avewright/proofwiki-math huggingface dataset:
Theorem: Regular Closed Set is Closed

Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$ be regular closed in $T$.
Then $H$ is an closed set of $T$.

Proof:
{{Recall|Regular Closed Set|regular closed set}}
{{:Definition:Regular Closed Set}}
{{Recall|Closure (Topology)|set closure}}
{{:Definition:Closure (Topology)/Definition 3}}
Hence:
:as $H$ equals the closure of its interior
the result follows from Topological Closure is Closed.
{{qed}}
Category:Regular Closed Sets
Category:Closed Sets
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Theorem: Continuity Defined by Closure/Necessary Condition

Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $f: T_1 \to T_2$ be a mapping.
Let $f$ be continuous.
Then:
:$\forall H \subseteq S_1: f \sqbrk {H^-} \subseteq \paren {f \sqbrk H}^-$
where $H^-$ denotes the closure of $H$ in $T_1$.
That is, the image of the closure is a subset of the closure of the image.

Proof:
Let $f$ be continuous.
Let $y \in f \sqbrk {\map \cl H}$.
Then:
:$\exists x \in \map \cl H: y = \map f x$
Let $U$ be an open set of $T_2$ such that $y \in U$.
Then by definition of continuous mapping:
:$f^{-1} \sqbrk U$ is an open set of $T_1$ such that:
::$x \in f^{-1} \sqbrk U$
Hence:
:$f^{-1} \sqbrk U \cap H \ne \O$
as $x \in \map \cl H$.
Hence:
{{begin-eqn}}
{{eqn | l = U \cap f \sqbrk H
      | o = \supseteq
      | r = f \sqbrk {f^{-1} \sqbrk U \cap H}
      | c = 
}}
{{eqn | o = \ne
      | r = \O
      | c = 
}}
{{end-eqn}}
and:
:$y \in \map \cl {f \sqbrk H}$
That is:
:$f \sqbrk {\map \cl H} \subseteq \map \cl {f \sqbrk H}$
Theorem: Regular Space is not necessarily Normal

A topological space which is a regular space is not necessarily also a normal space.

Proof:
Let $T$ be a Niemytzki plane.
From Niemytzki Plane is Completely Regular, $T$ is a completely regular space.
Hence by Completely Regular Space is Regular, $T$ is a regular space.
But by Niemytzki Plane is not Normal, $T$ is not a normal space.
Hence the result.
{{qed}}
Theorem: T2 Space is not necessarily Regular

A topological space which is a $T_2$ (Hausdorff) space is not necessarily also a regular space.

Proof:
Let $T$ be an irrational slope topological space.
From Irrational Slope Space is $T_2$, $T$ is a $T_2$ (Hausdorff) space.
But from Irrational Slope Space is not Regular, $T$ is not a regular space.
Hence the result.
{{qed}}
Theorem: T1 Space is not necessarily T2

A topological space which is a $T_1$ space is not necessarily also a $T_2$ (Hausdorff) space.

Proof:
Let $T = \struct {S, \tau}$ be the finite complement topology on an infinite set $S$.
From Finite Complement Space is $T_1$, $T$ is a $T_1$ space.
But from Finite Complement Space is not $T_2$, $T$ is not a $T_2$ (Hausdorff) space.
Hence the result.
{{qed}}
Theorem: T4 and T0 Space is not necessarily T3

A topological space which is both a $T_4$ space and a $T_0$ space is not necessarily also a $T_3$ space.

Proof:
Let $T$ be a Hjalmar Ekdal space.
From Hjalmar Ekdal Space is $T_4$, $T$ is a $T_4$ space.
From Hjalmar Ekdal Space is $T_0$, $T$ is a $T_0$ space.
But from Hjalmar Ekdal Space is not $T_3$, $T$ is not a $T_3$ space.
Hence the result.
{{qed}}
Theorem: T5 and T0 Space is not necessarily T3

A topological space which is both a $T_5$ space and a $T_0$ space is not necessarily also a $T_3$ space.

Proof:
Let $T$ be a Hjalmar Ekdal space.
From Hjalmar Ekdal Space is $T_5$, $T$ is a $T_5$ space.
From Hjalmar Ekdal Space is $T_0$, $T$ is a $T_0$ space.
But from Hjalmar Ekdal Space is not $T_3$, $T$ is not a $T_3$ space.
Hence the result.
{{qed}}
Theorem: Half-Disc Space is not Regular

Let $T$ be a half-disc space.
Then $T$ is not a regular space.

Proof:
From Half-Disc Space is not Semiregular, $T$ is not a semiregular space.
From Regular Space is Semiregular, it follows that $T$ cannot therefore be a regular space.
{{qed}}
Category:Half-Disc Topology
Category:Examples of Regular Spaces
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Theorem: Overlapping Interval Space fulfils no Separation Axioms but T0

Let $T$ be the overlapping interval space.
Then $T$ fulfils none of the Tychonoff separation axioms but the $T_0$ axiom.

Proof:
From Overlapping Interval Space is $T_0$ we note that indeed $T$ is a $T_0$ space.
From Overlapping Interval Space is not $T_1$, $T$ is not a $T_1$ space.
From Overlapping Interval Space is not $T_4$, $T$ is not a $T_4$ space.
The result follows from Sequence of Implications of Separation Axioms and the Rule of Transposition as follows:
We have that:
:$T$ is not a $T_4$ space.
From $T_5$ space is $T_4$:
:$T$ is not a $T_5$ space.
As a normal space is a $T_4$ space by definition:
:$T$ is not a normal space.
As a perfectly $T_4$ space is a $T_4$ space by definition:
:$T$ is not a perfectly $T_4$ space.
As a perfectly normal space is a perfectly $T_4$ space by definition:
:$T$ is not a perfectly normal space.
As a completely normal space is a $T_5$ space by definition:
:$T$ is not a completely normal space.
We have that:
:$T$ is not a $T_1$ space.
From $T_2$ Space is $T_1$:
:$T$ is not a $T_2$ space.
As a $T_2$ space is a semiregular space by definition:
:$T$ is not a semiregular space.
From $T_{2 \frac 1 2}$ Space is $T_2$:
:$T$ is not a $T_{2 \frac 1 2}$ space.
From both Regular Space is $T_{2 \frac 1 2}$ and Regular Space is Semiregular:
:$T$ is not a regular space.
From Urysohn Space is $T_{2 \frac 1 2}$:
:$T$ is not an Urysohn space.
From Completely Regular Space is Urysohn:
:$T$ is not a completely regular space.
{{AimForCont}} $T$ were a $T_3$ space.
As $T$ is a $T_0$ space, it follows that $T$ would be a regular space.
However, this would contradict the fact that $T$ is not a regular space.
Hence:
:$T$ is not a $T_3$ space.
From $T_{3 \frac 1 2}$ Space is $T_3$ Space:
:$T$ is not a $T_{3 \frac 1 2}$ space.
All the Tychonoff separation axioms have now been accounted for.
{{qed}}
Category:Overlapping Interval Topology
Category:Separation Axioms
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Theorem: Overlapping Interval Space is T0

Let $T$ be the overlapping interval space.
Then $T$ is a $T_0$ space.

Proof:
Let $a, b \in \closedint {-1} 1$ such that $a < b$.
Let $c \in \closedint {-1} 1$ such that $a < c < b$ but $c \ne 0$.
;Proof by Cases
$(1): \quad$ Let $c < 0$.
By Open Sets of Overlapping Interval Space, $U := \hointl c 1$ is an open set of $T$.
Thus:
:$a \notin U$
but
:$b \in U$
and we see that $U$ fulfils the conditions for $a$ and $b$ to be elements of a $T_0$ space.
$(2): \quad$ Let $c > 0$.
By Open Sets of Overlapping Interval Space, $V := \hointr {-1} c$ is an open set of $T$.
Thus:
:$a \in V$
but
:$b \notin V$
and again we see that $V$ fulfils the conditions for $a$ and $b$ to be elements of a $T_0$ space.
In both cases, $a$ and $b$ are arbitrary.
Hence the conditions are fulfilled for all $a, b \in T$.
The result follows.
{{qed}}
Theorem: Overlapping Interval Space is not T1

Let $T$ be the overlapping interval space.
Then $T$ is not a $T_1$ space.

Proof:
{{Recall|T1 Space|index = 1|$T_1$ space}}
{{:Definition:T1 Space/Definition 1}}
Let $x \in S$ be arbitrary such that $x \ne 0$.
Consider the element $0 \in S$.
Let $U \in \tau$ such that $U \ne \O$.
From Open Sets of Overlapping Interval Space, $U$ is of the form of one of:
{{begin-axiom}}
{{axiom | n = 1
        | m = \hointl a 1
        | rc= where $-1 \le a < 0$
}}
{{axiom | n = 2
        | m = \hointr {-1} b
        | rc= where $0 < b \le 1$
}}
{{axiom | n = 3
        | m = \openint a b
        | rc= where $-1 \le a < 0 < b \le 1$
}}
{{axiom | n = 4
        | m = \closedint {-1} 1
        | rc= 
}}
{{end-axiom}}
It is seen that in all cases:
:$0 \in U$
Hence there exists no $U \in \tau$ such that $x \in U$ but $0 \notin U$.
The result follows.
{{qed}}
Theorem: Overlapping Interval Space is not T4

Let $T$ be the overlapping interval space.
Then $T$ is not a $T_4$ space.

Proof:
{{Recall|T4 Space|index = 1|$T_4$ space}}
{{:Definition:T4 Space/Definition 1}}
Consider the sets $\set {-1}$ and $\set 1$, both of which are subsets of $S$.
We have that:
{{begin-eqn}}
{{eqn | l = \relcomp S {\set {-1} }
      | r = \hointl {-1} 1
}}
{{eqn | l = \relcomp S {\set 1}
      | r = \hointr {-1} 1
}}
{{end-eqn}}
From Open Sets of Overlapping Interval Space, both $\hointl {-1} 1$ and $\hointr {-1} 1$ are open sets of $T$.
Hence, by definition, $\set {-1}$ and $\set 1$ are both closed sets of $T$.
Also from Open Sets of Overlapping Interval Space:
:an open set of $T$ which contains $\set 1$ is of the form $\hointl a 1$ where $a < 0$
:an open set of $T$ which contains $\set {-1}$ is of the form $\hointr {-1} b$ where $b > 0$.
But:
:$\hointl a 1 \cap \hointr {-1} b = \openint a b \ne \O$
Therefore there are no disjoint open sets of $T$ which contain $\set {-1}$ and $\set 1$ respectively.
Hence the result.
{{qed}}
Theorem: Hjalmar Ekdal Space is not Regular

Let $T = \struct {S, \tau}$ be a Hjalmar Ekdal space.
Then $T$ is ''not'' a regular space.

Proof:
{{Recall|Regular Space|regular space|index = 2}}
{{:Definition:Regular Space/Definition 2}}
Let $T = \struct {S, \tau}$ be a Hjalmar Ekdal space.
But from Hjalmar Ekdal Space is not $T_1$, $T$ is not a $T_1$ space.
Hence the result.
{{qed}}
Theorem: Hjalmar Ekdal Space is not T3.5

Let $T$ be a Hjalmar Ekdal space.
Then $T$ is ''not'' a $T_{3 \frac 1 2}$ space.

Proof:
We have that a $T_{3 \frac 1 2}$ Space is $T_3$.
From Hjalmar Ekdal Space is not $T_3$, $T$ is not a $T_3$ space.
The result follows by applying the Rule of Transposition.
{{qed}}
Theorem: Hjalmar Ekdal Space is not T3

Let $T = \struct {S, \tau}$ be a Hjalmar Ekdal space.
Then $T$ is ''not'' a $T_3$ space.

Proof:
{{AimForCont}} $T = \struct {S, \tau}$ were a $T_3$ space.
From Hjalmar Ekdal Space is $T_0$, $T$ is a $T_0$ space.
Then, by definition, $T$ would then be a regular space.
But from Hjalmar Ekdal Space is not Regular, $T$ is not a regular space.
The result follows by Proof by Contradiction.
{{qed}}
Theorem: Hjalmar Ekdal Space is not T1

Let $T = \struct {S, \tau}$ be a Hjalmar Ekdal space.
Then $T$ is ''not'' a $T_1$ space.

Proof:
;Proof by Counterexample
Consider the integers $1$ and $2$, both of which are elements of $S$.
Let $U \in S$ such that $1 \in U$.
Then, by definition of $T$, $2 \in U$.
Hence there exists no $U \in S$ such that $1 \in U$ but $2 \notin U$.
Hence the result by definition of $T_1$ space.
{{qed}}
Theorem: Normal Space is T3.5

Let $\struct {S, \tau}$ be a normal space.
Then $\struct {S, \tau}$ is also a $T_{3 \frac 1 2}$ space.

Proof:
{{Recall|T3.5 Space|$T_{3 \frac 1 2}$ Space}}
{{:Definition:T3.5 Space}}
{{Recall|Normal Space|normal space|index = 1}}
{{:Definition:Normal Space/Definition 1}}
Let $F$ be an arbitrary closed set of $T$.
Let $y \in S$ such that $y \notin F$.
{{Recall|T1 Space|$T_1$ Space|index = 3}}
{{:Definition:T1 Space/Definition 3}}
Thus $\set y$ is a closed set of $T$ such that $F \cap \set y = \O$.
We have that $T$ is a $T_4$ space.
We have that $F$ and $\set y$ are disjoint closed sets of $T$.
Hence from Urysohn's Lemma, there exists an Urysohn function for $F$ and $\set y$ such that $y \notin F$.
This is the definition of a $T_{3 \frac 1 2}$ space.
{{qed}}
Theorem: Exponential of Category of Sets

Let $\mathbf{Set}$ be the category of sets.
Let $B$ and $C$ be sets.
Let $C^B$ be the set of all mappings from $B$ to $C$.
Define a mapping $\epsilon : C^B \times B \to C$ by:
:$\map \epsilon {f,b} = \map f b$
where $C^B \times B$ denotes the Cartesian product of $C^B$ and $B$.
Then $\struct {C^B,\epsilon}$ is an exponential of $C$ by $B$.
Moreover, the transpose $\tilde f : A \to C^B$ of $f : A\times B \to C$ is given by:
:$\map {\tilde f} a = \map f {a, \cdot}$

Proof:
{{ProofWanted}}
Category:Category of Sets
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Theorem: Tangent of 36 Degrees

:$\tan 36 \degrees = \tan \dfrac \pi 5 = 5^{\frac 1 4} \phi^{-\frac 3 2}$
where $\tan$ denotes the tangent function.

Proof:
{{begin-eqn}}
{{eqn | l = \tan 36 \degrees
      | r = \frac {\sin 36 \degrees} {\cos 36 \degrees}
      | c = Tangent is Sine divided by Cosine
}}
{{eqn | r = \frac {2^{-1} 5^{\frac 1 4} \phi^{-\frac 1 2} } {2^{-1} \phi}
      | c = {{sin|36}} and {{cos|36}}
}}
{{eqn | r = 5^{\frac 1 4} \phi^{-\frac 3 2}
      | c = 
}}
{{end-eqn}}
{{qed}}
Category:Golden Mean
Category:Tangent Function
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Theorem: Tangent of 72 Degrees

:$\tan 72 \degrees = \tan \dfrac {2 \pi} 5 = 5^{\frac 1 4} \phi^{\frac 3 2}$
where $\tan$ denotes the tangent function.

Proof:
{{begin-eqn}}
{{eqn | l = \tan 72 \degrees
      | r = \frac {\sin 72 \degrees} {\cos 72 \degrees}
      | c = Tangent is Sine divided by Cosine
}}
{{eqn | r = \frac {2^{-1} 5^{\frac 1 4} \phi^{\frac 1 2} } {2^{-1} \phi^{-1} }
      | c = {{sin|72}} and {{cos|72}}
}}
{{eqn | r = 5^{\frac 1 4} \phi^{\frac 3 2}
      | c = 
}}
{{end-eqn}}
{{qed}}
Category:Golden Mean
Category:Tangent Function
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Theorem: Cotangent of 36 Degrees

:$\cot 36 \degrees = \cot \dfrac \pi 5 = 5^{-\frac 1 4} \phi^{\frac 3 2}$
where $\cot$ denotes the cotangent function.

Proof:
{{begin-eqn}}
{{eqn | l = \cot 36 \degrees
      | r = \frac {\cos 36 \degrees} {\sin 36 \degrees}
      | c = Cotangent is Cosine divided by Sine
}}
{{eqn | r = \frac {2^{-1} \phi} {2^{-1} 5^{\frac 1 4} \phi^{-\frac 1 2} }
      | c = {{sin|36}} and {{cos|36}}
}}
{{eqn | r = 5^{-\frac 1 4} \phi^{\frac 3 2}
      | c = 
}}
{{end-eqn}}
{{qed}}
Category:Cotangent Function
Category:Golden Mean
0z2uhnv4anjv4p3bqlpohw5u1r0uq7v
Theorem: Cotangent of 72 Degrees

:$\cot 72 \degrees = \cot \dfrac {2 \pi} 5 = 5^{-\frac 1 4} \phi^{-\frac 3 2}$
where $\cot$ denotes the cotangent function.

Proof:
{{begin-eqn}}
{{eqn | l = \cot 72 \degrees
      | r = \frac {\cos 72 \degrees} {\sin 72 \degrees}
      | c = Tangent is Sine divided by Cosine
}}
{{eqn | r = \frac {2^{-1} \phi^{-1} } {2^{-1} 5^{\frac 1 4} \phi^{\frac 1 2} }
      | c = {{sin|72}} and {{cos|72}}
}}
{{eqn | r = 5^{-\frac 1 4} \phi^{-\frac 3 2}
      | c = 
}}
{{end-eqn}}
{{qed}}
Category:Cotangent Function
Category:Golden Mean
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Theorem: Digamma Function of One Fifth

:$\map \psi {\dfrac 1 5} = -\gamma - \dfrac 5 4 \ln 5 - \dfrac \pi 2 5^{-\frac 1 4} \phi^{\frac 3 2} + \paren {\dfrac 1 2 - \phi} \ln \phi$

Proof:
{{begin-eqn}}
{{eqn | l = \map \psi {\frac 1 5}
      | r = -\gamma - \ln 10 - \frac \pi 2 \map \cot {\frac 1 5 \pi} + 2 \sum_{n \mathop = 1}^{\ceiling {5 / 2} - 1} \map \cos {\frac {2 \pi n} 5} \map \ln {\map \sin {\frac {\pi n} 5} }
      | c = Gauss's Digamma Theorem
}}
{{eqn | r = -\gamma - \ln 2 - \ln 5 - \frac \pi 2 \times \paren {5^{-\frac 1 4} \phi^{\frac 3 2} } + 2 \paren {\paren {2^{-1} \phi^{-1} } \times \map \ln {2^{-1} 5^{\frac 1 4} \phi^{-\frac 1 2} } + \paren {-2^{-1} \phi} \times \map \ln {2^{-1} 5^{\frac 1 4} \phi^{\frac 1 2} } }
      | c = {{cot|36}}, {{cos|72}}, {{cos|144}}, {{sin|36}}, {{sin|72}} and Sum of Logarithms
}}
{{eqn | r = -\gamma - \ln 2 - \ln 5 - \frac \pi 2 5^{-\frac 1 4} \phi^{\frac 3 2} + \phi^{-1} \map \ln {2^{-1} 5^{\frac 1 4} \phi^{-\frac 1 2} } - \phi \map \ln {2^{-1} 5^{\frac 1 4} \phi^{\frac 1 2} }
      | c = 
}}
{{eqn | r = -\gamma - \ln 2 - \ln 5 - \frac \pi 2 5^{-\frac 1 4} \phi^{\frac 3 2} + \phi^{-1} \paren {-\ln 2 + \frac 1 4 \ln 5 - \frac 1 2 \ln \phi }  - \phi \paren {-\ln 2 + \frac 1 4 \ln 5 + \frac 1 2 \ln \phi }
      | c = Sum of Logarithms
}}
{{eqn | r = -\gamma - \ln 2 - \ln 5 - \frac \pi 2 5^{-\frac 1 4} \phi^{\frac 3 2} + \ln 2 \paren {\phi - \phi^{-1} } + \frac 1 4 \ln 5 \paren {\phi^{-1} - \phi} - \frac 1 2 \ln \phi \paren {\phi^{-1} + \phi}
      | c = 
}}
{{eqn | r = -\gamma - \ln 2 - \ln 5 - \frac \pi 2 5^{-\frac 1 4} \phi^{\frac 3 2} + \ln 2 \paren {\phi - \paren {\phi - 1} } + \frac 1 4 \ln 5 \paren {\paren {\phi - 1} - \phi} - \frac 1 2 \ln \phi \paren {\paren {\phi - 1} + \phi}
      | c = {{Defof|Golden Mean|index = 3}}
}}
{{eqn | r = -\gamma - \ln 2 - \ln 5 - \frac \pi 2 5^{-\frac 1 4} \phi^{\frac 3 2} + \ln 2 - \frac 1 4 \ln 5 - \frac 1 2 \ln \phi \paren {2\phi - 1}
      | c = 
}}
{{eqn | r = -\gamma - \frac 5 4 \ln 5 - \frac \pi 2 5^{-\frac 1 4} \phi^{\frac 3 2} - \frac {\sqrt 5} 2 \ln \phi
      | c = 
}}
{{eqn | r = -\gamma - \frac 5 4 \ln 5 - \frac \pi 2 5^{-\frac 1 4} \phi^{\frac 3 2} + \paren {\frac 1 2 - \phi} \ln \phi
      | c = 
}}
{{end-eqn}}
{{qed}}
Category:Digamma Function of One Fifth
Category:Examples of Digamma Function
Category:Euler-Mascheroni Constant
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Theorem: Digamma Function of Two Fifths

:$\map \psi {\dfrac 2 5} = -\gamma - \dfrac 5 4 \ln 5 - \dfrac \pi 2 5^{-\frac 1 4} \phi^{-\frac 3 2} + \paren {\phi - \dfrac 1 2} \ln \phi$

Proof:
{{begin-eqn}}
{{eqn | l = \map \psi {\frac 2 5}
      | r = -\gamma - \ln 10 - \frac \pi 2 \map \cot {\frac 2 5 \pi} + 2 \sum_{n \mathop = 1}^{\ceiling {5 / 2} - 1} \map \cos {\frac {4 \pi n} 5} \map \ln {\map \sin {\frac {\pi n} 5} }
      | c = Gauss's Digamma Theorem
}}
{{eqn | r = -\gamma - \ln 2 - \ln 5 - \frac \pi 2 \times \paren {5^{-\frac 1 4} \phi^{-\frac 3 2} } + 2 \paren {\paren {-2^{-1} \phi } \times \map \ln {2^{-1} 5^{\frac 1 4} \phi^{-\frac 1 2} } + \paren {2^{-1} \phi^{-1} } \times \map \ln {2^{-1} 5^{\frac 1 4} \phi^{\frac 1 2} } }
      | c = {{cot|72}}, {{cos|144}}, {{cos|72}}, {{sin|36}}, {{sin|72}}, Cosine Function is Even and Sum of Logarithms
}}
{{eqn | r = -\gamma - \ln 2 - \ln 5 - \frac \pi 2 5^{-\frac 1 4} \phi^{-\frac 3 2} - \phi \map \ln {2^{-1} 5^{\frac 1 4} \phi^{-\frac 1 2} } + \phi^{-1} \map \ln {2^{-1} 5^{\frac 1 4} \phi^{\frac 1 2} }
      | c = 
}}
{{eqn | r = -\gamma - \ln 2 - \ln 5 - \frac \pi 2 5^{-\frac 1 4} \phi^{-\frac 3 2} - \phi \paren {-\ln 2 + \frac 1 4 \ln 5 - \frac 1 2 \ln \phi }  + \phi^{-1} \paren {-\ln 2 + \frac 1 4 \ln 5 + \frac 1 2 \ln \phi }
      | c = Sum of Logarithms
}}
{{eqn | r = -\gamma - \ln 2 - \ln 5 - \frac \pi 2 5^{-\frac 1 4} \phi^{-\frac 3 2} + \ln 2 \paren {\phi - \phi^{-1} } + \frac 1 4 \ln 5 \paren {\phi^{-1} - \phi} + \frac 1 2 \ln \phi \paren {\phi^{-1} + \phi}
      | c = 
}}
{{eqn | r = -\gamma - \ln 2 - \ln 5 - \frac \pi 2 5^{-\frac 1 4} \phi^{-\frac 3 2} + \ln 2 \paren {\phi - \paren {\phi - 1} } + \frac 1 4 \ln 5 \paren {\paren {\phi - 1} - \phi} + \frac 1 2 \ln \phi \paren {\paren {\phi - 1} + \phi}
      | c = {{Defof|Golden Mean|index = 3}}
}}
{{eqn | r = -\gamma - \ln 2 - \ln 5 - \frac \pi 2 5^{-\frac 1 4} \phi^{-\frac 3 2} + \ln 2 - \frac 1 4 \ln 5 + \frac 1 2 \ln \phi \paren {2\phi - 1}
      | c = 
}}
{{eqn | r = -\gamma - \frac 5 4 \ln 5 - \frac \pi 2 5^{-\frac 1 4} \phi^{-\frac 3 2} + \frac {\sqrt 5} 2 \ln \phi
      | c = 
}}
{{eqn | r = -\gamma - \frac 5 4 \ln 5 - \frac \pi 2 5^{-\frac 1 4} \phi^{-\frac 3 2} + \paren {\phi - \frac 1 2} \ln \phi
      | c = 
}}
{{end-eqn}}
{{qed}}
Category:Examples of Digamma Function
Category:Euler-Mascheroni Constant
kzve52znz8j3ju0vxg1v2clc5t1uszk
Theorem: Digamma Function of Three Fifths

:$\map \psi {\dfrac 3 5} = -\gamma - \dfrac 5 4 \ln 5 + \dfrac \pi 2 5^{-\frac 1 4} \phi^{-\frac 3 2} + \paren {\phi - \dfrac 1 2} \ln \phi$

Proof:
From Digamma Reflection Formula. we have:
:$\map \psi z - \map \psi {1 - z} = -\pi \cot \pi z$
Therefore:
{{begin-eqn}}
{{eqn | l = \map \psi {\frac 3 5} - \map \psi {\frac 2 5}
      | r = -\pi \map \cot {\frac {3 \pi} 5}
      | c = 
}}
{{eqn | ll = \leadsto
      | l = \map \psi {\frac 3 5}
      | r = \map \psi {\frac 2 5} - \pi \map \cot {\frac {3 \pi} 5}
      | c = 
}}
{{eqn | r = \paren {-\gamma - \dfrac 5 4 \ln 5 - \dfrac \pi 2 5^{-\frac 1 4} \phi^{-\frac 3 2} + \paren {\phi - \dfrac 1 2} \ln \phi} - \pi \paren {-5^{-\frac 1 4} \phi^{-\frac 3 2} }
      | c = Digamma Function of Two Fifths, {{cot|72}}, Cotangent Function is Odd, Cotangent of Angle plus Straight Angle
}}
{{eqn | r = -\gamma - \dfrac 5 4 \ln 5 + \dfrac \pi 2 5^{-\frac 1 4} \phi^{-\frac 3 2} + \paren {\phi - \dfrac 1 2} \ln \phi
      | c = 
}}
{{end-eqn}}
{{qed}}
Category:Examples of Digamma Function
Category:Euler-Mascheroni Constant
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Theorem: Digamma Function of Four Fifths

:$\map \psi {\dfrac 4 5} = -\gamma - \dfrac 5 4 \ln 5 + \dfrac \pi 2 5^{-\frac 1 4} \phi^{\frac 3 2} + \paren {\dfrac 1 2 - \phi} \ln \phi$

Proof:
From Digamma Reflection Formula. we have:
:$\map \psi z - \map \psi {1 - z} = -\pi \cot \pi z$
Therefore:
{{begin-eqn}}
{{eqn | l = \map \psi {\frac 4 5} - \map \psi {\frac 1 5}
      | r = -\pi \map \cot {\frac {4 \pi} 5}
      | c = 
}}
{{eqn | ll = \leadsto
      | l = \map \psi {\frac 4 5}
      | r = \map \psi {\frac 1 5} - \pi \map \cot {\frac {4 \pi} 5}
      | c = 
}}
{{eqn | r = \paren {-\gamma - \dfrac 5 4 \ln 5 - \dfrac \pi 2 5^{-\frac 1 4} \phi^{\frac 3 2} + \paren {\dfrac 1 2 - \phi} \ln \phi} - \pi \paren {-5^{-\frac 1 4} \phi^{\frac 3 2} }
      | c = Digamma Function of One Fifth, {{cot|36}}, Cotangent Function is Odd, Cotangent of Angle plus Straight Angle
}}
{{eqn | r = -\gamma - \dfrac 5 4 \ln 5 + \dfrac \pi 2 5^{-\frac 1 4} \phi^{\frac 3 2} + \paren {\dfrac 1 2 - \phi} \ln \phi
      | c = 
}}
{{end-eqn}}
{{qed}}
Category:Examples of Digamma Function
Category:Euler-Mascheroni Constant
2pur9cjjkgysbqytjqu6uuqv2f0k6pu
Theorem: Product Space is T0 iff Factor Spaces are T0/Necessary Condition

Let $\SS = \family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be an indexed family of non-empty topological spaces where $I$ is an arbitrary index set.
Let $\ds T = \struct {S, \tau} = \prod \struct {S_\alpha, \tau_\alpha}$ be the product space of $\SS$.
Let each of $\struct {S_\alpha, \tau_\alpha}$ be a $T_0$ space.
Then $T$ is a $T_0$ space.

Proof:
Let each of $\struct {S_\alpha, \tau_\alpha}$ be a $T_0$ space.
{{Recall|T0 Space|$T_0$ space|1}}
{{:Definition:T0 Space/Definition 1}}
{{AimForCont}} $T$ is not a $T_0$ space.
Then $\exists a, b \in S, a \ne b$ such that for all $U \in \tau$, either $a, b \in U$ or $a, b \notin U$.
Then $a$ and $b$ are different in at least one coordinate.
Suppose that $a_m = p, b_m = q$ for some coordinate $m$.
Then either $a_m, b_m \in U_m$ or $a_m, b_m \notin U_m$.
It follows that $\struct {S_m, \tau_m}$ is not a $T_0$ space.
Hence by Proof by Contradiction it follows that $T$ is a $T_0$ space.
Theorem: Product Space is T0 iff Factor Spaces are T0/Sufficient Condition

Let $\SS = \family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be an indexed family of non-empty topological spaces where $I$ is an arbitrary index set.
Let $\ds T = \struct {S, \tau} = \prod \struct {S_\alpha, \tau_\alpha}$ be the product space of $\SS$.
Let $T$ be a $T_0$ space.
Then each of $\struct {S_\alpha, \tau_\alpha}$ is a $T_0$ space.

Proof:
Let $T$ be a $T_0$ space.
As $S_\alpha \ne \O$ we also have $S \ne \O$ by the axiom of choice.
Let $\alpha \in I$.
From Subspace of Product Space is Homeomorphic to Factor Space, $\struct {S_\alpha, \tau_\alpha}$ is homeomorphic to a subspace $T_\alpha$ of $T$.
From $T_0$ Property is Hereditary, $T_\alpha$ is a $T_0$ space.
From $T_0$ Property is Preserved under Homeomorphism, $\struct {S_\alpha, \tau_\alpha}$ is a $T_0$ space.
As $\alpha \in I$ was arbitrary, it follows that each of $\struct {S_\alpha, \tau_\alpha}$ is a $T_0$ space.