hcju/proofwikiqa huggingface dataset:
=== Sufficient Condition === Let $X$ be order generating. Let $x, y \in S$ such that :$y \npreceq x$ By Order Generating iff Every Element is Infimum: :$\exists P \subseteq X: x = \inf P$ By definition of infimum: :$y$ is lower bound for $P \implies y \preceq x$ By definition of lower bound: :$\exists p \in P: y \npreceq p$ Thus by definition of subset: :$p \in X$ By definition of infimum: :$x$ is lower bound for $P$. Thus by definition of lower bound: :$x \preceq p$ Thus :$y \npreceq p$ {{qed|lemma}} === Necessary Condition === Suppose that :$\forall x, y \in S: \left({ y \npreceq x \implies \exists p \in X: x \preceq p \land y \npreceq p}\right)$ Let $x \in S$. Thus by definition of complete lattice: :$x^\succeq \cap X$ admits an infimum. Define $y := \inf \left({x^\succeq \cap X}\right)$ By definition of infimum: :$y$ is lower bound for $x^\succeq \cap X$ We will prove that :$x$ is lower bound for $x^\succeq \cap X$ Let $a \in x^\succeq \cap X$ By definition of intersection: :$a \in x^\succeq$ Thus by definition of upper closure of element: :$x \preceq a$ {{qed|lemma}} By definition of infimum: :$x \preceq y$ By definition of antisymmetry: :$y \nprec x$ It remains to prove that :$y = x$ Aiming for a contradiction suppose that :$y \ne x$ By definition of $\prec$ :$y \npreceq x$ or $y = x$ Case $y \npreceq x$ By assumption: :$\exists p \in X: x \preceq p \land y \npreceq p$ By definition of upper closure of element: :$p \in x^\succeq$ By definition of intersection: :$p \in x^\succeq \cap X$ By definition of lower bound: :$y \preceq p$ Thus this contradicts $y \npreceq p$ Case $y = x$ Thus this contradicts $y \ne x$ {{qed}}
We have that Permutation Induces Equivalence Relation. The equivalence classes of that equivalence relation are the cycles of $\pi$. Hence the result by definition of equivalence class. {{qed}} Category:Symmetric Groups Category:Equivalence Relations r1yzoalv7uf4e0a7t9ld6j0e4d4r65j
Let $\left\lvert{g}\right\rvert = n$. Then from Order of Power of Group Element: :$\forall m \in \Z: \left\lvert{g^m}\right\rvert = \dfrac n {\gcd \left\{ {m, n}\right\} }$ where $\gcd \left\{ {m, n}\right\}$ denotes the greatest common divisor of $m$ and $n$. The result follows from Greatest Common Divisor is at least 1. {{qed}}
Let $g^n = e$. Let $h = g^m$. Then: :$h^n = g^{m n} = \paren {g^n}^m = e^m = e$ Hence by definition of order of group element: :$\order h \le n$ Hence the result. {{qed}}
Let $\order g = n$. Then from Order of Power of Group Element: :$\forall m \in \Z: \order {g^m} = \dfrac n {\gcd \set {m, n}}$ where $\gcd \set {m, n}$ denotes the greatest common divisor of $m$ and $n$. Thus: :$\order {g^m} = \order g \iff \gcd \set {m, n} = 1$ The result follows by definition of coprime integers. {{qed}}
Let $\order g = n$. === Necessary Condition === Let $m \perp n$. Then by Bézout's Lemma: :$\exists x, y \in \Z: x m + y n = 1$ Let $h = g^m$. Then: :$h^x = g^{m x} = g^{1 - y n} = g g^{- y n} = g e = g$ and so $g$ is also a power of $h$. Hence from Order of Group Element not less than Order of Power: :$\order g \le \order h \le \order g$ and it follows that: :$\order g = \order h$ {{qed|lemma}} === Sufficient Condition === We prove the contrapositive. Suppose $m \not \perp n$. Then: :$\exists d \in \Z_{>1}: \exists a, b \in \Z: m = a d \land n = b d$ Let $h = g^m$. Then: :$h^b = \paren {g^m}^{n/d} = g^{a n} = \paren {g^n}^a = e^a = e$ Hence: :$\order h \le b < n$ and it follows that: :$\order g \ne \order h$ {{qed}}
Let $g_1 = g^n$ and $g_2 = g^m$. By Powers of Group Element Commute, $g_1$ and $g_2$ commute. We have: :${g_1}^m = g^{n m} = e$ :${g_2}^n = g^{m n} = e$ It follows that: :the order of $g_1$ is $m$ :the order of $g_2$ is $n$ otherwise if either $g_1$ or $g_2$ were of a smaller order then $g$ would also be of a smaller order. By Bézout's Lemma: :$\exists u, v \in \Z: u n + v m = 1$ as $m \perp n$. Thus: :$g = g^{u n + v m} = \paren {g^n}^u \paren {g^m}^v = {g_1}^u {g_2}^v$ Also by Bézout's Lemma: :$u \perp m$ and: :$v \perp n$ Thus by Order of Group Element equals Order of Coprime Power: :$\order { {g_1}^u} = m$ and: :$\order { {g_2}^v} = n$ We have that $g_1$ and $g_2$ commute. So by Commutativity of Powers in Group, ${g_1}^u$ and ${g_2}^v$ also commute. Putting $a = g_1^u$ and $b = g_2^v$, we have $a$ and $b$ which satisfy the required conditions. It remains to prove uniqueness. Suppose that: :$(1): \quad g = r_1 r_2 = s_1 s_2$ where: :$r_1$ and $r_2$ commute :$s_1$ and $s_2$ commute :$\order {r_1} = \order {s_1} = m$ :$\order {r_2} = \order {s_2} = n$ Raising $(1)$ to the $n u$th power: :$g^{n u} = {r_1}^{n u} {r_2}^{n u} = {s_1}^{n u} {s_2}^{n u}$ and so: {{begin-eqn}} {{eqn | l = {r_1}^{n u} {r_2}^{n u}       | r = {s_1}^{n u} {s_2}^{n u}       | c =  }} {{eqn | l = {r_1}^{n u} \paren { {r_2}^n}^u       | r = {s_1}^{n u} \paren { {s_2}^n}^u       | c =  }} {{eqn | l = {r_1}^{n u} e^u       | r = {s_1}^{n u} e^u       | c = as $\order {r_2} = \order {s_2} = n$ }} {{eqn | ll= \leadsto       | l = {r_1}^{n u}       | r = {s_1}^{n u}       | c =  }} {{eqn | ll= \leadsto       | l = {r_1}^{1 - m v}       | r = {s_1}^{1 - m v}       | c =  }} {{eqn | ll= \leadsto       | l = r_1 \paren { {r_1}^m}^{-v}       | r = s_1 \paren { {s_1}^m}^{-v}       | c =  }} {{eqn | ll= \leadsto       | l = r_1 e^{-v}       | r = s_1 e^{-v}       | c =  }} {{eqn | ll= \leadsto       | l = r_1       | r = s_1       | c =  }} {{end-eqn}} It follows directly from $(1)$ that $r_2 = s_2$. Hence the result. {{qed}}
Let $\order a = n$. From List of Elements in Finite Cyclic Group: :$G = \set {a_0, a_1, \ldots, a_{n - 1} }$ Hence the result. {{qed}}
The subgroup of $G$ generated by $g$ is: :$\gen g = \set {g^0, g^1, g^2, \ldots, g^{n - 1} }$ This contains $n$ elements. Thus: :$\gen g = G$ and the result follows by definition of cyclic group. {{qed}}
It can be seen by inspection that $S \subseteq \Q_{>0}$. That is, all the elements of $S$ are strictly positive rational numbers. From Strictly Positive Rational Numbers under Multiplication form Countably Infinite Abelian Group: :$\struct {\Q_{>0}, \times}$ is an infinite abelian group. It is noted that $S$ is an infinite set and so trivially $S \ne \varnothing$. Let $a, b \in S$. Then: :$\exists k_1 \in \Z: a = 2^{k_1}$ :$\exists k_2 \in \Z: b = 2^{k_2}$ We have that: :$\dfrac 1 {2^{k_2} } 2^{k_2} = 1$ and so $\dfrac 1 {2^{k_2} } = 2^{-k_2} = b^{-1}$ is the inverse of $b \in \struct {\Q_{>0}, \times}$. Then we have: {{begin-eqn}} {{eqn | l = a \times b^{-1}       | r = 2^{k_1} \times \dfrac 1 {2^{k_2} }       | c =  }} {{eqn | r = 2^{k_1 - k_2}       | c =  }} {{end-eqn}} We have that: :$k_1 + k_2 \in \Z$ and so $a \times b^{-1} \in \S$. Hence by the One-Step Subgroup Test, $\struct {S, \times}$ is a subgroup of $\struct {\Q_{>0}, \times}$. It has been established that $S$ is an infinite set. Hence by definition $\struct {S, \times}$ is an infinite group. Finally, from Subgroup of Abelian Group is Abelian, $\struct {S, \times}$ is an abelian group. {{qed}}
Let $k \in \Z$. Then $1 + 2 k \ne 0$. Thus: :$\forall x \in S: x \in \Q_{\ne 0}$ Thus by definition of subset: :$S \subseteq \Q_{\ne 0}$ From Non-Zero Rational Numbers under Multiplication form Abelian Group: :$\struct {\Q_{\ne 0}, \times}$ is an infinite abelian group. It is noted that $S$ is an infinite set and so trivially $S \ne \varnothing$. Consider $a = \dfrac {1 + 2 m} {1 + 2 n} \in S$. We have that: :$\dfrac {1 + 2 m} {1 + 2 n} \times \dfrac {1 + 2 n} {1 + 2 m} = 1$ and so $\dfrac {1 + 2 n} {1 + 2 m}$ is the inverse of $b \in \struct {\Q_{>0}, \times}$. By inspection it can be seen that $b^{-1} \in S$. Let $a, b \in S$. Then: :$\exists m_1, n_1 \in \Z: a = \dfrac {1 + 2 m_1} {1 + 2 n_1}$ :$\exists m_2, n_2 \in \Z: b = \dfrac {1 + 2 m_2} {1 + 2 n_2}$ Then we have: {{begin-eqn}} {{eqn | l = a \times b       | r = \dfrac {1 + 2 m_1} {1 + 2 n_1} \dfrac {1 + 2 m_2} {1 + 2 n_2}       | c =  }} {{eqn | r = \dfrac {\paren {1 + 2 m_1} \paren {1 + 2 m_2} } {\paren {1 + 2 n_1} \paren {1 + 2 n_2} }       | c =  }} {{eqn | r = \dfrac {1 + 2 m_1 + 2 m_2 + 4 m_1 m_2} {1 + 2 n_1 + 2 n_2 + 4 n_1 n_2}       | c =  }} {{eqn | r = \dfrac {1 + 2 \paren {m_1 + m_2 + 2 m_1 m_2} } {1 + 2 \paren {n_1 + n_2 + 2 n_1 n_2} }       | c =  }} {{eqn | r = \dfrac {1 + 2 m} {1 + 2 n}       | c = where $m = m_1 + m_2 + 2 m_1 m_2, n = n_1 + n_2 + 2 n_1 n_2$ }} {{end-eqn}} Hence by the Two-Step Subgroup Test, $\struct {S, \times}$ is a subgroup of $\struct {\Q_{>0}, \times}$. It has been established that $S$ is an infinite set. Hence by definition $\struct {S, \times}$ is an infinite group. Finally, from Subgroup of Abelian Group is Abelian, $\struct {S, \times}$ is an abelian group. {{qed}}
We will prove that :$\forall x, y \in S: \left({ y \npreceq x \implies \exists p \in X: x \preceq p \land y \npreceq p }\right)$ Let $x, y \in S$ such that :$y \npreceq x$ By Not Preceding implies There Exists Meet Irreducible Element Not Preceding :$\exists p \in S: p$ is meet irreducible and $x \preceq p$ and $y \npreceq p$ By definition of greatest element: :$p \ne \top$ and $p \in \mathit{IRR}\left({L}\right)$ By definitions of difference and singleton: :$p \in X$ Thus :$\exists p \in X: x \preceq p \land y \npreceq p$ {{qed|lemma}} Hence by Order Generating iff Not Preceding implies There Exists Element Preceding and Not Preceding: :$X$ is order generating. {{qed}}
Let $x \in S$. Thus by definition of complete lattice: :$x^\succeq \cap Y$ admits an infimum. By definition of complete lattice: :$x^\succeq$ admits an infimum and :$x^\succeq \cap X$ admits an infimum. By Intersection is Subset: :$x^\succeq \cap Y \subseteq x^\succeq$ By Infimum of Subset: :$\inf \left({x^\succeq}\right) \preceq \inf \left({x^\succeq \cap Y}\right)$ By Infimum of Upper Closure of Element: :$x \preceq \inf \left({x^\succeq \cap Y}\right)$ By Set Intersection Preserves Subsets/Corollary: :$x^\succeq \cap X \subseteq x^\succeq \cap Y$ By Infimum of Subset: :$\inf \left({x^\succeq \cap Y}\right) \preceq \inf \left({x^\succeq \cap X}\right)$ By definition of order generating: :$\inf \left({x^\succeq \cap Y}\right) \preceq  x$ Thus by definition of antisymmetry: :$x = \inf \left({x^\succeq \cap Y}\right)$ {{qed}}
By definition of polar form of complex numbers, the elements of $S$ are also elements of the circle group $\struct {K, \times}$: :$K = \set {z \in \C: \cmod z = 1}$ $S$ is infinite by construction. Thus $S \subseteq C$ and trivially $S \ne \O$. Let $a, b \in S$. Then: :$a = \cos \theta_1 + i \sin \theta_1$ and: :$b = \cos \theta_2 + i \sin \theta_2$ for some $\theta_1, \theta_2 \in \Q$. We have that: {{begin-eqn}} {{eqn | l = a \times b       | r = \paren {\cos \theta_1 + i \sin \theta_1} \paren {\cos \theta_2 + i \sin \theta_2}       | c =  }} {{eqn | r = \map \cos {\theta_1 + \theta_2} + i \map \sin {\theta_1 + \theta_2}       | c =  }} {{eqn | o = \in       | r = S       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = b^{-1}       | r = \cos \theta_2 - i \sin \theta_2       | c =  }} {{eqn | o = \in       | r = S       | c =  }} {{end-eqn}} Hence by the Two-Step Subgroup Test, $\struct {S, \times}$ is a subgroup of $\struct {K, \times}$. It has been established that $S$ is an infinite set. Hence by definition $\struct {S, \times}$ is an infinite group. Finally, from Subgroup of Abelian Group is Abelian, $\struct {S, \times}$ is an abelian group. {{qed}}
From Inversion Mapping is Permutation, $\iota$ is a permutation. It remains to be shown that $\iota$ has the morphism property {{iff}} $G$ is abelian. === Sufficient Condition === Suppose $\iota$ is an automorphism. Then: {{begin-eqn}} {{eqn | l = \forall x, y \in G: \map \iota {x \circ y}       | r = \map \iota x \circ \map \iota y       | c = {{Defof|Morphism Property}} }} {{eqn | ll= \leadsto       | l = \paren {x \circ y}^{-1}       | r = x^{-1} \circ y^{-1}       | c = Definition of $\iota$ }} {{end-eqn}} Thus from Inverse of Commuting Pair, $x$ commutes with $y$. This holds for all $x, y \in G$. So $\struct {G, \circ}$ is abelian by definition. {{qed|lemma}} === Necessary Condition === Let $\struct {G, \circ}$ be abelian. {{begin-eqn}} {{eqn | l = \forall x, y \in G: \paren {x \circ y}^{-1}       | r = x^{-1} \circ y^{-1}       | c = Inverse of Commuting Pair }} {{eqn | ll= \leadsto       | l = \map \iota {x \circ y}       | r = \map \iota x \circ \map \iota y       | c = Definition of $\iota$ }} {{end-eqn}} Thus $\iota$ has the morphism property and is therefore an automorphism. {{qed}}
Let $S = \set {x \in G: \order x > 2}$ be the set of all elements of $G$ whose order is strictly greater than $2$. Let $h \in S$. Then from Order of Group Element equals Order of Inverse: :$h^{-1} \in S$. Because $\order h > 2$ it follows that $h^{-1} \ne h$ by definition of self-inverse. Thus every element in $S$ can be paired off with its inverse which is also in $S$. It follows that $S$ has an even number of elements. Thus $G \setminus S$ also contains an even number of elements. From Identity is Only Group Element of Order 1, there is precisely $1$ element of $G$ whose order is less than $2$. This leaves $G$ with an odd number of elements of order $2$. {{qed}}
By definition of primitive complex $n$th root of unity: :$\omega = e^{2 m i \pi k}$ for some $m \in \Z: 1 \le m < n$. Thus: {{begin-eqn}} {{eqn | l = \paren {w \omega^k}^n       | r = w^n \paren {e^{2 m i \pi k / n} }^n       | c =  }} {{eqn | r = z e^{2 m i \pi k}       | c =  }} {{eqn | r = z \paren {e^{2 i \pi} }^{m k}       | c =  }} {{eqn | r = z \times 1^{m k}       | c =  }} {{eqn | r = z       | c =  }} {{end-eqn}} This demonstrates that $w \omega^k$ is one of the complex $n$th roots of $z$. All of the complex $n$th roots of unity are represented by powers of $\omega$. Thus it follows from Roots of Complex Number that: :$z^{1 / n} = \set {w \omega^k: k \in \set {1, 2, \ldots, n - 1} }$ are the $n$ complex $n$th roots of $z$. {{qed}}
Let $x, y \in S$ such that :$x \wedge y \preceq \top$ Thus by definition of greatest element: :$x \preceq \top$ or $y \preceq \top$ {{qed}}
=== Sufficient Condition === Let $p$ be prime element. Let $A$ be non-empty finite subsets of $S$. Define :$\map P X : \equiv X \ne \O \land \inf X \preceq p \implies \exists x \in X: x \preceq p$ where $X \subseteq S$. We will prove that :$\forall x \in A, B \subseteq A: \map P B \implies \map P {B \cup \set x}$ Let $x \in A, B \subseteq A$ such that :$\map P B$ and :$B \cup \set x \ne \O$ and $\map \inf {B \cup \set x} \preceq p$ Case $B = \O$. By Union with Empty Set: :$B \cup \set x = \set x$ By Infimum of Singleton: :$\inf \set x = x$ By definition of singleton: :$x \in \set x$ Thus :$\exists z \in B \cup \set x: z \preceq p$ {{qed|lemma}} Case $B \ne \O$. By Subset of Finite Set is Finite: :$B$ is finite. By Existence of Non-Empty Finite Infima in Meet Semilattice: :$B, \set x$ admit infima. By Infimum of Infima: :$\map \inf {B \cup \set x} = \inf B \wedge \inf \set x$ By definition of prime element: :$\inf B \preceq p$ or $x \preceq p$ Case $\inf B \preceq p$. By assumption: :$\exists z \in B: z \preceq p$ By definition of union: :$z \in B \cup \set x$ Thus :$\exists z \in B \cup \set x: z \preceq p$ {{qed|lemma}} Case $x \preceq p$. By definitions of union and singleton: :$x \in B \cup \set x$ Thus :$\exists z \in B \cup \set x: z \preceq p$ {{qed|lemma}} By definition of empty set: :$\map P \O$ By Induction of Finite Set: :$\map P A$ Thus :if $\inf A \preceq p$, then there exists element $x$ of $A$ such that $x \preceq p$. {{qed|lemma}} === Necessary Condition === Suppose that :for all non-empty finite subsets $A$ of $S$: ::if $\inf A \preceq p$, then there exists element $x$ of $A$ such that $x \preceq p$. Let $x, y \in S$ such that :$x \wedge y \preceq p$ Define $A := \set {x, y}$ By definitions of non-empty set and finite set: :$A$ is non-empty finite subset of $S$. By definition of meet: :$\inf A = x \wedge y$ By assumption: :$\exists z \in A: z \preceq p$ Thus by definition of unordered tuple: :$x \preceq p$ or $y \preceq p$ {{qed}}
Establish the mapping $\phi: S_3 \to G$ as follows: {{begin-eqn}} {{eqn | l = \map \phi e       | r = z }} {{eqn | l = \map \phi {\paren {123} }       | r = \dfrac 1 {1 - z} }} {{eqn | l = \map \phi {\paren {132} }       | r = \dfrac {z - 1} z }} {{eqn | l = \map \phi {\paren {23} }       | r = \dfrac 1 z }} {{eqn | l = \map \phi {\paren {13} }       | r = 1 - z }} {{eqn | l = \map \phi {\paren {12} }       | r = \dfrac z {z - 1} }} {{end-eqn}} From Isomorphism by Cayley Table, the two Cayley tables can be compared by eye to ascertain that $\phi$ is an isomorphism: === Cayley Table of Symmetric Group on $3$ Letters === The Cayley table for $S_3$ is as follows: {{:Symmetric Group on 3 Letters/Cayley Table}}</onlyinclude> === Group Generated by $1 / z$ and $1 - z$ === The Cayley table for $S$ is as follows: {{:Group Generated by Reciprocal of z and 1 minus z/Cayley Table}}
Let $p$ be a prime element. Let $x, y \in S$ such that :$p = x \wedge y$ By definition of reflexivity: :$x \wedge y \preceq p$ By definition of prime element: :$x \preceq p$ or $y \preceq p$ By Meet Precedes Operands: :$p \preceq x$ and $p \preceq y$ Thus by definition of antisymmetry: :$p = x$ or $p = y$ {{qed}}
By Prime Element is Meet Irreducible: :if $p$ is a prime element, then $p$ is meet irreducible. Assume that: :$p$ is meet irreducible. Let $x, y \in S$ such that: :$x \wedge y \preceq p$ {{begin-eqn}} {{eqn | l = p       | r = p \vee \paren {x \wedge y}       | c = Preceding iff Join equals Larger Operand }} {{eqn | r = \paren {p \vee x} \wedge \paren {p \vee y}       | c = {{Defof|Distributive Lattice}} }} {{end-eqn}} By definition of meet irreducible: :$p = p \vee x$ or $p = p \vee y$ Thus by Preceding iff Join equals Larger Operand: :$x \preceq p$ or $y \preceq p$ {{qed}}
At each outcome, the total payoff of $G$ will be an amount which will (for at least one outcome) not be zero Let an $n + 1$th player be introduced to $G$ who has one move: :$(1): \quad$ Select any player $m$. :$(2): \quad$ If the total payoff of $G$ is $+k$, receive in payment $k$ from player $m$. :$(3): \quad$ If the total payoff of $G$ is $-k$, pay $k$ to player $m$. This new game is the same as $G$ but with an extra (dummy) player, and is now zero-sum. {{qed}}
A cooperative game is one where players form coalitions against the other players. If the players in a two-person zero-sum game were to form a coalition, there would be no other players to form it against. Further, as the total payoff is zero, there would be no benefit in collaborating on using one strategy over another, as when they pool their payoffs they are back where they started. {{qed}}
Since $G$ is simple, it can have at most one edge between each two vertices. Therefore, the mapping: :$v: \map E G \to \powerset {\map V G}, \map v e = \set {a_e, b_e}$ which assigns to each edge its endvertices $a_e, b_e$, is an injection. From Power Set of Finite Set is Finite and Domain of Injection Not Larger than Codomain, it follows that $\map E G$ is finite. Hence $G$ is finite. {{qed}} Category:Graph Theory Category:Simple Graphs itanc072i6il71lvytg9hxwb3ysbn34
By the definition of a complex number, we have: :$z = \map \Re z + i \map \Im z$ Then: {{begin-eqn}} {{eqn | l = \cmod z       | r = \sqrt {\paren {\map \Re z}^2 + \paren {\map \Im z}^2}       | c = {{Defof|Complex Modulus}} }} {{eqn | o = \ge       | r = \sqrt {\paren {\map \Re z}^2 }       | c = Square of Real Number is Non-Negative, as $\map \Im z$ is real }} {{eqn | r = \cmod {\map \Re z}       | c = Square of Real Number is Non-Negative, as $\map \Re z$ is real }} {{end-eqn}} {{qed}}
By the definition of a complex number, we have: :$z = \map \Re z + i \map \Im z$ {{begin-eqn}} {{eqn | l = \cmod z       | r = \sqrt {\paren {\map \Re z}^2 + \paren {\map \Im z}^2}       | c = {{Defof|Complex Modulus}} }} {{eqn | o = \ge       | r = \sqrt {\paren {\map \Im z}^2 }       | c = Square of Real Number is Non-Negative, as $\map \Re z$ is real }} {{eqn | r = \size {\map \Im z}       | c = Square of Real Number is Non-Negative, as $\map \Im z$ is real }} {{end-eqn}} {{qed}}
=== Sufficient Condition === Suppose $\exists \beta: \struct {S_\beta, \tau_\beta}$ is not a $T_0$ space. Then $\exists a, b \in S_\beta$ such that $\forall U_\beta \in \tau_\beta$, either $a, b \in U_\beta$ or $a, b \notin U_\beta$. Consider the elements $y, z \in S$ defined as: :$y = \family {x_\alpha} : x_\alpha = \begin {cases} s_\alpha & : \alpha \ne \beta \\ a & : \alpha = \beta \end {cases}$ :$z = \family {x_\alpha} : x_\alpha = \begin {cases} s_\alpha & : \alpha \ne \beta \\ b & : \alpha = \beta \end {cases}$ That is, $y$ and $z$ match (arbitrarily) on all ordinates except that for $\beta$. Let $H \subseteq S: y \in H$. Then $z \in H$ as $\forall U_\beta \in \map {\pr_\beta} H : b \in U_\beta$ Similarly, let $K \subseteq S: z \in K$. Then $y \in K$ as $\forall U_\beta \in \map {\pr_\beta} H : a \in U_\beta$ So $T$ is not a $T_0$ (Kolmogorov) space. {{qed|lemma}} === Necessary Condition === Suppose $T$ is not a $T_0$ (Kolmogorov) 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 ordinate. Suppose, $a_m = p, b_m = q$ for some ordinate $m$. 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 $T_0$ (Kolmogorov) space. {{qed}}
A regular space is a topological space which is both a $T_0$ (Kolmogorov) space and a $T_3$ space. Hence from: :Product Space is $T_0$ iff Factor Spaces are $T_0$ and :Product Space is $T_3$ iff Factor Spaces are $T_3$ it follows that $T$ is a regular space {{iff}} each of $\struct {S_i, \tau_i}$ is a regular space. {{qed|lemma}} A Tychonoff (completely regular) space is a topological space which is both a $T_0$ (Kolmogorov) space and a $T_{3 \frac 1 2}$ space. Hence from: :Product Space is $T_0$ iff Factor Spaces are $T_0$ and :Product Space is $T_{3 \frac 1 2}$ iff Factor Spaces are $T_{3 \frac 1 2}$ it follows that $T$ is a Tychonoff space {{iff}} each of $\struct {S_i, \tau_i}$ is a Tychonoff space. {{qed|lemma}} A normal space is a topological space which is both a $T_1$ (Fréchet) space and a $T_4$ space. Hence from: :Product Space is $T_1$ iff Factor Spaces are $T_1$ and :Factor Spaces are $T_4$ if Product Space is $T_4$ it follows that if $T$ is a normal space then each of $\struct {S_i, \tau_i}$ is a normal space. {{qed|lemma}} A completely normal space is a topological space which is both a $T_1$ (Fréchet) space and a $T_5$ space. Hence from: :Product Space is $T_1$ iff Factor Spaces are $T_1$ and :Factor Spaces are $T_5$ if Product Space is $T_5$ it follows that if $T$ is a completely normal space then each of $\struct {S_i, \tau_i}$ is a completely normal space. {{qed}}
Let $\alpha \in \R$, $n \in \Z$. From Real Number is between Floor Functions: :$\floor \alpha \le \alpha < \floor \alpha + 1$ For any $n < \floor \alpha \le \alpha$: {{begin-eqn}} {{eqn | l = \size {n - \alpha}       | r = \alpha - n       | c = {{Defof|Absolute Value}} }} {{eqn | o = >       | r = \alpha - \floor \alpha }} {{eqn | r = \size {\alpha - \floor \alpha}       | c = {{Defof|Absolute Value}} }} {{end-eqn}} For any $n > \floor \alpha + 1 > \alpha$: {{begin-eqn}} {{eqn | l = \size {n - \alpha}       | r = n - \alpha       | c = {{Defof|Absolute Value}} }} {{eqn | o = >       | r = \floor \alpha + 1 - \alpha       | c =  }} {{eqn | r = \size {\floor \alpha + 1 - \alpha}       | c = {{Defof|Absolute Value}} }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \min \set {\size {n - \alpha}: n \in \Z}       | r = \min \set {\size {\floor \alpha - \alpha}, \size {\floor \alpha + 1 - \alpha} }       | c = Other $n$'s are disregarded by above }} {{eqn | r = \min \set {\alpha - \floor \alpha, \floor \alpha + 1 - \alpha} }} {{eqn | r = \min \set {\set \alpha, 1 - \set \alpha}       | c = {{Defof|Fractional Part}} }} {{end-eqn}} which shows that the definitions are indeed equivalent. {{qed}} Category:Real Analysis mc6dwrx70vuuipe3s97te11mdqlyvuo