avewright/proofwiki-math huggingface dataset:
Theorem: Solenoidal Vector Field is Curl

Let $\mathbf V$ be a solenoidal vector field on a simply connected region of $\mathbb R^3$.
Then there exists a vector field $\mathbf A$ such that:
:$\mathbf V = \curl \mathbf A$

Proof:
{{Recall|Solenoidal Vector Field|solenoidal vector field|index = 1}}
{{:Definition:Solenoidal Vector Field/Definition 1}}
We seek any vector field $\mathbf A$ such that:
{{begin-eqn}}
{{eqn | l = \mathbf V
      | r = \nabla \times \mathbf A
}}
{{eqn | r = \paren {\frac {\partial A_z} {\partial y} - \frac {\partial A_y} {\partial z} } \mathbf i + \paren {\frac {\partial A_x} {\partial z} - \frac {\partial A_z} {\partial x} } \mathbf j + \paren {\frac {\partial A_y} {\partial x} - \frac {\partial A_x} {\partial y} } \mathbf k
      | c = {{Defof|Curl Operator}}
}}
{{end-eqn}}
Let $A_x = 0$.
Then:
{{begin-eqn}}
{{eqn | l = V_y
      | r = -\dfrac {\partial A_z} {\partial x}
}}
{{eqn | l = V_z
      | r = \dfrac {\partial A_y} {\partial x}
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = A_y
      | r = \int V_z \rd x + \map f {y, z}
}}
{{eqn | l = A_z
      | r = -\int V_y \rd x + \map g {y, z}
}}
{{end-eqn}}
for some scalar functions $f$ and $g$ which act as constants of integration.
So:
:$\ds \mathbf A = \paren {\int V_z \rd x + \map f {y, z} } \mathbf j - \paren {\int V_y \rd x + \map g {y, z} } \mathbf k$
We can check that this definition of $\mathbf A$ is correct by computing $V_x$:
{{begin-eqn}}
{{eqn | l = V_x
      | r = \frac {\partial A_z} {\partial y} - \frac {\partial A_y} {\partial z}
      | c = $\mathbf V = \nabla \times \mathbf A$
}}
{{eqn | r = \frac \partial {\partial y} \paren {-\int V_y \rd x + \map g {y, z} } - \frac \partial {\partial z} \paren {\int V_z \rd x + \map f {y, z} }
}}
{{eqn | r = \int \paren {-\frac {\partial V_y} {\partial y} - \frac {\partial V_z} {\partial z} } \d x + \map h {y, z}
      | c = $\map h {y, z} = \int \paren {\frac {\partial g} {\partial y} - \frac {\partial g} {\partial z} } \d x$
}}
{{eqn | r = \int \frac {\partial V_x} {\partial x} \rd x + \map h {y, z}
      | c = $\nabla \cdot \mathbf V = 0$
}}
{{end-eqn}}
This last expression equals $V_x$ with the proper choice of $\map h {y, z}$.
{{qed}}
Theorem: Vector Fields with Equal Curls Differ by Gradient

Let $\mathbf A$ and $\mathbf B$ be vector fields over ordinary space.
Then:
:$\nabla \times \mathbf A = \nabla \times \mathbf B$
{{iff}}
:$\mathbf B = \mathbf A + \grad f$
where $f$ is a scalar function with continuous partial derivatives.

Proof:
=== Sufficient Condition ===
We have:
{{begin-eqn}}
{{eqn | l = \nabla \times \mathbf B
      | r = \nabla \times \mathbf A
}}
{{eqn | ll= \leadsto
      | l = \nabla \times \mathbf B - \nabla \times \mathbf A
      | r = \mathbf 0
}}
{{eqn | ll= \leadsto
      | l = \nabla \times \paren {\mathbf B - \mathbf A}
      | r = \mathbf 0
      | c = Vector Cross Product Operator is Bilinear
}}
{{end-eqn}}
{{Recall|Conservative Vector Field|conservative vector field|index = 1}}
{{:Definition:Conservative Vector Field/Definition 1}}
Therefore, $\mathbf B - \mathbf A$ is conservative.
{{Recall|Conservative Vector Field|conservative vector field|index = 5}}
{{:Definition:Conservative Vector Field/Definition 5}}
Therefore, there exists a scalar function $f$ such that:
:$\mathbf B - \mathbf A = \grad f$
And thus:
:$\mathbf B = \mathbf A + \grad f$
{{qed|lemma}}
Theorem: Open Unit Interval on Rational Number Space is not Compact

Let $T = \struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Then:
:$\openint 0 1 \cap \Q$ is not compact in $T$
where $\openint 0 1$ is the open unit interval.

Proof:
From the Heine-Borel Theorem on a metric space, $\openint 0 1 \cap \Q$ is compact {{iff}} it is both totally bounded and complete.
From Rational Number Space is not Complete Metric Space it follows that $\openint 0 1 \cap \Q$ is not compact.
{{qed}}
Theorem: Totally Bounded Metric Space is not necessarily Compact

A metric space which is totally bounded is not necessarily also compact.

Proof:
Let $T = \struct {\closedint 0 1 \cap \Q, \tau_d}$ be the closed unit interval of rational numbers under the usual (Euclidean) topology.
From Open Unit Interval on Rational Number Space is Totally Bounded, $T$ is a totally bounded space.
But from Open Unit Interval on Rational Number Space is not Compact, $T$ is not a compact space.
Hence the result.
{{qed}}
Theorem: Second-Countable Metric Space is not necessarily Totally Bounded

A metric space which is second-countable is not necessarily also totally bounded.

Proof:
{{ProofWanted|somewhere down in the depths of item $134$}}
Theorem: Complete Metric Space is not necessarily Compact

Let $M = \struct {A, d}$ be a complete metric space.
Then it is not necessarily the case that $M$ is also compact.

Proof:
Let $M = \struct {\R, d}$ denote the real number line under the Euclidean metric.
From Real Number Line is Complete Metric Space, $M$ is a complete metric space.
From Real Number Line is not Compact, $M$ is not compact.
{{qed}}
Theorem: Completely Metrizable Space is Non-Meager

Let $T = \struct {S, \tau}$ be a completely metrizable topological space.
Then $T$ is non-meager.

Proof:
Let $T = \struct {S, \tau}$ be completely metrizable.
By Complete Metric Space is Non-Meager, it follows that $T$ is non-meager.
{{qed}}
{{ADC|Complete Metric Space is Non-Meager}}
Theorem: Regular Lindelöf Space is Paracompact

Let $T = \struct {S, \tau}$ be a regular space which is also Lindelöf.
Then $T$ is paracompact.

Proof:
Let $T = \struct {S, \tau}$ be a regular Lindelöf space.
{{Recall|Regular Space|index = 1}}
{{:Definition:Regular Space/Definition 1}}
Hence $T$ is {{afortiori}} a $T_3$ space.
The result follows from $T_3$ Lindelöf Space is Paracompact.
{{qed}}
Category:Regular Spaces
Category:Lindelöf Spaces
Category:Paracompact Spaces
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Theorem: Radial Interval Space is Completely Normal

The radial interval space is completely normal.

Proof:
{{ProofWanted}}
Category:Radial Interval Topology
Category:Examples of Completely Normal Spaces
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Theorem: Radial Interval Space is Paracompact

The radial interval space is paracompact.

Proof:
{{ProofWanted}}
Category:Radial Interval Topology
Category:Examples of Paracompact Spaces
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Theorem: Radial Interval Space is not Metrizable

The radial interval space is not metrizable.

Proof:
{{ProofWanted}}
Category:Radial Interval Topology
Category:Examples of Metrizable Spaces
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Theorem: Quasiuniformity Inducing Topology may not be Unique

Let $T = \struct {S, \tau}$ be a topological space.
Let $\UU$ be a quasiuniformity on a set $S$ such that $T$ is the topology induced by $\UU$.
Then it is not necessarily the case that $\UU$ is the only quasiuniformity that induces $T$.

Proof:
{{ProofWanted|we look at item $44$ in S&S}}
Theorem: Spectral Theorem for Hermitian Matrices

Let $\mathbf A$ be a square matrix over $\mathbb C$.
Then $\mathbf A$ is Hermitian {{Iff}} it is diagonalizable to a real diagonal matrix via a unitary transformation.
That is, we can write:
:$\mathbf A = \mathbf U \mathbf D \mathbf U^\dagger$
where:
:$\mathbf D$ is a real diagonal matrix
:$\mathbf U$ is a unitary matrix
:$\mathbf U^\dagger$ is the Hermitian conjugate of $\mathbf U$.

Proof:
=== Necessary Case ===
Let $\mathbf A$ be diagonalizable to a real diagonal matrix via a unitary transformation.
Then:
{{begin-eqn}}
{{eqn | l = \mathbf A^\dagger
      | r = \paren {\mathbf U \mathbf D \mathbf U^\dagger}^\dagger
}}
{{eqn | r = \mathbf U \mathbf D^\dagger \mathbf U^\dagger
      | c = Hermitian Conjugate of Matrix Product and Hermitian Conjugate is Involution
}}
{{eqn | r = \mathbf U \mathbf D \mathbf U^\dagger
      | c = $\mathbf D$ is real and diagonal
}}
{{eqn | r = \mathbf A
}}
{{end-eqn}}
Therefore, $\mathbf A$ is Hermitian.
{{qed|lemma}}
Theorem: Set of Real Orthogonal Matrices is Compact in Real Square Matrix Vector Space

let $n$ be a positive integer.
Then, $\mathrm O (n, \mathbb R)$ is a compact subset of the vector space $\mathcal M _n(\mathbb R)$

Proof:
{{Proofread}}
Theorem: Definition:QR Factorization:Existance Of QR Factorization For Square Matrices

Let $\mathbf A$ be a real square matrix.
Then $\mathbf A$ has a QR factorization, that is we can write
: $\mathbf A = \mathbf Q \mathbf R $
where
: $\mathbf Q$ is orthogonal
: $\mathbf R$ is upper triangular

Proof:
{{ProofWanted}}
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Theorem: Product Formula for Cosine

:$\ds \map \cos {n z} = \paren {-1}^{\floor {n/2} } 2^{n - 1} \prod_{k \mathop = 0}^{n - 1} \map \cos {z + \frac {\paren {\map {\cos^2} {\frac {n \pi} 2} + 2 k} \pi} {2 n} }$

Proof:
=== Lemma 1 ===
{{:Product Formula for Cosine/Lemma 1}}{{qed|lemma}}
Theorem: Trivial Topological Space has no Limit Point

Let $T = \struct {S, \tau}$ be a trivial topological space.
Then $T$ has no limit points.

Proof:
By definition of trivial topological space, $S$ is a singleton: $S = \set s$, say.
{{Recall|Limit Point of Set/Definition from Open Neighborhood|limit point of $H \subseteq S$}}
{{:Definition:Limit Point of Set/Definition from Open Neighborhood}}
As $S$ is a singleton, the only non-empty subset of $S$ is $S$ itself.
As $s$ is the only point of $S$, there is trivially no point of $S$ distinct from $s$.
Hence the result.
{{qed}}
Category:Trivial Topological Spaces
Category:Examples of Limit Points
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Theorem: Finite Particular Point Space is Weakly Countably Compact

Let $T = \struct {S, \tau_p}$ be a finite particular point space.
Then $T$ is weakly countably compact.

Proof:
Let $T = \struct {S, \tau_p}$ be a finite particular point space.
By definition, $T$ is a finite topological space.
The result follows from Finite Space Satisfies All Compactness Properties.
{{qed}}
Theorem: Finite Particular Point Space is Compact

Let $T = \struct {S, \tau_p}$ be a finite particular point space.
Then $T$ is  compact.

Proof:
By definition, $T$ is a finite topological space.
The result follows from Finite Topological Space is Compact.
{{qed}}
Theorem: Open Extension of Non-Trivial Particular Point Space is T0

$T^*_{\bar q}$ is a $T_0$ space.

Proof:
We have that a Particular Point Space is $T_0$.
Then from Condition for Open Extension Space to be $T_0$, it follows that $T^*_{\bar q}$ is a $T_0$ space.
{{qed}}
Theorem: Open Extension of Non-Trivial Particular Point Space is T4

$T^*_{\bar q}$ is a $T_4$ space.

Proof:
We have directly that an Open Extension Topology is $T_4$.
{{qed}}
Theorem: Open Extension of Non-Trivial Particular Point Space is not T1

$T^*_{\bar q}$ is not a $T_1$ space.

Proof:
We have directly that an Open Extension Topology is not $T_1$.
{{qed}}
Theorem: Open Extension of Non-Trivial Particular Point Space is not T5

$T^*_{\bar q}$ is not a $T_5$ space.

Proof:
A Particular Point Topology with three points or more is not $T_4$.
From $T_5$ Space is $T_4$, it follows that $T$ is not a $T_5$ space.
It follows from Condition for Open Extension Space to be $T_5$ that $T^*_{\bar q}$ is not a $T_5$ space.
{{qed}}
Theorem: Product Formula for Tangent

:$\ds \map \tan {n z} = \begin{cases} \paren {-1}^r \ds \prod_{k \mathop = 0}^{n - 1} \dfrac {\map \sin {z + \dfrac {k \pi} n} } {\map \cos {z + \dfrac {\paren {2 k + 1 } \pi} {2 n} } } & : n = 2 r & r \in \N \\ \paren {-1}^r \ds \prod_{k \mathop = 0}^{n - 1} \map \tan {z + \dfrac {k \pi} n} & : n = 2 r + 1 & r \in \N \end{cases}$

Proof:
From Product Formula for Sine, we have:
:$\ds \map \sin {n z} = 2^{n - 1} \prod_{k \mathop = 0}^{n - 1} \map \sin {z + \frac {k \pi} n}$
From Product Formula for Cosine Lemma 1, we have:
When $n \equiv 1 \pmod 4$
Then: 
:$\ds \map \cos {n z} = 2^{n - 1} \prod_{k \mathop = 0}^{n - 1} \map \cos {z + \dfrac {2 k \pi} {2 n} }$
From Product Formula for Cosine Lemma 2, we have:
When $n \equiv 3 \pmod 4$
Then: 
:$\ds \map \cos {n z} = -2^{n - 1} \prod_{k \mathop = 0}^{n - 1} \map \cos {z + \dfrac {2 k \pi} {2 n} }$
Therefore, when $n$ is odd $\paren {n = 2 r + 1}$, we have:
:$\ds \map \tan {n z} = \paren {-1}^r \ds \prod_{k \mathop = 0}^{n - 1} \map \tan {z + \frac {k \pi} n}$
From Product Formula for Cosine Lemma 3, we have:
When $n \equiv 2 \pmod 4$
Then: 
:$\ds \map \cos {n z} = -2^{n - 1} \prod_{k \mathop = 0}^{n - 1} \map \cos {z + \frac {\paren {2 k + 1 } \pi} {2 n} }$
From Product Formula for Cosine Lemma 4, we have:
When $n \equiv 0 \pmod 4$
Then: 
:$\ds \map \cos {n z} = 2^{n - 1} \prod_{k \mathop = 0}^{n - 1} \map \cos {z + \frac {\paren {2 k + 1 } \pi} {2 n} }$
Therefore, when $n$ is even $\paren {n = 2 r }$, we have:
:$\ds \map \tan {n z} = \paren {-1}^r \ds \prod_{k \mathop = 0}^{n - 1} \dfrac {\map \sin {z + \dfrac {k \pi} n} } {\map \cos {z + \dfrac {\paren {2 k + 1 } \pi} {2 n} } }$
Combining these results, we obtain:
:$\ds \map \tan {n z} = \begin{cases} \paren {-1}^r \ds \prod_{k \mathop = 0}^{n - 1} \dfrac {\map \sin {z + \dfrac {k \pi} n} } {\map \cos {z + \dfrac {\paren {2 k + 1 } \pi} {2 n} } } & : n = 2 r & r \in \N \\ \paren {-1}^r \ds \prod_{k \mathop = 0}^{n - 1} \map \tan {z + \dfrac {k \pi} n} & : n = 2 r + 1 & r \in \N \end{cases}$
{{qed}}
Theorem: Countable Complement Space is not T3

Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ is not a $T_3$ space.

Proof:
Let $S$ be an uncountable set.
Let $T = \struct {S, \tau}$ be the countable complement space on $S$.
We have that:
:Countable Complement Space is a $T_1$ space
From $T_1$ Space is $T_0$ Space, $T$ is a $T_0$ space.
{{AimForCont}} $T$ is a $T_3$ space.
From Space which is $T_3$ and $T_0$ is also $T_2$, $T$ is a $T_2$ space.
But from Countable Complement Space is not $T_2$, $T$ is not a $T_2$ space.
From this contradiction it follows that $T$ is not a $T_3$ space.
{{qed}}
Theorem: Countable Complement Space is not T4

Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ is not a $T_4$ space.

Proof:
We have that a Normal Space is $T_3$.
But from Countable Complement Space is not $T_3$, $T$ cannot be a normal space.
By definition, a normal space is a space that is both a $T_1$ space and a $T_4$ space.
But $T$ is a $T_1$ space and not a normal space.
So it follows that $T$ cannot be a $T_4$ space.
{{qed}}
Theorem: Countable Complement Space is not T5

Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ is not a $T_5$ space.

Proof:
A $T_5$ Space is $T_4$.
But from Countable Complement Space is not $T_4$, $T$ cannot be a $T_5$ space.
{{qed}}
Theorem: Compact Complement Space is not T2

Let $T = \struct {\R, \tau}$ be the compact complement space on $\R$.
Then $T$ is not a $T_2$ (Hausdorff) space.

Proof:
We have:
:Compact Complement Topology is Irreducible
:Irreducible Hausdorff Space is Singleton
Hence the result.
{{qed}}
Theorem: Compact Complement Space is not T2.5

Let $T = \struct {\R, \tau}$ be the compact complement topology on $\R$.
Then $T$ is not a $T_{2 \frac 1 2}$ space.

Proof:
From Compact Complement Space is not $T_2$, $T$ is not a $T_2$ space.
So from $T_{2 \frac 1 2}$ Space is $T_2$, it cannot be the case that $T$ is a $T_{2 \frac 1 2}$ space.
{{qed}}
Theorem: Compact Complement Space is not T3

Let $T = \struct {\R, \tau}$ be the compact complement topology on $\R$.
Then $T$ is not a $T_3$ space.

Proof:
We have that a Compact Complement Space is a $T_0$ space.
From Compact Complement Space is not Regular, $T$ is not a regular space.
By definition, a regular space is a space that is both a $T_0$ space and a $T_3$ space.
But $T$ is a $T_0$ space and not a regular space.
So it follows that $T$ cannot be a $T_3$ space.
{{qed}}
Theorem: Compact Complement Space is not T4

Let $T = \struct {\R, \tau}$ be the compact complement space on $\R$.
Then $T$ is not a $T_4$ space.

Proof:
We have that a Compact Complement Space is a $T_1$ space.
By definition, a normal space is a space that is both a $T_1$ space and a $T_4$ space.
But from Compact Complement Space is not Normal, $T$ is not a normal space.
So it follows that $T$ cannot be a $T_4$ space.
{{qed}}
Theorem: Compact Complement Space is not T5

Let $T = \struct {\R, \tau}$ be the compact complement topology on $\R$.
Then $T$ is not a $T_5$ space.

Proof:
We have that a $T_5$ Space is $T_4$.
We have that a Compact Complement Space is not a $T_4$ space.
So as $T$ is not a $T_4$ space, $T$ cannot be a $T_5$ space.
{{qed}}
Theorem: Compact Complement Space is T0

Let $T = \struct {\R, \tau}$ be the compact complement space on $\R$.
Then $T$ is a $T_0$ space.

Proof:
We have that a Compact Complement Space is a $T_1$ space.
From $T_1$ Space is $T_0$ Space, $T$ is a $T_0$ space.
{{qed}}
Category:Compact Complement Topology
Category:Examples of T0 Spaces
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Theorem: Compact Complement Space is not Regular

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

Proof:
{{Recall|Regular Space|regular space}}
{{:Definition:Regular Space/Definition 3}}
From Compact Complement Space is not $T_2$, $T$ is not a $T_2$ space.
Hence the result.
{{qed}}
Category:Compact Complement Topology
Category:Examples of Regular Spaces
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Theorem: Compact Complement Space is not Normal

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

Proof:
We have that a Compact Complement Space is $T_1$.
We also have that a Normal Space is a $T_3$.
But as $T$ is not a $T_3$ space, $T$ cannot be a normal space.
{{qed}}
Category:Compact Complement Topology
Category:Examples of Normal Spaces
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Theorem: Similar Matrices have Same Characteristic Polynomial

Let $\mathbf A$ and $\mathbf B$ be similar $n$ by $n$ square matrices.
Then $\mathbf A$ and $\mathbf B$ have the same characteristic polynomial.

Proof:
Because $\mathbf A$ and $\mathbf B$ are similar, there exists an invertible matrix $\mathbf P$ such that:
:$\mathbf A = \mathbf P \mathbf B \mathbf P^{-1}$
Then:
{{begin-eqn}}
{{eqn | l = \map \det {X\mathbf I_n - \mathbf A}
      | r = \map \det {X \mathbf I_n - \mathbf P \mathbf B \mathbf P^{-1} }
}}
{{eqn | r = \map \det {X \mathbf I_n \mathbf P \mathbf P^{-1} - \mathbf P \mathbf B \mathbf P^{-1} }
}}
{{eqn | r = \map \det {\mathbf P X \mathbf I_n \mathbf P^{-1} - \mathbf P \mathbf B \mathbf P^{-1} }
}}
{{eqn | r = \map \det {\mathbf P \paren {X \mathbf I_n - \mathbf B} \mathbf P^{-1} }
}}
{{end-eqn}}
By Determinant of Matrix Product:
:$\map \det {\mathbf P \paren {X \mathbf I_n - \mathbf B} \mathbf P^{-1} } = \map \det {\mathbf P} \map \det {X \mathbf I_n - \mathbf B} \map \det {\mathbf P^{-1} }$
By Determinant of Inverse Matrix:
:$\map \det {\mathbf P} \map \det {\mathbf P^{-1} } = 1$
Hence:
:$\map \det {X \mathbf I_n - \mathbf A} = \map \det {X \mathbf I_n - \mathbf B}$
{{qed}}
Category:Characteristic Polynomial of Matrix
Category:Matrix Similarity
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Theorem: Limit of Image of Sequence/Topology

Let $\struct {X, \tau_X}$ and $\struct {Y, \tau_Y}$ be topological spaces. 
Let $f: X \to Y$ be a continuous at (the point) $x$.
Let $\sequence {x_n}$ be a convergent sequence to the limit $x$.
Then the sequence $\sequence {\map f {x_n}}$ is convergent to the limit $f(x)$.

Proof:
Let $V$ be an arbitrary open neighborhood of $\map f x$. 
By continuous at (the point) $f(x)$ there exists an open neighborhood $U$ of $x$ such that $f[U] \subseteq V$. 
By convergence of $\sequence {x_n}$ to $x$ there exists $N \in \R_{\ge 0}$ such that:
:$\forall n \in \N : n \ge N \implies x_n \in U$
Thus we found $N\in \R_{\ge 0}$ such that:
:$\forall n \in N: n \ge N \implies \map f {x_n} \in V$
Because $V$ was taken as an arbitary open neighborhood of $\map f x$ we got $\sequence {\map f {x_n}}$ is convergent to the limit $f(x)$ by definition. 
{{qed}}
Category:Continuous Mappings (Topology)
Category:Convergent Sequences (Topology)
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Theorem: Fort Space is T2

Let $T = \struct {S, \tau_p}$ be a Fort space on an infinite set $S$.
Then $T$ is a $T_2$ (Hausdorff) space.

Proof:
From Fort Space is $T_{2 \frac 1 2}$, $T$ is a $T_{2 \frac 1 2}$ space.
The result follows from $T_{2 \frac 1 2}$ Space is $T_2$.
{{qed}}
Category:Fort Spaces
Category:Examples of Hausdorff Spaces
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Theorem: Fort Space is T2.5

Let $T = \struct {S, \tau_p}$ be a Fort space on an infinite set $S$.
Then $T$ is a $T_{2 \frac 1 2}$ space.

Proof:
From Fort Space is Regular, $T$ is a regular space.
The result follows from Regular Space is $T_{2 \frac 1 2}$.
{{qed}}
Category:Fort Spaces
Category:Examples of T2.5 Spaces
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Theorem: Fortissimo Space is T0

Let $T = \struct {S, \tau_p}$ be a Fortissimo space.
Then $T$ is a $T_0$ space.

Proof:
Follows directly from:
:Fortissimo Space is $T_1$
:$T_1$ Space is $T_0$ Space
{{qed}}
Category:Fortissimo Spaces
Category:Examples of T0 Spaces
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Theorem: Arens-Fort Space is T2

Let $T = \struct {S, \tau}$ be the Arens-Fort space.
Then $T$ is a $T_2$ (Hausdorff) space.

Proof:
We have:
:Fort Space is $T_2$
:Arens-Fort Space is Expansion of Fort Space.
The result follows from $T_2$ Property is Preserved under Expansion.
{{qed}}
Theorem: Arens-Fort Space is T0

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

Proof:
We have that the Arens-Fort Space is $T_1$.
The result follows from $T_1$ Space is $T_0$ Space.
{{qed}}
Theorem: Arens-Fort Space is Normal

Let $T = \struct {S, \tau}$ be the Arens-Fort space.
Then $T$ is a normal space.

Proof:
We have:
:Arens-Fort Space is Completely Normal
:Completely Normal Space is Normal.
{{qed}}
Category:Arens-Fort Space
Category:Examples of Normal Spaces
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Theorem: Arens-Fort Space is Completely Regular

Let $T = \struct {S, \tau}$ be the Arens-Fort space.
Then $T$ is a completely regular space.

Proof:
We have:
:Arens-Fort Space is Normal
:Normal Space is Completely Regular.
{{qed}}
Category:Arens-Fort Space
Category:Examples of Completely Regular Spaces
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Theorem: Arens-Fort Space is Regular

Let $T = \struct {S, \tau}$ be the Arens-Fort space.
Then $T$ is a regular space.

Proof:
We have:
:Arens-Fort Space is Completely Regular
:Completely Regular Space is Regular.
{{qed}}
Category:Arens-Fort Space
Category:Examples of Regular Spaces
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Theorem: Arens-Fort Space is T3

Let $T = \struct {S, \tau}$ be the Arens-Fort space.
Then $T$ is a $T_3$ space.

Proof:
We have  Arens-Fort Space is Regular.
{{Recall|Regular Space|index = 1}}
{{:Definition:Regular Space/Definition 1}}
The result follows.
{{qed}}
Category:Arens-Fort Space
Category:Examples of T3 Spaces
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Theorem: Modified Fort Space is T0

Let $T = \struct {S, \tau_{a, b} }$ be a modified Fort space.
Then $T$ is a $T_0$ space.

Proof:
We have that a Modified Fort Space is $T_1$.
From $T_1$ Space is $T_0$ Space, $T$ is a $T_0$ space.
{{qed}}
Category:Modified Fort Spaces
Category:Examples of T0 Spaces
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Theorem: Modified Fort Space is not T3

Let $T = \struct {S, \tau_{a, b} }$ be a modified Fort space.
Then $T$ is not a $T_3$ space.

Proof:
We have that a Modified Fort Space is $T_0$.
We then have that a Modified Fort Space is not Regular.
{{Recall|Regular Space|index = 1}}
{{:Definition:Regular Space/Definition 1}}
But $T$ is a $T_0$ space and not a regular space.
So it follows that $T$ cannot be a $T_3$ space.
{{qed}}
Theorem: Modified Fort Space is not Normal

Let $T = \struct {S, \tau_{a, b} }$ be a modified Fort space.
Then $T$ is not a normal space.

Proof:
We have:
:Normal Space is $T_3$
:Modified Fort Space is not $T_3$
Hence the result.
{{qed}}
Category:Modified Fort Spaces
Category:Examples of Normal Spaces
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Theorem: Modified Fort Space is not Regular

Let $T = \struct {S, \tau_{a, b} }$ be a modified Fort space.
Then $T$ is not a regular space.

Proof:
{{Recall|Regular Space|regular space}}
{{:Definition:Regular Space/Definition 3}}
We have:
:Modified Fort Space is not $T_2$
Hence the result.
{{qed}}
Category:Modified Fort Spaces
Category:Examples of Regular Spaces
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