avewright/proofwiki-math huggingface dataset:
Theorem: Closure of Hadamard Product

Let $\struct {S, \cdot}$ be an algebraic structure.
Let $\map {\MM_S} {m, n}$ be a $m \times n$ matrix space over $S$.
For $\mathbf A, \mathbf B \in \map {\MM_S} {m, n}$, let $\mathbf A \circ \mathbf B$ be defined as the Hadamard product of $\mathbf A$ and $\mathbf B$.
The operation $\circ$ is closed on $\map {\MM_S} {m, n}$ {{iff}} $\cdot$ is closed on $\struct {S, \cdot}$.

Proof:
=== Necessary Condition ===
Let the operation $\cdot$ be closed on $\struct {S, \cdot}$.
Let $\mathbf A = \sqbrk a_{m n}$ and $\mathbf B = \sqbrk b_{m n}$ be elements of $\map {\MM_S} {m, n}$.
Let $\sqbrk c_{m n} = \sqbrk a_{m n} \cdot \sqbrk b_{m n}$.
Then:
:$\forall i \in \closedint 1 m, j \in \closedint 1 n: c_{i j} = a_{i j} \circ b_{i j}$
Thus:
:$\struct {S, \cdot}$ is closed {{iff}} $c_{i j} \in S$.
From the definition of Hadamard product, $\sqbrk c_{m n}$ has the same order as both $\sqbrk a_{m n}$ and $\sqbrk b_{m n}$.
Thus it follows that:
:$\sqbrk c_{m n} \in \map {\MM_S} {m, n}$
Thus $\struct {\map {\MM_S} {m, n}, \circ}$, as it is defined, is closed.
{{qed|lemma}}
=== Sufficient Condition ===
Suppose $\struct {S, \cdot}$ is such that $\cdot$ is not closed on $\struct {S, \cdot}$.
Then there exists $a$ and $b$ such that:
:$a \cdot b \notin S$
Let $\mathbf A$ and $\mathbf B$ be elements of $\map {\MM_S} {m, n}$ such that:
:$a_{i j} = a$, $b_{i j} = b$
where:
:$a_{i j}$ is the $\tuple {i, j}$th element of $\mathbf A$
:$b_{i j}$ is the $\tuple {i, j}$th element of $\mathbf B$
Then:
:$a_{i j} \cdot b_{i j} \notin S$
That is:
:$\mathbf A \circ \mathbf B \notin \map {\MM_S} {m, n}$
because (at least) one element of $\mathbf A \circ \mathbf B$ is not an element of $S$.
That is, $\circ$ is not closed on $\map {\MM_S} {m, n}$.
{{qed}}
Theorem: Associativity of Hadamard Product

Let $\struct {S, \cdot}$ be an algebraic structure.
Let $\map {\MM_S} {m, n}$ be a $m \times n$ matrix space over $S$.
For $\mathbf A, \mathbf B \in \map {\MM_S} {m, n}$, let $\mathbf A \circ \mathbf B$ be defined as the Hadamard product of $\mathbf A$ and $\mathbf B$.
The operation $\circ$ is associative on $\map {\MM_S} {m, n}$ {{iff}} $\cdot$ is associative on $\struct {S, \cdot}$.

Proof:
=== Necessary Condition ===
Let the operation $\cdot$ be associative on $\struct {S, \cdot}$.
Let $\mathbf A = \sqbrk a_{m n}$, $\mathbf B = \sqbrk b_{m n}$ and $\mathbf C = \sqbrk c_{m n}$ be elements of the $m \times n$ matrix space over $S$.
Then:
{{begin-eqn}}
{{eqn | l = \paren {\mathbf A \circ \mathbf B} \circ \mathbf C
      | r = \paren {\sqbrk a_{m n} \circ \sqbrk b_{m n} } \circ \sqbrk c_{m n}
      | c = Definition of $\mathbf A$, $\mathbf B$ and $\mathbf C$
}}
{{eqn | r = \sqbrk {a \cdot b}_{m n} \circ \sqbrk c_{m n}
      | c = {{Defof|Hadamard Product}}
}}
{{eqn | r = \sqbrk {\paren {a \cdot b} \cdot c}_{m n}
      | c = {{Defof|Hadamard Product}}
}}
{{eqn | r = \sqbrk {a \cdot \paren {b \cdot c} }_{m n}
      | c = as $\cdot$ is associative
}}
{{eqn | r = \sqbrk a_{m n} \circ \sqbrk {b \cdot c}_{m n}
      | c = {{Defof|Hadamard Product}}
}}
{{eqn | r = \sqbrk a_{m n} \circ \paren {\sqbrk b_{m n} \circ \sqbrk c_{m n} }
      | c = {{Defof|Hadamard Product}}
}}
{{eqn | r = \mathbf A \circ \paren {\mathbf B \circ \mathbf C}
      | c = Definition of $\mathbf A$, $\mathbf B$ and $\mathbf C$
}}
{{end-eqn}}
That is, $\circ$ is associative on $\map {\MM_S} {m, n}$.
{{qed|lemma}}
Theorem: Commutativity of Hadamard Product

Let $\struct {S, \cdot}$ be an algebraic structure.
Let $\map {\MM_S} {m, n}$ be a $m \times n$ matrix space over $S$.
For $\mathbf A, \mathbf B \in \map {\MM_S} {m, n}$, let $\mathbf A \circ \mathbf B$ be defined as the Hadamard product of $\mathbf A$ and $\mathbf B$.
The operation $\circ$ is commutative on $\map {\MM_S} {m, n}$ {{iff}} $\cdot$ is commutative on $\struct {S, \cdot}$.

Proof:
=== Necessary Condition ===
Let the operation $\cdot$ be commutative on $\struct {S, \cdot}$.
Let $\mathbf A = \sqbrk a_{m n}$ and $\mathbf B = \sqbrk b_{m n}$ be elements of the $m \times n$ matrix space over $S$.
Then:
{{begin-eqn}}
{{eqn | l = \mathbf A \circ \mathbf B
      | r = \sqbrk a_{m n} \circ \sqbrk b_{m n}
      | c = Definition of $\mathbf A$ and $\mathbf B$
}}
{{eqn | r = \sqbrk {a \cdot b}_{m n}
      | c = {{Defof|Hadamard Product}}
}}
{{eqn | r = \sqbrk {b \cdot a}_{m n}
      | c = as $\cdot$ is commutative
}}
{{eqn | r = \sqbrk b_{m n} \circ \sqbrk a_{m n}
      | c = {{Defof|Hadamard Product}}
}}
{{eqn | r = \mathbf B \circ \mathbf A
      | c = Definition of $\mathbf A$ and $\mathbf B$
}}
{{end-eqn}}
That is, $\circ$ is commutative on $\map {\MM_S} {m, n}$.
{{qed|lemma}}
Theorem: Matrix Entrywise Addition over Ring is Closed

Let $\struct {R, +, \circ}$ be a ring.
Let $\map {\MM_R} {m, n}$ be a $m \times n$ matrix space over $R$.
For $\mathbf A, \mathbf B \in \map {\MM_R} {m, n}$, let $\mathbf A + \mathbf B$ be defined as the matrix entrywise sum of $\mathbf A$ and $\mathbf B$.
The operation $+$ is closed on $\map {\MM_R} {m, n}$.

Proof:
Let $\mathbf A = \sqbrk a_{m n}$ and $\mathbf B = \sqbrk b_{m n}$ be elements of $\map {\MM_R} {m, n}$.
Let $\sqbrk c_{m n} = \sqbrk a_{m n} + \sqbrk b_{m n}$.
By definition of matrix entrywise addition:
:$\forall i \in \closedint 1 m, j \in \closedint 1 n: a_{i j} + b_{i j} = c_{i j}$
By {{Ring-axiom|A0}}, $R$ is closed under addition.
Hence:
:$\forall i \in \closedint 1 m, j \in \closedint 1 n: c_{i j} \in R$
From the definition of matrix entrywise addition, $\sqbrk c_{m n}$ has the same order as both $\sqbrk a_{m n}$ and $\sqbrk b_{m n}$.
Thus it follows that:
:$\sqbrk c_{m n} \in \map {\MM_R} {m, n}$
Thus $\struct {\map {\MM_R} {m, n}, +}$, as it is defined, is closed.
{{qed}}
Theorem: Matrix Entrywise Addition over Ring is Closed

Let $\struct {R, +, \circ}$ be a ring.
Let $\map {\MM_R} {m, n}$ be a $m \times n$ matrix space over $R$.
For $\mathbf A, \mathbf B \in \map {\MM_R} {m, n}$, let $\mathbf A + \mathbf B$ be defined as the matrix entrywise sum of $\mathbf A$ and $\mathbf B$.
The operation $+$ is closed on $\map {\MM_R} {m, n}$.

Proof:
By definition, matrix entrywise addition is the '''Hadamard product''' of $\mathbf A$ and $\mathbf B$ with respect to ring addition.
We have from {{Ring-axiom|A0}} that ring addition is closed.
The result then follows directly from Closure of Hadamard Product.
{{qed}}
Theorem: Points are Path-Connected iff Contained in Path-Connected Set

Let $T = \struct {S, \tau}$ be a topological space.
Let $x, y \in S$
Then:
:$x, y$ are path-connected points in $T$ {{iff}} there exists a path-connected set of $T$ containing $x$ and $y$.

Proof:
=== Necessary Condition ===
Let $x, y$ be path-connected points in $T$.
Let $\gamma: \closedint 0 1 \to T$ be a path from $x$ to $y$.
From Image of Path is Path-Connected Set, $\Img \gamma$ is a path-connected set of $T$ containing $x$ and $y$.
The result follows.
{{qed|lemma}}
Theorem: Equivalence of Definitions of Path Component/Union of Path-Connected Sets is Maximal Path-Connected Set

Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in T$.
Let $\CC_x = \leftset {A \subseteq S : x \in A \land A}$ is path-connected in $\rightset T$
Let $C = \bigcup \CC_x$
Then $C$ is a maximal path-connected set of $T$.

Proof:
Let $\tilde C$ be an arbitrary path-connected set such that:
:$C \subseteq \tilde C$
Then $x \in \tilde C$.
Hence $\tilde C \in \CC_x$.
From Set is Subset of Union, 
:$\tilde C \subseteq C$.
Hence $\tilde C = C$.
It follows that $C$ is a maximal path-connected set of $T$ by definition.
Theorem: Equivalence of Definitions of Path Component/Maximal Path-Connected Set is Union of Path-Connected Sets

Let $\tilde C$ be a maximal path-connected set of $T$ that contains $x$.

Proof:
By definition:
:$\tilde C \in \CC_x$
From Set is Subset of Union:
:$\tilde C \subseteq C$
By maximality of $\tilde C$:
:$\tilde C = C$
Theorem: Power Set is Closed under Set Complement

Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Then:
:$\forall A \in \powerset S: \relcomp S A \in \powerset S$

Proof:
Let $A \in \powerset S$.
Then by the definition of power set, $A \subseteq S$.
By definition of relative complement:
:$\relcomp S A = \set {x \in S: x \notin A}$
Hence $\relcomp S A$ is a subset of $S$.
That is:
:$\relcomp S A \in \powerset S$
and closure is proved.
{{qed}}
Theorem: Normal Subgroup of Symmetric Group on More than 4 Letters is Alternating Group

Let $n \in \N$ be a natural number such that $n > 4$.
Let $S_n$ denote the symmetric group on $n$ letters.
Let $A_n$ denote the alternating group on $n$ letters.
$A_n$ is the only proper non-trivial normal subgroup of $S_n$.

Proof:
From Alternating Group is Normal Subgroup of Symmetric Group, $A_n$ is seen to be normal in $S_n$.
It remains to be shown that $A_n$ is the only such normal subgroup of $S_n$.
{{AimForCont}} $N$ is a proper non-trivial normal subgroup of $S_n$ such that $N$ is a proper subset of $A_n$.
From Intersection with Normal Subgroup is Normal, we have that:
:$A_n \cap N$ is a normal subgroup of $A_n$.
As $N \subseteq A_n$, from Intersection with Subset is Subset we have that:
:$A_n \cap N = N$
That is:
:$N$ is a normal subgroup of $A_n$.
But from Alternating Group is Simple except on 4 Letters, $A_n$ has no proper non-trivial normal subgroup.
So $N$ cannot be a normal subgroup of $A_n$.
Hence by Proof by Contradiction it follows that $N$ cannot be a normal subgroup of $S_n$.
So $S_n$ has no proper non-trivial normal subgroup apart from $A_n$.
{{qed}}
Theorem: Path Components are Open iff Union of Open Path-Connected Sets/Path Components are Open implies Space is Union of Open Path-Connected Sets

Let the path components of $T$ be open sets.

Proof:
By definition, the path components of $T$ are a partition of $S$.
Hence $S$ is the union of the open path components of $T$.
Since a path component is a maximal path-connected set by definition, then $S$ is a union of open path-connected sets of $T$.
Theorem: Path Components are Open iff Union of Open Path-Connected Sets/Space is Union of Open Path-Connected Sets implies Path Components are Open

Let $T = \struct {S, \tau}$ be a topological space.
Let $S$ be the union of open path-connected sets of $T$.
Then:
:The path components of $T$ are open sets.

Proof:
Let $S = \ds \bigcup \set {U \subseteq S : U \in \tau \text { and } U \text { is path-connected} }$.
Let $C$ be a path component of $T$.
=== Lemma ===
{{:Path Components are Open iff Union of Open Path-Connected Sets/Lemma}}{{qed|lemma}}
Then:
{{begin-eqn}}
{{eqn | l = C
      | r = C \cap S
      | c = Intersection with Subset is Subset
}}
{{eqn | r = C \cap \bigcup \set {U \subseteq S : U \in \tau \text { and } U \text { is path-connected} }
}}
{{eqn | r = \bigcup \set {C \cap U : U \in \tau \text { and } U \text { is path-connected} }
      | c = Intersection Distributes over Union
}}
{{eqn | r = \bigcup \set {C \cap U : U \in \tau, U \cap C \ne \O \text { and } U \text { is path-connected} }
      | c = Union with Empty Set
}}
{{eqn | r = \bigcup \set {U \subseteq C : U \in \tau, U \ne \O \text { and } U \text { is path-connected} }
      | c = {{Lemma|Path Components are Open iff Union of Open Path-Connected Sets}}
}}
{{end-eqn}} 
Hence $C$ is the union of open sets.
By definition of a topology then $C$ is an open set.
The result follows.
Theorem: Path Components are Open iff Union of Open Path-Connected Sets/Lemma

Let $U$ be a path-connected set of $T$.
Then:
:$U \cap C \ne \O$ {{iff}} $U \ne \O$ and $U \subseteq C$

Proof:
=== Necessary Condition ===
Let $U \cap C \ne \O$.
From Union of Path-Connected Sets with Common Point is Path-Connected, $U \cup C$ is a path-connected set of $T$.
From Set is Subset of Union, $C \subseteq U \cup C$.
By definition of a path component, $C$ is a maximal path-connected set.
Hence $C = U \cup C$.
From Union with Superset is Superset, $U \subseteq C$.
Then:
{{begin-eqn}}
{{eqn | r = U \cap C
      | l = U
      | c = Intersection with Subset is Subset
}}
{{eqn | o = \ne
      | r = \O
      | c = {{hypothesis}}
}}
{{end-eqn}}
The result follows.
{{qed|lemma}}
Theorem: Components are Open iff Union of Open Connected Sets/Components are Open implies Space is Union of Open Connected Sets

Let $T = \struct {S, \tau}$ be a topological space.
Let the components of $T$ be open sets.
Then:
:$S$ is a union of open connected sets of $T$.

Proof:
Let the components of $T$ be open.
By definition, the components of $T$ are a partition of $S$.
Hence $S$ is the union of the open components of $T$.
Since a component is a maximal connected set by definition, then $S$ is a union of open connected sets of $T$.
Theorem: Components are Open iff Union of Open Connected Sets/Space is Union of Open Connected Sets implies Components are Open

Let $T = \struct {S, \tau}$ be a topological space.
Let $S$ be a union of open connected sets of $T$.
Then:
:The components of $T$ are open sets.

Proof:
Let $S = \ds \bigcup \set {U \subseteq S : U \in \tau \text { and } U \text { is connected} }$.
Let $C$ be a component of $T$.
=== Lemma ===
:For any connected set $U$ then:
{{:Components are Open iff Union of Open Connected Sets/Lemma 1}}{{qed|lemma}}
Then:
{{begin-eqn}}
{{eqn | l = C
      | r = C \cap S
      | c = Intersection with Subset is Subset
}}
{{eqn | r = C \cap \bigcup \set { U \subseteq S : U \in \tau \text { and } U \text { is connected} }
}}
{{eqn | r = \bigcup \set {C \cap U : U \in \tau \text { and } U \text { is connected} }
      | c = Intersection Distributes over Union
}}
{{eqn | r = \bigcup \set {C \cap U : U \in \tau, U \cap C \ne \O \text { and } U \text { is connected} }
      | c = Union with Empty Set
}}
{{eqn | r = \bigcup \set {U \subseteq C : U \in \tau \text { and } U \text { is connected} }
      | c = {{Lemma|Components are Open iff Union of Open Connected Sets|1}}
}}
{{end-eqn}} 
Hence $C$ is the union of open sets.
By definition of a topology then $C$ is an open set.
The result follows.
Theorem: Components are Open iff Union of Open Connected Sets/Lemma 1

::$U \cap C \ne \O$ {{iff}} $U \ne \O$ and $U \subseteq C$

Proof:
=== Necessary Condition ===
Let $U \cap C \ne \O$.
From Union of Connected Sets with Common Point is Connected, $U \cup C$ is a connected set of $T$.
From Set is Subset of Union, $C \subseteq U \cup C$.
By definition of a component, $C$ is a maximal connected set.
Hence $C = U \cup C$.
From Union with Superset is Superset, $U \subseteq C$.
Then:
{{begin-eqn}}
{{eqn | r = U \cap C
      | l = U
      | c = Intersection with Subset is Subset
}}
{{eqn | o = \ne
      | r = \O
      | c = Assumption
}}
{{end-eqn}}
The result follows.
{{qed|lemma}}
Theorem: Integral to Infinity of Square of Sine p x over x Squared

:$\ds \int_0^\infty \paren {\frac {\sin p x} x}^2 \rd x = \frac {\pi \size p} 2$

Proof:
We have: 
{{begin-eqn}}
{{eqn | l = \map {\frac \d {\d x} } {\sin^2 p x}
      | r = \map {\frac \d {\d x} } {\sin p x} \map {\frac \d {\map \d {\sin p x} } } {\sin^2 p x}
      | c = Chain Rule for Derivatives
}}
{{eqn | r = 2 p \cos p x \sin p x
      | c = Derivative of Sine of a x, Derivative of Power
}}
{{eqn | r = p \sin 2 p x
      | c = Double Angle Formula for Sine
}}
{{end-eqn}}
We also have, by Primitive of Power: 
:$\ds \int \frac {\d x} {x^2} = -\frac 1 x + C$
So: 
{{begin-eqn}}
{{eqn | l = \int_0^\infty \paren {\frac {\sin p x} x}^2 \rd x
      | r = \intlimits {-\frac {\sin^2 p x} x} 0 \infty - \int_0^\infty \paren {-\frac {p \sin 2 p x} x} \rd x
      | c = Integration by Parts
}}
{{eqn | r = -\lim_{x \mathop \to \infty} \paren {\frac {\sin^2 p x} x} + \lim_{x \mathop \to 0} \paren {\frac {\sin^2 p x} x} + p \int_0^\infty \frac {\sin 2 p x} x \rd x
}}
{{end-eqn}}
We have, from Real Sine Function is Bounded, that: 
:$0 \le \sin^2 p x \le 1$
for all real $x$. 
Therefore: 
:$0 \le \dfrac {\sin^2 p x} x \le \dfrac 1 x$
for all strictly positive real $x$.
We have: 
:$\ds \lim_{x \mathop \to \infty} \frac 1 x = 0$
Therefore, by the Squeeze Theorem, we have: 
:$\ds \lim_{x \mathop \to \infty} \paren {\frac {\sin^2 p x} x} = 0$
As to the other limit, note that: 
{{begin-eqn}}
{{eqn | l = \lim_{x \mathop \to 0} \frac {\sin p x} x
      | r = \lim_{p x \mathop \to 0} \frac {\sin p x} {\frac {p x} p}
}}
{{eqn | r = p \lim_{u \mathop \to 0} \frac {\sin u} u
      | c = letting $u = p x$
}}
{{eqn | r = p
      | c = Limit of $\dfrac {\sin x} x$ at Zero
}}
{{end-eqn}}
So:
{{begin-eqn}}
{{eqn | l = \lim_{x \mathop \to 0} \frac {\sin^2 p x} x
      | r = \paren {\lim_{x \mathop \to 0} \sin p x} \paren {\lim_{x \mathop \to 0} \frac {\sin p x} x}
      | c = Product Rule for Limits of Real Functions
}}
{{eqn | r = p \map \sin {0 p}
}}
{{eqn | r = 0
      | c = Sine of Zero is Zero
}}
{{end-eqn}}
So: 
:$\ds \int_0^\infty \paren {\frac {\sin p x} x}^2 \rd x = p \int_0^\infty \frac {\sin 2 p x} x \rd x$
By Integral to Infinity of Sine p x over x, we have: 
:$\ds \int_0^\infty \frac {\sin 2 p x} x \rd x = \begin {cases} \dfrac \pi 2 & : p > 0 \\ \\ 0 & : p = 0 \\ \\ -\dfrac \pi 2 & : p < 0 \end{cases}$
So:
:$\ds p \int_0^\infty \frac {\sin 2 p x} x \rd x = \begin{cases} \dfrac \pi 2 p & : p > 0 \\ \\ 0 & : p = 0 \\ \\ -\dfrac \pi 2 p & : p < 0 \end{cases}$
Hence:
:$\ds p \int_0^\infty \frac {\sin 2 p x} x \rd x = \frac {\pi \size p} 2$
Hence the result.
{{qed}}
Theorem: Integral to Infinity of One minus Cosine p x over x Squared

:$\ds \int_0^\infty \frac {1 - \cos p x} {x^2} \rd x = \frac {\pi \size p} 2$

Proof:
{{begin-eqn}}
{{eqn	| l = \int_0^\infty \frac {1 - \cos p x} {x^2} \rd x
	| r = \int_0^\infty \frac {2 \sin^2 \paren {\frac {p x} 2} } {x^2} \rd x
	| c = Square of Sine
}}
{{eqn	| r = 2 \times \frac {\pi \size p} {2 \times 2}
	| c = Integral to Infinity of $\paren {\dfrac {\sin p x} x}^2$
}}
{{eqn	| r = \frac {\pi \size p} 2
}}
{{end-eqn}}
{{qed}}
Theorem: 1 can be Expressed as Sum of 4 Distinct Unit Fractions in 6 Ways

The number $1$ can be expressed as the sum of $4$ distinct unit fractions in $6$ different ways:
{{begin-eqn}}
{{eqn | l = 1
      | r = \frac 1 2 + \frac 1 3 + \frac 1 7 + \frac 1 {42}
}}
{{eqn | r = \frac 1 2 + \frac 1 3 + \frac 1 8 + \frac 1 {24}
}}
{{eqn | r = \frac 1 2 + \frac 1 3 + \frac 1 9 + \frac 1 {18}
}}
{{eqn | r = \frac 1 2 + \frac 1 3 + \frac 1 {10} + \frac 1 {15}
}}
{{eqn | r = \frac 1 2 + \frac 1 4 + \frac 1 5 + \frac 1 {20}
}}
{{eqn | r = \frac 1 2 + \frac 1 4 + \frac 1 6 + \frac 1 {12}
}}
{{end-eqn}}

Proof:
From Sum of 4 Unit Fractions that equals 1:
{{:Sum of 4 Unit Fractions that equals 1}}
This includes repeated unit fractions.
The full list is:
{{begin-eqn}}
{{eqn | n = 1
      | r = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 7 + \dfrac 1 {42}
      | o = 
}}
{{eqn | n = 2
      | r = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 8 + \dfrac 1 {24}
      | o = 
}}
{{eqn | n = 3
      | r = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 9 + \dfrac 1 {18}
      | o = 
}}
{{eqn | n = 4
      | r = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {10} + \dfrac 1 {15}
      | o = 
}}
{{eqn | n = 5
      | r = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {12} + \dfrac 1 {12}
      | o = 
}}
{{eqn | n = 6
      | r = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 5 + \dfrac 1 {20}
      | o = 
}}
{{eqn | n = 7
      | r = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 6 + \dfrac 1 {12}
      | o = 
}}
{{eqn | n = 8
      | r = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 8 + \dfrac 1 8
      | o = 
}}
{{eqn | n = 9
      | r = \dfrac 1 2 + \dfrac 1 5 + \dfrac 1 5 + \dfrac 1 {10}
      | o = 
}}
{{eqn | n = 10
      | r = \dfrac 1 2 + \dfrac 1 6 + \dfrac 1 6 + \dfrac 1 6
      | o = 
}}
{{eqn | n = 11
      | r = \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 {12}
      | o = 
}}
{{eqn | n = 12
      | r = \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 6 + \dfrac 1 {6}
      | o = 
}}
{{eqn | n = 13
      | r = \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 6
      | o = 
}}
{{eqn | n = 14
      | r = \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4
      | o = 
}}
{{end-eqn}}
The result follows from inspection: solutions $(1)$, $(2)$, $(3)$, $(4)$, $(6)$ and $(7)$ are the only solutions of the above such that the denominators of the summands are distinct.
{{qed}}
Theorem: Equivalence of Definitions of Weakly Locally Connected at Point

Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in S$.
{{TFAE|def = Weakly Locally Connected at Point}}

Proof:
=== Definition 1 implies Definition 2 ===
{{:Equivalence of Definitions of Weakly Locally Connected at Point/Definition 1 implies Definition 2}}{{qed|lemma}}
Theorem: Integral to Infinity of Cosine p x minus Cosine q x over x

:$\ds \int_0^\infty \frac {\cos p x - \cos q x} x \rd x = \ln \frac q p$

Proof:
{{begin-eqn}}
{{eqn | l = \int_0^\infty \frac {\cos p x - \cos q x} x \rd x
      | r = \int_0^\infty \map {\laptrans {\cos p t - \cos q t} } x \rd x
      | c = Integral to Infinity of Function over Argument, assuming both integrals converge
}}
{{eqn | r = \int_0^\infty \frac x {x^2 + p^2} - \frac x {x^2 + q^2} \rd x
      | c = Laplace Transform of Cosine
}}
{{eqn | r = \intlimits {\frac {\map \ln {x^2 + p^2} - \map \ln {x^2 + q^2} } 2} 0 \infty
      | c = Primitive of $\dfrac x {x^2 + a^2}$
}}
{{eqn | r = \lim_{x \mathop \to \infty} \frac 1 2 \map \ln {\frac {x^2 + p^2} {x^2 + q^2} } - \frac 1 2 \map \ln {\frac {0 + p^2} {0 + q^2} }
      | c = inserting limit values
}}
{{eqn | r = \lim_{x \mathop \to \infty} \frac 1 2 \map \ln {\frac {x^2 + p^2} {x^2 + q^2} } + \ln \frac q p
      | c = simplifying
}}
{{eqn | r = \lim_{x \mathop \to \infty} \frac 1 2 \map \ln {\frac {1 + \frac {p^2} {x^2} } {1 + \frac {q^2} {x^2} } } + \ln \frac q p
      | c = multiplying numerator and denominator by $\dfrac 1 {x^2}$
}}
{{eqn | r = \frac 1 2 \map \ln {\frac {1 + p^2 \cdot 0^2} {1 + q^2 \cdot 0^2} } +  \ln \frac q p
      | c = Limit of $\dfrac 1 x$ at $\infty$
}}
{{eqn | r = \ln \frac q p
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Empty Set is Subset of Power Set

The empty set is a subset of all power sets:
:$\forall S: \O \subseteq \powerset S$

Proof:
Follows directly from Empty Set is Subset of All Sets.
{{qed}}
Theorem: Equivalence of Definitions of Weakly Locally Connected at Point/Definition 1 implies Definition 2

Let $x$ have a neighborhood basis consisting of connected sets.

Proof:
Let $U$ be an open neighborhood of $x$.
By assumption there exists a connected neighborhood $C$ of $x$ such that $C \subseteq U$.
By definition of a neighborhood, there exists an open neighborhood $V$ of $x$ such that $V \subseteq C$.
From Subset Relation is Transitive, $V \subseteq U$.
By definition of a subset:
:$\forall y, z \in V: y, z \in C$
The result follows.
Theorem: Equivalence of Definitions of Weakly Locally Connected at Point/Definition 2 implies Definition 1

Let every open neighborhood $U$ of $x$ contain an open neighborhood $V$ of $x$ such that every two points of $V$ lie in some connected subset of $U$.

Proof:
Let $\BB = \set {B \subseteq S: B \text{ is a connected neighborhood of } x}$
It will be shown that $\BB$ is a neighborhood basis consisting of connected sets.
Let $N$ be any neighborhood of $x$.
By definition of a neighborhood there exists an open neighborhood $U$ of $x$ such that $U \subseteq N$
By assumption there exists an open neighborhood $V$ of $x$ such that every two points of $V$ lie in some connected subset of $U$.
For each $y \in V, y \ne x$, let $C_y$ be a connected subset of $U$ such that $x, y \in C_y \subseteq U$.
Let $C = \set x \cup \bigcup \set {C_y:y \in V, y \ne x}$
From Singleton is Connected in Topological Space, $\set x$ is connected.
From Union of Connected Sets with Common Point is Connected, $C$ is a connected subset.
From Union is Smallest Superset, $C \subseteq U \subseteq N$.
Now $x \in C$ and for each $y \in V, y \ne x$ then $y \in C_y \subseteq C$.
Hence $V \subseteq C$.
By definition of a neighborhood, $C$ is a connected neighborhood of $x$.
That is, $C \in \BB$.
Since $N$ was an arbitrary neighborhood of $x$, it follows that $\BB$ is a neighborhood basis of $x$ consisting of connected sets.