hcju/proofwikiqa huggingface dataset:
Let $\mathbf a = \operatorname{Card} \left({A}\right)$, $\mathbf b = \operatorname{Card} \left({B}\right)$ and $\mathbf c = \operatorname{Card} \left({C}\right)$ for some sets $A$, $B$ and $C$. Let $A, B, C$ be pairwise disjoint, that is: :$A \cap B = \varnothing$ :$B \cap C = \varnothing$ :$A \cap C = \varnothing$ Then we can define: :$A \sqcup B := A \cup B$ :$B \sqcup C := B \cup C$ :$A \sqcup C := A \cup C$ where $A \sqcup B$ denotes the disjoint union of $A$ and $B$. Then we have: :$\mathbf a + \mathbf b = \operatorname{Card} \left({A \sqcup B}\right) = \operatorname{Card} \left({A \cup B}\right)$ :$\mathbf b + \mathbf c = \operatorname{Card} \left({B \sqcup C}\right) = \operatorname{Card} \left({B \cup C}\right)$ Then: {{begin-eqn}} {{eqn | l=\left({A \cup B}\right) \cap C       | r=\left({A \cap C}\right) \cup \left({B \cap C}\right)       | c=Intersection Distributes over Union }} {{eqn | r=\varnothing \cup \varnothing       | c=as $A \cap C = \varnothing$ and $B \cap C = \varnothing$ }} {{eqn | r=\varnothing       | c=Union with Empty Set }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l=\operatorname{Card} \left({\left({A \cup B}\right) \cup C}\right)       | r=\operatorname{Card} \left({A \cup B}\right) + \operatorname{Card} \left({C}\right)       | c=as $\left({A \cup B}\right) \cap C = \varnothing$ from above }} {{eqn | r=\left({\mathbf a + \mathbf b}\right) + \mathbf c       | c=by definition of sum of cardinals }} {{end-eqn}} Similarly: {{begin-eqn}} {{eqn | l=A \cap \left({B \cup C}\right)       | r=\left({A \cap B}\right) \cup \left({A \cap C}\right)       | c=Intersection Distributes over Union }} {{eqn | r=\varnothing \cup \varnothing       | c=as $A \cap B = \varnothing$ and $A \cap C = \varnothing$ }} {{eqn | r=\varnothing       | c=Union with Empty Set }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l=\operatorname{Card} \left({A \cup \left({B \cup C}\right)}\right)       | r=\operatorname{Card} \left({A}\right) + \operatorname{Card} \left({B \cup C}\right)       | c=as $A \cap \left({B \cup C}\right) = \varnothing$ from above }} {{eqn | r=\mathbf a + \left({\mathbf b + \mathbf c}\right)       | c=by definition of sum of cardinals }} {{end-eqn}} Finally note that from Union is Associative: :$A \cup \left({B \cup C}\right) = \left({A \cup B}\right) \cup C$ {{qed}}
=== Necessary Condition === Let $x, y \in X$: {{begin-eqn}} {{eqn | l = x \circ \left({x \circ y} \right)       | r = \left({x \circ \left(0 \circ y \right)} \right) \circ x       | c = Identity: $x \circ \left({y \circ z}\right) = \left({x \circ \left({0 \circ z}\right)}\right) \circ y$ }} {{eqn | r = \left ({y \circ \left({0 \circ x} \right)} \right) \circ x       | c = $\left({X, \circ}\right)$ is commutative }} {{eqn | r = y \circ \left({x \circ x} \right)       | c = Axiom $(A3)$ for $B$-algebras }} {{eqn | r = y \circ 0       | c = Axiom $(A1)$ for $B$-algebras }} {{eqn | r = y       | c = Axiom $(A2)$ for $B$-algebras }} {{end-eqn}} {{qed|lemma}} === Sufficient Condition === Let $x, y \in X$: {{begin-eqn}} {{eqn | l = \left ({x \circ \left({0 \circ y} \right)} \right) \circ y       | r = x \circ \left ({y \circ \left ({0 \circ \left ({0 \circ y} \right)} \right)} \right)       | c = Axiom $(A3)$ for $B$-algebras }} {{eqn | r = x \circ \left ({y \circ y} \right)       | c = Identity: $x \circ y = x \circ \left({0 \circ \left({0 \circ y}\right)}\right)$ }} {{eqn | r = y \circ \left({x \circ x} \right)       | c = Axiom $(A3)$ for $B$-algebras }} {{eqn | r = x \circ 0       | c = Axiom $(A2)$ for $B$-algebras }} {{eqn | r = x       | c = Axiom $(A1)$ for $B$-algebras }} {{eqn | r = y \circ \left({y \circ x} \right)       | c = by hypothesis }} {{eqn | r = y \circ \left({y \circ \left({0 \circ \left({0 \circ x} \right)}\right)} \right)       | c = Identity: $x \circ y = x \circ \left({0 \circ \left({0 \circ y}\right)}\right)$ }} {{eqn | r = \left ({y \circ \left({0 \circ x} \right)} \right) \circ y       | c = Axiom $(A3)$ for $B$-algebras }} {{end-eqn}} Hence: :$\left({x \circ \left({0 \circ y} \right)} \right) \circ y = \left ({y \circ \left({0 \circ x} \right)} \right) \circ y$ From $B$-Algebra is Right Cancellable, we have: :$x \circ \left({0 \circ y} \right) = y \circ \left({0 \circ x} \right)$ and hence commutativity. {{qed}} Category: B-Algebras 02to894cq4ucmt7iphzwgtbvividpuc
Let $\mathbf a = \operatorname{Card} \left({A}\right)$ for some set $A$. From Union with Empty Set we have $A \cup \varnothing = A$. From Intersection with Empty Set we have $A \cap \varnothing = \varnothing$. So $A$ and $\varnothing$ are disjoint and so: {{begin-eqn}} {{eqn | l=\mathbf a + \mathbf 0       | r=\operatorname{Card} \left({A \cup \varnothing}\right)       | c=by definition of sum of cardinals }} {{eqn | r=\operatorname{Card} \left({A}\right) + \operatorname{Card} \left({\varnothing}\right)       | c=Cardinality of Set Union/Corollary }} {{eqn | r=\operatorname{Card} \left({A}\right)       | c=Cardinality of Empty Set }} {{eqn | r=\mathbf a       | c= }} {{end-eqn}} {{qed}}
Let $x, y, z \in X$. Then: {{begin-eqn}} {{eqn | l = \paren {z \circ y} \circ \paren {z \circ x}       | r = \paren {\paren {z \circ y} \circ \paren {0 \circ x} } \circ z       | c = Identity: $x \circ \paren {y \circ z} = \paren {x \circ \paren {0 \circ z} } \circ y$ }} {{eqn | r = \paren {x \circ \paren {0 \circ \paren {z \circ y} } } \circ z       | c = {{Defof|Commutative B-Algebra}} }} {{eqn | r = x \circ \paren {z \circ \paren {0 \circ \paren {0 \circ \paren {z \circ y} } } }       | c = Axiom $(A3)$ for $B$-algebras }} {{eqn | r = x \circ \paren {z \circ \paren {z \circ y} }       | c = Identity: $x \circ y = x \circ \paren {0 \circ \paren {0 \circ y} }$  }} {{eqn | r = x \circ y       | c = $B$-Algebra is Commutative iff $x \circ \paren {x \circ y} = y$ }} {{end-eqn}} {{qed}} Category:B-Algebras 27pjfzfr4w2ugzh76gv6jfqnfpc4kpz
Let $\mathbf a = \operatorname{Card} \left({A}\right)$, $\mathbf b = \operatorname{Card} \left({B}\right)$ and $\mathbf c = \operatorname{Card} \left({C}\right)$ for some sets $A$, $B$ and $C$. Let $B$ and $C$ be pairwise disjoint, that is: :$B \cap C = \varnothing$ Then we can define: :$B \sqcup C := B \cup C$ where $B \sqcup C$ denotes the disjoint union of $B$ and $C$. Then we have: :$\mathbf b + \mathbf c = \operatorname{Card} \left({B \sqcup C}\right) = \operatorname{Card} \left({B \cup C}\right)$ Then: {{begin-eqn}} {{eqn | l=\left({A \times B}\right) \cap \left({A \times C}\right)       | r=A \times \left({B \cap C}\right)       | c=Cartesian Product of Intersections/Corollary 1 }} {{eqn | r=A \times \varnothing       | c=as $B \cap C = \varnothing$ }} {{eqn | r=\varnothing       | c=Cartesian Product is Empty iff Factor is Empty }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l=\operatorname{Card} \left({\left({A \times B}\right) \cup \left({A \times C}\right)}\right)       | r=\operatorname{Card} \left({A \times B}\right) + \operatorname{Card} \left({A \times C}\right)       | c=as $A \times B$ and $A \times C$ are disjoint from above }} {{eqn | r=\mathbf a \mathbf b + \mathbf a \mathbf c       | c=by definition of product of cardinals }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l=\operatorname{Card} \left({\left({A \times B}\right) \cup \left({A \times C}\right)}\right)       | r=\operatorname{Card} \left({A \times \left({B \cup C}\right)}\right)       | c=Cartesian Product Distributes over Union }} {{eqn | r=\mathbf a \left({\mathbf b + \mathbf c}\right)       | c=by definition of product of cardinals and sum of cardinals }} {{end-eqn}} {{qed}}
Let $\mathbf a = \operatorname{Card} \left({A}\right)$, $\mathbf b = \operatorname{Card} \left({B}\right)$ and $\mathbf c = \operatorname{Card} \left({C}\right)$ for some sets $A$, $B$ and $C$. Let $\mathbf a \le \mathbf b$. Then by definition of cardinal, there exists an injection $f: A \to B$. Then the mapping $g: A \times C \to B \times C$ defined as: :$\forall \left({a, c}\right) \in A \times C: g \left({a, c}\right) = \left({f \left({a}\right), c}\right)$ Let $\left({a_1, c_1}\right) \in A \times C$ and $\left({a_2, c_2}\right) \in A \times C$ such that: :$g \left({a_1, c_1}\right) = g \left({a_2, c_2}\right)$ That is: :$\left({f \left({a_1}\right), c_1}\right) = \left({f \left({a_2}\right), c_2}\right)$ By Equality of Ordered Pairs: :$f \left({a_1}\right) = f \left({a_2}\right), c_1 = c_2$ and so as $f$ is an injection: :$a_1 = a_2, c_1 = c_2$ By Equality of Ordered Pairs: :$\left({a_1, c_1}\right) = \left({a_2, c_2}\right)$ demonstrating that $g$ is an injection. So, by definition of product of cardinals: :$\mathbf a \mathbf c \le \mathbf b \mathbf c$ {{qed}}
Let $\mathbf a = \map \Card A$, $\mathbf b = \map \Card B$ and $\mathbf c = \map \Card C$ for some sets $A$, $B$ and $C$. Let $C$ be chosen such that $A \cap C = \O = B \cap C$. Let $\mathbf a \le \mathbf b$. Then by definition of cardinal, there exists an injection $f: A \to B$. Then the mapping $h: A \cup C \to B \cup C$ defined as: :$\forall x \in A \cup C: \map h x = \begin{cases} \map f x & : x \in A \\ x & : x \in C \end{cases}$ Let $a_1 \in A \cup C$ and $a_2 \in A \cup C$ such that: :$\map h {a_1} = \map h {a_2}$ Then: {{begin-eqn}} {{eqn | l = a_1       | o = \in       | r = C       | c =  }} {{eqn | ll= \leadsto       | l = a_1       | r = \map h {a_1}       | c = Definition of $h$ }} {{eqn | ll= \leadsto       | l = a_1       | r = \map h {a_2}       | c = as $\map h {a_1} = \map h {a_2}$ }} {{eqn | ll= \leadsto       | l = a_1       | r = a_2       | c = as $\map h {a_2}$ must also be in $C$ }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = a_1       | o = \notin       | r = C       | c =  }} {{eqn | ll= \leadsto       | l = \map h {a_1}       | r = \map f {a_1}       | c = Definition of $h$ }} {{eqn | ll= \leadsto       | l = \map h {a_2}       | r = \map f {a_1}       | c = as $\map h {a_1} = \map h {a_2}$ }} {{eqn | ll= \leadsto       | l = \map h {a_2}       | r = \map f {a_2}       | c = as $\map h {a_2}$ must also not be in $C$ }} {{eqn | ll= \leadsto       | l = \map f {a_1}       | r = \map f {a_2}       | c = as $\map h {a_1} = \map h {a_2}$ }} {{eqn | ll= \leadsto       | l = a_1       | r = a_2       | c = as $f$ is an injection }} {{end-eqn}} So: :$\map h {a_1} = \map h {a_2} \implies a_1 = a_2$ demonstrating that $h$ is an injection. So, by definition of sum of cardinals: :$\mathbf a + \mathbf c \le \mathbf b + \mathbf c$ {{qed}}
Let $\left({G, \circ}\right)$ be a group such that $\forall g \in G: g = g^{-1}$. From Group Induces B-Algebra, the algebraic structure $\left({G, *}\right)$ where $*$ is defined as: :$\forall a, b \in G: a * b := a \circ b^{-1}$ is a $B$-algebra. But as $b = b^{-1}$, it follows that: :$\forall a, b \in G: a * b := a \circ b$ and so $\left({G, \circ}\right)$ is itself a $B$-algebra. Now suppose it is not the case that $\forall g \in G: g = g^{-1}$. Suppose $\left({G, \circ}\right)$ were a $B$-algebra. By axiom $A2$ of the definition of a $B$-algebra, the identity element $e$ would be the $0$, as: :$\forall a \in G: a \circ 0 = a$ As an Identity is Unique, there can be no other element in $G$ which can satisfy the conditions to be the $0$ of a $B$-algebra. Now as not every element is self-inverse: :$\exists x \in G: x \circ x \ne e$ Then by axiom $A1$ of the definition of a $B$-algebra it follows that $\left({G, \circ}\right)$ is not a $B$-algebra. Hence the result. {{qed}} Category:Group Theory Category:B-Algebras 6fj3ex1vmigumuqfug1wohxxdnln4it
Let $\mathbf a = \operatorname{Card} \left({A}\right)$ and $\mathbf b = \operatorname{Card} \left({B}\right)$ for some sets $A$ and $B$. From Equivalence of Definitions of Dominate (Set Theory), there exists a bijection from $A$ onto a subset $E$ of $B$. Thus: : $\mathop a = \operatorname{Card} \left({E}\right)$ Let $F = \complement_B \left({E}\right)$ From Set with Relative Complement forms Partition: : $B = E \cup F$ : $E \cap F = \varnothing$ By definition of sum of cardinals, it follows that: {{begin-eqn}} {{eqn | l=\mathbf b       | r=\operatorname{Card} \left({B}\right)       | c= }} {{eqn | r=\operatorname{Card} \left({E \cup F}\right)       | c= }} {{eqn | r=\operatorname{Card} \left({E}\right) + \operatorname{Card} \left({F}\right)       | c= }} {{eqn | r=\mathop a + \operatorname{Card} \left({F}\right)       | c= }} {{end-eqn}} By defining $\mathbf c := \operatorname{Card} \left({F}\right)$ it follows that: :$\mathbf a + \mathbf c = \mathbf b$ for the $\mathbf c$ that has been demonstrated to exist. Now suppose there exists a cardinal $\mathbf c$ such that $\mathbf a + \mathbf c = \mathbf b$. Then by definition of sum of cardinals: :$\operatorname{Card} \left({B}\right) = \operatorname{Card} \left({A \cup C}\right)$ for some sets $A$, $B$ and $C$ such that $A \cap C = \varnothing$. Let $f: A \to B$ be an injection, proved to exist by Equivalence of Definitions of Dominate (Set Theory). Then it follows that $\mathbf a \le \mathbf b$. {{qed}}
Suppose $\mathbf a + \mathbf 1 = \mathbf b + \mathbf 1$. Then from Condition for Existence of Cardinal Sum there exists some cardinal $\mathbf c$ such that  :$\mathbf a + \mathbf 1 = \mathbf b + \mathbf 1 = \mathbf c$ By definition of cardinal there exists a set $C$ such that $\mathbf c = \map \Card C$. $C$ also has subsets $A$ and $B$ such that there also exist elements of $\alpha, \beta \in C$ such that: :$\relcomp C A = \set \alpha$ :$\relcomp C B = \set \beta$ If $\alpha = \beta$ then $A = B$ and so $\mathbf a = \mathbf b$. On the other hand, if $\alpha \ne \beta$ then $\alpha \in B$ and $\beta \in A$ and so: :$\set \beta \cup \paren {A \cap B} = A$ :$\set \alpha \cup \paren {A \cap B} = B$ Since: :$\set \beta \cap A \cap B = \O = \set \alpha \cap A \cap B$ it follows that: :$\mathbf a = \map \Card A = \mathbf 1 + \map \Card {A \cap B} = \map \Card B = \mathbf b$ {{qed}}
Let $f$ be an injection. From Injection to Image is Bijection, $S$ is equivalent to the image $\Img f$ of $f$. We are given that $S$ is finite. It follows from Infinite Set is Equivalent to Proper Subset that $\Img f = S$. It follows by definition that $f$ is surjective. {{qed|lemma}} Let $f$ be a surjection. Then by Surjection iff Right Inverse there exists a mapping $g: S \to S$ such that: :$f \circ g = I_S$ where $I_S$ is the identity mapping. By Right Inverse Mapping is Injection, $g$ is an injection. By the above, it follows that $g$ is also a surjection. Thus $g$ is a bijection. It follows that $f = g^{-1}$ is also a bijection and so by definition an injection. {{qed}}
From Finite Cardinals form Infinite Set we have that the cardinals form a set which is infinite. {{proof wanted|Show that the set of finite cardinals satisfy Peano's axioms.}}
A proof of Tartaglia's Formula will be found in a proof of the Cayley-Menger Determinant as a tetrahedron is a 3-simplex. {{proof wanted}} {{Namedfor|Niccolò Fontana Tartaglia|cat = Tartaglia}}
Let $x, y \in X$ and let $0 \circ x = 0 \circ y$. Then: {{begin-eqn}} {{eqn | l = 0       | r = x \circ x       | c = Axiom $(A1)$ for $B$-algebras }} {{eqn | r = \left({x \circ x}\right) \circ 0       | c = Axiom $(A2)$ for $B$-algebras }} {{eqn | r = x \circ \left({0 \circ \left({0 \circ x} \right)}\right)       | c = Axiom $(A3)$ for $B$-algebras }} {{eqn | r = x \circ \left({0 \circ \left({0 \circ y} \right)}\right)       | c = by hypothesis }} {{eqn | r = \left ({x \circ y}\right) \circ 0       | c = Axiom $(A3)$ for $B$-algebras }} {{eqn | r = x \circ y       | c = Axiom $(A2)$ for $B$-algebras }} {{end-eqn}} So we have shown: :$\forall x, y \in X: 0 \circ x = 0 \circ y \implies x \circ y = 0$ From $B$-Algebra Identity: $x \circ y = 0 \iff x = y$: :$\forall x, y \in X: 0 \circ x = 0 \circ y \implies x = y$ Hence the result. {{qed}} Category:B-Algebras swrg0vu35d423sa4pt66djcw95w3zmf
Let $x \in X$. Then: {{begin-eqn}} {{eqn | l = 0 \circ x       | r = \paren {0 \circ x}\circ 0       | c = Axiom $\text A 2$ for $B$-algebra }} {{eqn | r = 0 \circ \paren {0 \circ \paren {0 \circ x} }       | c = Axiom $\text A 3$ for $B$-algebra }} {{end-eqn}} From 0 in B-Algebra is Left Cancellable Element, we conclude: :$x = 0 \circ \paren {0 \circ x}$ as desired. {{qed}} Category:B-Algebras ofehh0hbrk4tm1kp3e62xh9ypbuq4u7
Let $x, y, z \in X$: We will show that $\struct {X, *}$ satisfies each of the group axioms in turn: === G0: Closure === By definition of $*$, we have: :$x * y = x \circ \paren {0 \circ y}$ By Axiom $(AC)$ for $B$-algebras: :$x \circ \paren {0 \circ y} \in X$ Whence $x * y \in X$, and so $\paren {X, *}$ is a closed structure. {{qed|lemma}} === G1: Associativity === {{begin-eqn}} {{eqn | l = \paren {x * y} * z       | r = \paren {x \circ \paren {0 \circ y} } \circ \paren {0 \circ z}       | c = by definition of $*$ }} {{eqn | r = x \circ \paren {\paren {0 \circ z} \circ \paren {0 \circ \paren {0 \circ y} } }       | c = Axiom $(A3)$ for $B$-algebras }} {{eqn | r = x \circ \paren {\paren {0 \circ z} \circ y}       | c = Identity: $0 \circ \paren {0 \circ x} = x$ }} {{eqn | r = x \circ \paren {0 \circ \paren {y \circ \paren {0 \circ z} } }       | c = Axiom $(A3)$ for $B$-algebras }} {{eqn | r = x * \paren {y * z}       | c = by definition of $*$ }} {{end-eqn}} Thus it is seen that $*$ is associative. {{qed|lemma}} === G2: Identity === Let $e := 0$; we will show that it is an identity element of $\struct {X, *}$. {{begin-eqn}} {{eqn | l = x * e       | r = x \circ \paren {0 \circ 0}       | c = by definition of $*$ and $e$ }} {{eqn | r = x \circ 0       | c = Axiom $(A1)$ for $B$-algebras }} {{eqn | r = x       | c = Axiom $(A2)$ for $B$-algebras }} {{end-eqn}} {{begin-eqn}} {{eqn | l = e * x       | r = 0 \circ \paren {0 \circ x}       | c = by definition of $*$ and $e$ }} {{eqn | r = x       | c = Identity: $0 \circ \paren {0 \circ x} = x$ }} {{end-eqn}} Hence $0$ is an identity for $*$. {{qed|lemma}} === G3: Invertibility === Let us prove that for all $x \in X$, $0 \circ x$ is an inverse element to $x$. {{begin-eqn}} {{eqn | l = x * \paren {0 \circ x}       | r = x \circ \paren {0 \circ \paren {0 \circ x} }       | c = by definition of $*$ }} {{eqn | r = x \circ x       | c = Identity: $ 0 \circ \paren {0 \circ x} = x$ }} {{eqn | r = 0       | c = Axiom $(A1)$ for $B$-algebras }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \paren {0 \circ x} \circ x       | r = \paren {0 \circ x} \circ \paren {0 \circ x}       | c = by definition of $*$ }} {{eqn | r = 0       | c = Axiom $(A1)$ for $B$-algebras }} {{end-eqn}} That is, each $x \in X$ has a unique inverse element $x^{-1}$ under $*$. This inverse element is $0 \circ x$. {{qed|lemma}} It follows that: {{begin-eqn}} {{eqn | l = a * b^{-1}       | r = a \circ \paren {0 \circ b^{-1} }       | c = by definition of $*$ }} {{eqn | r = a \circ \paren {0 \circ \paren {0 \circ b} }       | c = by definition of $b^{-1}$ }} {{eqn | r = a \circ b       | c = Identity: $0 \circ \paren {0 \circ x} = x$ }} {{end-eqn}} {{qed|lemma}} All the axioms have been shown to hold and the result follows. {{qed}}
From Unit Matrix is Unity of Ring of Square Matrices, $I$ can be identified as the unit matrix of order $2$. Then: :$A^2 = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix}  1 & 0 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix}  1 & 0 \\ 0 &  1 \end{bmatrix} = I$ :$A B = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 &  1 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} = C$ :$A C = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 &  1 \end{bmatrix} = B$ :$B^2 = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 &  1 \end{bmatrix} = \begin{bmatrix}  1 & 0 \\ 0 &  1 \end{bmatrix} = I$ :$B A = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix}  1 & 0 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} = C$ :$B C = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix}  1 & 0 \\ 0 & -1 \end{bmatrix} = A$ :$C^2 = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix}  1 & 0 \\ 0 &  1 \end{bmatrix} = I$ :$C A = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix}  1 & 0 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 &  1 \end{bmatrix} = B$ :$C B = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 &  1 \end{bmatrix} = \begin{bmatrix}  1 & 0 \\ 0 & -1 \end{bmatrix} = A$ Putting this together into a Cayley table: :$\begin{array}{c|cccccc}   & I & A & B & C \\ \hline I & I & A & B & C \\ A & A & I & C & B \\ B & B & C & I & A \\ C & C & B & A & I \\ \end{array}$ it is verified by inspection that this is an instance of the Klein four-group. {{qed}}
By definition of $F_\sigma$ set, there exist sequences $\sequence {C_n}_{n \mathop \in \N}$ and $\sequence {C'_n}_{n \mathop \in \N}$ of closed sets of $T$ such that: :$F = \displaystyle \bigcup_{n \mathop \in \N} C_n$ :$F' = \displaystyle \bigcup_{n \mathop \in \N} C'_n$ By General Distributivity of Set Union, we have: :$F \cup F' = \displaystyle \bigcup_{n \mathop \in \N} \paren {C_n \cup C'_n}$ By Finite Union of Closed Sets is Closed in Topological Space, $C_n \cup C'_n$ is closed, for all $n \in \N$. Thus $F \cup F'$ is seen to be an $F_\sigma$ set. {{qed}}
By definition of $F_\sigma$ set, there exist sequences $\left({C_n}\right)_{n \in \N}$ and $\left({C'_n}\right)_{n \in \N}$ of closed sets of $T$ such that: :$F = \displaystyle \bigcup_{n \mathop \in \N} C_n$ :$F' = \displaystyle \bigcup_{n \mathop \in \N} C'_n$ Now compute: {{begin-eqn}} {{eqn|l = F \cap F'      |r = \bigcup_{n \mathop \in \N} \left({C_n \cap F'}\right)      |c = Intersection Distributes over Union: General Result }} {{eqn|r = \bigcup_{n \mathop \in \N} \bigcup_{m \mathop \in \N} \left({C_n \cap C'_m}\right)      |c = Intersection Distributes over Union: General Result }} {{end-eqn}} By Intersection of Closed Sets is Closed in Topological Space, $C_n \cap C'_m$ is closed, for all $n, m \in \N$. By Cartesian Product of Countable Sets is Countable, $\N \times \N$ is countable. Thus $F \cap F'$ is seen to be a countable union of closed sets, hence a $F_\sigma$ set. {{qed}}
By definition of $G_\delta$ set, there exist sequences $\left({U_n}\right)_{n \in \N}$ and $\left({U'_n}\right)_{n \in \N}$ of open sets of $T$ such that: :$G = \displaystyle \bigcap_{n \mathop \in \N} U_n$ :$G' = \displaystyle \bigcap_{n \mathop \in \N} U'_n$ Now compute: {{begin-eqn}} {{eqn|l = G \cup G'      |r = \bigcap_{n \mathop \in \N} \left({U_n \cup G'}\right)      |c = Intersection Distributes over Union: General Result }} {{eqn|r = \bigcap_{n \mathop \in \N} \bigcap_{m \mathop \in \N} \left({U_n \cup U'_m}\right)      |c = Intersection Distributes over Union: General Result }} {{end-eqn}} By axiom $(1)$ of a topology, $U_n \cup U'_m$ is open, for all $n, m \in \N$. By Cartesian Product of Countable Sets is Countable, $\N \times \N$ is countable. Thus $G \cup G'$ is seen to be a $G_\delta$ set. {{qed}}
By definition of $G_\delta$ set, there exist sequences $\sequence {U_n}_{n \mathop \in \N}$ and $\sequence {U'_n}_{n \mathop \in \N}$ of open sets of $T$ such that: :$G = \displaystyle \bigcap_{n \mathop \in \N} U_n$ :$G' = \displaystyle \bigcap_{n \mathop \in \N} U'_n$ By General Distributivity of Intersection, we have: :$G \cap G' = \displaystyle \bigcap_{n \mathop \in \N} \paren {U_n \cap U'_n}$ By Intersection of Closed Sets is Closed in Topological Space, $U_n \cap U'_m$ is closed, for all $n, m \in \N$. Thus $G \cap G'$ is seen to be a $G_\delta$ set. {{qed}}
We may construct a sequence $\left\langle{p_n}\right\rangle$ using finite recursion. Let $p_0$ be some arbitrary point in $E$ not equal to $p$. $p_{n+1}$ is some arbitrary point in $E \cap B_{\frac{d \left({p, p_n}\right)} 2} \left({p}\right)$ where $B_{\frac{d \left({p, p_n}\right)} 2} \left({p}\right)$ denotes the open $\dfrac{d \left({p, p_n}\right)} 2$-ball of $p$. We can be assured that such a point exists by the definition of a limit point. Then: : $d \left({p_n, p}\right) < 2^{-n} d \left({p_0, p}\right)$ Since $d \left({p_0, p}\right)$ is fixed: : $d \left({p_n, p}\right) < 2^{-n} d \left({p_0, p}\right) < r$ for some $n$ because the natural numbers are Archimedean. Therefore $p$ is the limit of the sequence $\left\langle{p_n}\right\rangle$ by definition. {{qed}}