U : U \cup \set x \notin \mathscr I} \implies V \notin \mathscr I$
Let:
:$\forall x \in V \setminus U: U \cup \set x \notin \mathscr I$
That is, by definition of $\mathscr I$:
:$\forall x \in V \setminus U : \map \rho {U \cup \set x} \ne \card {U \cup \set x}$
We have:
Hence:
:$V \notin \mathscr I$
It follows:
:$M$ satisfies matroid axiom $(\text I 3)$.
Category:Formulation 1 Rank Axioms Implies Rank Function of Matroid
Formulation 1 Rank Axioms Implies Rank Function of Matroid/Lemma 7

== Theorem ==
Let $S$ be a finite set.
Let $\rho : \powerset S \to \Z$ be a mapping from the power set of $S$ to the integers.
Let $\rho$ satisfy formulation 1 of the rank axioms:
Let $M = \struct{S, \mathscr I}$ where:
:$\mathscr I = \set{X \subseteq S : \map \rho X = \card X}$
Then $M$ satisfies matroid axiom $(\text I 4)$.
=== Lemma 1 ===
=== Lemma 3 ===
== Proof ==
It is now proved that $\mathscr I$ satisifes the matroid Axiom $(\text I 4)$:
Let $X, Y \in \mathscr I$ such that $\size Y = \size X + 1$.
Let:
:$X = \set {x_1, x_2, \ldots, x_q, z_{q + 1}, \ldots, z_k}$
and
:$Y = \set {x_1, x_2, \ldots, x_q, y_{q + 1}, \ldots, y_k, y_{k + 1}}$
where $\forall i \in \set {q + 1, \ldots, k}$ and $\forall j \in \set {q + 1, \ldots, k + 1}$
:$z_i \neq y_j$
:
:$\forall j \in \set {q + 1, \ldots, k + 1}: X \cup \set{y_j} \notin \mathscr I$
We have:
Now:
:$Y \subseteq X \cup \set{y_{q + 1}, \ldots, y_{k + 1} }$
From Lemma 1:
:$\map \rho Y \le \map \rho {X \cup \set{y_{q + 1}, \ldots, y_{k + 1} } } < \size Y$
This contradicts the assumption that $Y \in \mathscr I$.
Hence:
:$\exists j \in \set {q + 1, \ldots, k + 1}: X \cup \set{y_j} \in \mathscr I$
It follows that $\mathscr I$ satisfies matroid Axiom $(\text I 4)$.
Category:Formulation 1 Rank Axioms Implies Rank Function of Matroid
Formulation 1 Rank Axioms Implies Rank Function of Matroid/Lemma 8

== Theorem ==
Let $S$ be a finite set.
Let $\rho : \powerset S \to \Z$ be a mapping from the power set of $S$ to the integers.
Let $\rho$ satisfy formulation 1 of the rank axioms:
Let $M = \struct{S, \mathscr I}$ be the matroid where:
:$\mathscr I = \set{X \subseteq S : \map \rho X = \card X}$
Let $\rho_M$ be the rank function of the matroid $M = \struct{S, \mathscr I}$.
Then $\rho_M = \rho$.
=== Lemma 3 ===
== Proof ==
Let $X \subseteq S$.
By definition of the rank function:
:$\map {\rho_M} X = \max \set{\card Y : Y \subseteq X, Y \in \mathscr I}$
Let $Y_0 \subseteq X$:
:$\card {Y_0} = \max \set{\card Y : Y \subseteq X, Y \in \mathscr I}$
Then:
So it remains to show:
:$\map \rho {Y_0} = \map \rho X$
===== Case 1 : $Y_0 = X$ =====
Let $Y_0 = X$.
Then:
:$\map \rho {Y_0} = \map \rho X$
===== Case 2 : $Y_0 \ne X$ =====
Let $Y_0 \ne X$.
Then:
:$Y_0 \subsetneq X$
From Set Difference with Proper Subset:
:$X \setminus Y_0 \ne \O$
By choice of $Y_0$:
:$\forall y \in X \setminus Y_0 : Y_0 \cup \set y \notin \mathscr I$
That is:
:$\forall y \in X \setminus Y_0 : \map \rho {Y_0 \cup \set y} \ne \card {Y_0 \cup \set y}$
From Lemma 3:
:$\map \rho {Y_0} = \map \rho {Y_0 \cup X} = \map \rho X$
In either case:
: $\map \rho {Y_0} = \map \rho X$
It follows that:
:$\forall X \subseteq S : \map {\rho_M} X = \map \rho X$
Hence $\rho$ is the rank function of the matroid $M = \struct{S, \mathscr I}$.
Category:Formulation 1 Rank Axioms Implies Rank Function of Matroid
Formulation 1 Rank Axioms Implies Rank Function of Matroid/Proof 1

== Theorem ==
Let $S$ be a finite set.
Let $\rho : \powerset S \to \Z$ be a mapping from the power set of $S$ to the integers.
Let $\rho$ satisfy formulation 1 of the rank axioms:
Then $\rho$ is the rank function of a matroid on $S$.
=== Lemma 1 ===
=== Lemma 2 ===
=== Lemma 3 ===
== Proof ==
Let:
:$\mathscr I = \set{X \subseteq S : \map \rho X = \card X}$
It is to be shown that:
:* $\quad \mathscr I$ satisfies the matroid axioms
and
:* $\quad \rho$ is the rank function of the matroid $M = \struct{S, \mathscr I}$
==== Matroid Axioms ====
===== Matroid Axiom $(\text I 1)$ =====
===== Matroid Axiom $(\text I 2)$ =====
===== Matroid Axiom $(\text I 3)$ =====
==== $\rho$ is Rank Function ====
Let $\rho_M$ be the rank function of the matroid $M = \struct{S, \mathscr I}$.
Category:Formulation 1 Rank Axioms Implies Rank Function of Matroid
Formulation 1 Rank Axioms Implies Rank Function of Matroid/Proof 2

== Theorem ==
Let $S$ be a finite set.
Let $\rho : \powerset S \to \Z$ be a mapping from the power set of $S$ to the integers.
Let $\rho$ satisfy formulation 1 of the rank axioms:
Then $\rho$ is the rank function of a matroid on $S$.
===== Lemma 1 =====
===== Lemma 2 =====
===== Lemma 3 =====
== Proof ==
Let:
:$\mathscr I = \set{X \subseteq S : \map \rho X = \card X}$
It is to be shown that:
:* $\quad \mathscr I$ satisfies the matroid axioms
and
:* $\quad \rho$ is the rank function of the matroid $M = \struct{S, \mathscr I}$
==== Matroid Axioms ====
===== Matroid Axiom $(\text I 1)$ =====
===== Matroid Axiom $(\text I 2)$ =====
===== Matroid Axiom $(\text I 4)$ =====
We have proven that $\mathscr I$ satisifies the matroid axioms.
It follows that $M = \struct{S, \mathscr I}$ is a matroid by definition.
==== $\rho$ is Rank Function ====
Let $\rho_M$ be the rank function of the matroid $M = \struct{S, \mathscr I}$.
== Sources ==
* Chapter $1.$ $\S 6.$ Properties of the rank function, Proof of Theorem $2.2$
Category:Formulation 1 Rank Axioms Implies Rank Function of Matroid
Fort Space is Compact

== Theorem ==
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.
== Sources ==
* : Part $\text {II}$: Counterexamples: $23 \text { - } 24$. Fort Space: $4$
Category:Compact Spaces
Category:Fort Spaces
Fort Space is Completely Normal

== Theorem ==
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.
See Sequence of Implications of Separation Axioms for confirmation of the statement about weaker separation axioms.
== Sources ==
* : Part $\text {II}$: Counterexamples: $23 \text { - } 24$. Fort Space: $2$
Category:Completely Normal Spaces
Category:Fort Spaces
Fort Space is Excluded Point Space with Finite Complement Space

== Theorem ==
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.
== Sources ==
* : Part $\text {II}$: Counterexamples: $23 \text { - } 24$. Fort Space: $1$
Category:Fort Spaces
Category:Excluded Point Topology
Category:Finite Complement Topology
Fort Space is Regular

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a Fort space.
Then $T$ is a regular space.
== Proof ==
We have that the Fort Space is Completely Normal.
The result follows from tracing the relevant implications on Sequence of Implications of Separation Axioms.
== Sources ==
* : Part $\text {II}$: Counterexamples: $23 \text { - } 24$. Fort Space: $6$
Category:Regular Spaces
Category:Fort Spaces
Fort Space is Scattered/Proof 1

== Theorem ==
== Proof ==
Let $H \subseteq T$ such that $H \ne \O$ and $H \ne \set p$.
Then $\exists x \in H: x \ne p$.
From Clopen Points in Fort Space, every point of $T$ apart from $p$ is open in $T$.
So $\set x$ is an open set of $T$.
So $H \cap \set x = \set x$ and so $x$ is isolated in $H$.
Thus $H$ contains at least one point which is isolated in $H$.
On the other hand, suppose $H = \set p$.
From Singleton Point is Isolated, $p$ is an isolated point in $\set p = H$.
So again $H$ contains at least one point which is isolated in $H$.
So for all $H \subseteq S$ we have that $H$ contains at least one point which is isolated in $H$.
Hence the result, by definition of scattered space.
Category:Fort Space is Scattered
Fort Space is Scattered/Proof 2

== Theorem ==
== Proof ==
Suppose that $H \subseteq T$ has no isolated points of $H$.
So, by definition, $H$ is dense in itself.
We have that:
:a Fort Space is $T_1$
:a Dense-in-itself Subset of $T_1$ Space is Infinite.
So $H$ is infinite, and so contains more than one point.
So $\exists q \in H: q \ne p$.
But, from Clopen Points in Fort Space, $\set q$ is open in $T$.
So $H \cap \set q = \set q$ and so by definition $q$ is isolated in $H$.
From this contradiction it follows that $H$ is not dense in itself and contains at least one isolated point.
Hence the result, by definition of scattered space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $23 \text { - } 24$. Fort Space: $8$
Category:Fort Space is Scattered
Fort Space is Sequentially Compact

== Theorem ==
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.
== Sources ==
* : Part $\text {II}$: Counterexamples: $23 \text { - } 24$. Fort Space: $4$
Category:Sequentially Compact Spaces
Category:Fort Spaces
Fort Space is T0

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a Fort space on an infinite set $S$.
Then $T$ is a $T_0$ (Kolmogorov) space.
== Proof ==
Follows directly from:
:Fort Space is $T_1$
:$T_1$ (Fréchet) Space is $T_0$ (Kolmogorov) Space
Category:T0 Spaces
Category:Fort Spaces
Fort Space is T1

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a Fort space on an infinite set $S$.
Then $T$ is a $T_1$ (Fréchet) 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 Separation Properties Preserved by Expansion we have that as a finite complement space is a $T_1$ space, then so is $T$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $23 \text { - } 24$. Fort Space: $1$
Category:T1 Spaces
Category:Fort Spaces
Fort Space is T5

== Theorem ==
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.
, suppose that $p \in A$.
Then $p \notin B$ so $B$ is open.
Now suppose $B$ were 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 $p \in A$, by hypothesis.
So $A \cap B^- \ne \O$ and so $A$ and $B$ are not 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.
== Sources ==
* : Part $\text {II}$: Counterexamples: $23 \text { - } 24$. Fort Space: $2$
Category:T5 Spaces
Category:Fort Spaces
Fort Space is Totally Separated

== Theorem ==
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 $T$.
, 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.
== Sources ==
* : Part $\text {II}$: Counterexamples: $23 \text { - } 24$. Fort Space: $7$
Category:Totally Separated Spaces
Category:Fort Spaces
Fort Space is Zero Dimensional

== Theorem ==
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.
== Sources ==
* : Part $\text {II}$: Counterexamples: $23 \text { - } 24$. Fort Space: $9$
Category:Zero Dimensional Spaces
Category:Fort Spaces
Fort Space is not Extremally Disconnected

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a Fort space on an infinite set $S$.
Then $T$ is not an extremally disconnected space.
== Proof ==
We note from Fort Space is Completely Normal that $T$ is a $T_2$ (Hausdorff) space.
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 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 that $H^-$ is not open in $T$.
So by definition $T$ is not an extremally disconnected space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $23 \text { - } 24$. Fort Space: $7$
Category:Extremally Disconnected Spaces
Category:Fort Spaces