nyuuzyou/wikis huggingface dataset:
Finite Lexicographic Order on Well-Ordered Sets is Well-Ordering

== Theorem ==
Let $S$ be a set which is well-ordered by $\preccurlyeq$.
Let $\preccurlyeq_l$ be the lexicographic order on the set $T_n$ of all ordered $n$-tuples of $S$:
:$\tuple {x_1, x_2, \ldots, x_n} \prec \tuple {y_1, y_2, \ldots, y_n}$ :
::$\exists k: 1 \le k \le n$ such that $\forall 1 \le j < k: x_j = y_j$ but $x_k \prec y_k$ in $S$.
Then for a given $n \in \N_{>0}$, $\preccurlyeq_l$ is a well-ordering on $T_n$.
== Proof ==
It is straightforward to show that $\preccurlyeq_l$ is a total ordering on $T_n$.
It remains to investigate the nature of $\preccurlyeq_l$ as to whether or not it is a well-ordering.
Consider $T_n$ where $n \in \N_{>0}$.
It is clear that $\struct {T_1, \preccurlyeq_l}$ is order isomorphic to $\struct {S, \preccurlyeq}$.
Thus as $\preccurlyeq$ is a well-ordering on $S$, $\preccurlyeq_l$ is a well-ordering on $T_1$.
Now, let us assume that $\preccurlyeq_l$ is a well-ordering on $T_k$ for some $k \in \N: k \ge 1$.
Let $A$ be a non-empty subset of $T_{k + 1}$.
Let $A_1$ be the set of all of the first components of the ordered $n$-tuples that comprise $A$.
Since $A_1$ is a non-empty subset of $S$, and $S$ is itself well-ordered by $\preccurlyeq$, it follows that $A_1$ contains a minimal element $x$ by $\preccurlyeq_l$.
Let $A_x$ be the subset of $A$ in which the first component equals $x$.
We may consider $A_x$ to be a subset of $T_k$ where this first component $x$ has been suppressed.
But we assumed that $T_k$ is well-ordered by $\preccurlyeq_l$.
So $A_x$ contains a minimal element $\tuple {x, x_2, x_3, \ldots, x_{k + 1} }$ by $\preccurlyeq_l$.
This element $\tuple {x, x_2, x_3, \ldots, x_{k + 1} }$ is the minimal element of $A$ by $\preccurlyeq_l$.
Hence, by definition, $T_{k + 1}$ is well-ordered by $\preccurlyeq_l$.
The result follows by induction.
== Sources ==
* : $\S 17$: Well Ordering
* : $\S 1.2.1$: Mathematical Induction: Exercise $15 \ \text{(d)}$
Category:Lexicographic Order
Category:Well-Orderings
Finite Main Lemma of Propositional Tableaux

== Lemma ==
Let $\mathbf H$ be a finite set of WFFs of propositional logic.
Either $\mathbf H$ has a tableau confutation or $\mathbf H$ has a model.
== Proof ==
Let $\mathbf H$ be a finite set of WFFs of propositional logic which does not have a tableau confutation.
By the Tableau Extension Lemma, the tableau which consists only of a root node with hypothesis set $\mathbf H$ can be extended into a finite finished tableau $T$.
The tableau $T$ still has root $\mathbf H$.
Since $T$ is not a confutation, it has a finished branch $\Gamma$.
By , $\Gamma$ has a model, $\MM$, say.
In particular, $\MM$ is a model of $\mathbf H$ as required.
== Comment ==
From Tableau Confutation implies Unsatisfiable, we already know that $\mathbf H$ can not have both a tableau confutation and a model.
This result gives us that $\mathbf H$ has a tableau confutation $\mathbf H$ does not have a model.
== Sources ==
* : $\S 1.10$: Completeness: Lemma $1.10.1$
Category:Propositional Tableaux
Finite Monoid with Left Cancellable Operation is Group

== Theorem ==
Let $\struct {S, \circ}$ be a finite monoid.
Let $\circ$ be a left cancellable operation.
Then $\struct {S, \circ}$ is a group.
== Proof ==
, and are satisfied by dint of $\paren {S, \circ}$ being a monoid.
Recall the definition of left cancellable operation:
:$\forall a, b, c \in S: c \circ a = c \circ b \implies a = b$
Let $\lambda_c: S \to S$ be the left regular representation of $\struct {S, \circ}$ with respect to $c$.
By Left Cancellable iff Left Regular Representation Injective, $\lambda_c$ is an injection.
By Left Regular Representation wrt Left Cancellable Element on Finite Semigroup is Bijection, $\lambda_c$ is a bijection.
Thus $a \circ b = e$ has a unique solution for all $a \in S$.
That is, holds on $S$.
== Sources ==
* : $\S 3.2$: Groups; the axioms: Proposition $2$
Category:Monoids
Category:Finite Groups
Finite Monoid with Right Cancellable Operation is Group

== Theorem ==
Let $\struct {S, \circ}$ be a finite monoid.
Let $\circ$ be a right cancellable operation.
Then $\struct {S, \circ}$ is a group.
== Proof ==
, and are satisfied by dint of $\struct {S, \circ}$ being a monoid.
Recall the definition of right cancellable operation:
:$\forall a, b, c \in S: a \circ c = b \circ c \implies a = b$
Let $\rho_c: S \to S$ be the right regular representation of $\struct {S, \circ}$ with respect to $c$.
By Right Cancellable iff Right Regular Representation Injective, $\rho_c$ is an injection.
By Right Regular Representation wrt Right Cancellable Element on Finite Semigroup is Bijection, $\rho_c$ is a bijection.
Thus $a \circ b = e$ has a unique solution for all $a \in S$.
That is, holds on $S$.
== Sources ==
* : $\S 3.2$: Groups; the axioms: Proposition $2$
Category:Monoids
Category:Finite Groups
Finite Multiplicative Subgroup of Field is Cyclic

== Theorem ==
Let $\struct {F, +, \times}$ be a field.
Let $\struct {F^*, \times}$ denote the multiplicative group of $F$.
Let $C$ be a finite subgroup of $\struct {F^*, \times}$.
Then $C$ is cyclic.
== Proof ==
We have that $\struct {F^*, \times}$ is an abelian group.
From Subgroup of Abelian Group is Abelian, $C$ is a finite abelian group.
From Fundamental Theorem of Finite Abelian Groups, $C$ is the internal group direct product of cyclic groups $H_1, H_2, \ldots, H_r$ whose orders are given as:
:$\order {H_i} = {p_i}^{e_i}$
for some prime numbers $p_1, p_2, \ldots, p_r$ and positive integers $e_1, e_2, \ldots, e_r$.
From Internal and External Group Direct Products are Isomorphic:
:$C \cong H_1 \times H_2 \times \cdots \times H_r$
where $\cong$ denotes (group) isomorphism.
As Powers of Coprime Numbers are Coprime, the orders of $H_1, \ldots, H_r$ are coprime.
By Group Direct Product of Cyclic Groups, $H_1 \times \ldots \times H_r$ is cyclic.
Hence $C$ is cyclic.
Category:Finite Multiplicative Subgroup of Field is Cyclic
Category:Field Theory
Category:Cyclic Groups
Finite Non-Empty Subset of Ordered Set has Maximal and Minimal Elements

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $T \subseteq S$ be a finite, non-empty subset of $S$.
Then $T$ has a maximal element and a minimal element.
=== Corollary ===
== Proof ==
We will show that each finite subset of $S$ has a minimal element.
The existence of a maximal element then follows from duality.
Proof by induction:
For all $n \in \Z_{\ge 1}$, let $\map P n$ be the proposition:
:Every set with $n$ elements has a minimal element.
=== Basis for the Induction ===
Let $T$ have exactly one element $x$.
Since $x \nprec x$ it follows that $x$ is minimal.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:Every set with $k$ elements has a minimal element.
from which it is to be shown that:
:Every set with $k + 1$ elements has a minimal element.
=== Induction Step ===
Suppose that every subset of $S$ with $k$ elements has a minimal element.
Let $T$ have $k + 1$ elements.
Then:
:$T = T_0 \cup \set x$
where $T_0$ has $k$ elements and $x \in T \setminus T_0$.
Then $T_0$ has a minimal element $m_0$.
If $m_0$ is not a minimal element of $T$, then:
:$x \prec m_0$
Thus $x$ is a minimal element of $T$.
Thus either $m_0$ or $x$ is a minimal element of $T$.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:For every finite, non-empty subset $T$ of $S$, $T$ has a maximal element and a minimal element 
The result follows by the Principle of Mathematical Induction.
== Sources ==
* : $\S\text I.3$: Theorem $3$
Category:Minimal Elements
Category:Maximal Elements
Finite Non-Empty Subset of Ordered Set has Maximal and Minimal Elements/Corollary

== Theorem ==
Let $\struct {S, \preceq}$ be a finite ordered set.
Let $x \in S$.
Then there exists a maximal element $M \in S$ and a minimal element $m \in S$ such that:
:$m \preceq x \preceq M$
== Proof ==
Let $T = \set{y : x \preceq y}$.
By the reflexivity of the ordering $\preceq$:
:$x \preceq x$
So $x \in T$ and $T$ is non-empty.
From Finite Non-Empty Subset of Ordered Set has Maximal and Minimal Elements:
:$\struct {T, \preceq}$ has a maximal element $M \in T$
We now show that $M$ is a maximal element in $\struct{S, \preceq}$. 
Let $y \in S$ such that:
:$M \preceq y$
By the transitiviy of the ordering $\preceq$:
:$x \prec y$
So $y \in T$.
By the definition of a maximal element:
:$y = M$
Similarly for $T' = \set {y : y \preceq x}$:
:$\struct {T', \preceq}$ has a minimal element $m \in T'$
and $m$ is a minimal element in $\struct {S, \preceq}$.
Category:Minimal Elements
Category:Maximal Elements
Finite Non-Empty Subset of Totally Ordered Set has Smallest and Greatest Elements

== Theorem ==
Let $\struct {S, \preceq}$ be a totally ordered set.
Then every finite $T$ such that $\O \subset T \subseteq S$ has both a smallest and a greatest element.
== Proof 1 ==
== Proof 2 ==
Category:Total Orderings
Category:Finite Non-Empty Subset of Totally Ordered Set has Smallest and Greatest Elements
Finite Non-Empty Subset of Totally Ordered Set has Smallest and Greatest Elements/Proof 1

== Theorem ==
== Proof ==
Let $A \subseteq \N_{>0}$ such that every $B \subseteq S$ such that $\card B = n$ has a greatest and a smallest element.
Any $B \subseteq S$ such that $\card B = 1$ has $1$ element, $b \in B$ say.
Then $b$ is both the greatest and smallest element of $B$.
So $1 \in A$.
Let $n \in A$.
Let $B \subseteq S$ such that $\card B = n + 1$.
Then $\exists b \in B$, and $\card {B \setminus \set b} = n$ elements by Cardinality Less One.
So, by the induction hypothesis, $B \setminus \set b$ has a greatest element $c$ and a smallest element $a$, as $n \in A$.
Note that $b \ne c$ as $c \in B \setminus \set b$ but $b \notin B \setminus \set b$.
We have that $\struct {S, \preceq}$ is a totally ordered set.
Hence by the Trichotomy Law, either $b \prec c$ or $c \prec b$ as $b \ne c$.
If $b \prec c$ then $c$ is the greatest element of $B$, otherwise it's $b$.
Similarly, either $b \prec a$ or $a \prec b$, and thus either $a$ or $b$ is the smallest element of $B$.
Either way, $B$ has both a greatest and a smallest element, and therefore $n + 1 \in A$.
Therefore, by the Principle of Mathematical Induction, $A = \N_{>0}$ and the proof is complete.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 17$: Finite Sets: Theorem $17.9$
Category:Finite Non-Empty Subset of Totally Ordered Set has Smallest and Greatest Elements
Category:Proofs by Induction
Finite Non-Empty Subset of Totally Ordered Set has Smallest and Greatest Elements/Proof 2

== Theorem ==
== Proof ==
The result follows from:
: Finite Non-Empty Subset of Ordered Set has Maximal and Minimal Elements
: Minimal Element of Chain is Smallest Element
: Maximal Element of Chain is Greatest Element
== Sources ==
* : $\S\text I.3$
Category:Finite Non-Empty Subset of Totally Ordered Set has Smallest and Greatest Elements
Finite Number of Groups of Given Finite Order

== Theorem ==
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Then there exists a finite number of types of group of order $n$.
== Proof ==
For any group $\struct {G, \circ}$ of order $n$ and any set of $n$ elements, $X$ can be the underlying set of a group which is isomorphic to $\struct {G, \circ}$, as follows:
Choose a bijection $\phi: G \to X$.
Define the group operation $*$ on $X$ by the rule:
:$\map \phi {g_1} * \map \phi {g_2} = \map \phi {g_1 \circ g_2}$
for all $g_1, g_2 \in G$.
By definition, $\phi$ is an isomorphism.
From Isomorphism Preserves Groups it follows that $\struct {X, *}$ is a group.
Thus all groups of order $n$ of all possible types appear among all possible arrangements of binary operation.
But from Count of Binary Operations on Set, this count is $n^{\paren {n^2} }$.
This is the upper bound to the number of types of group of order $n$.
== Sources ==
* : $1$: Introduction to Finite Group Theory: $1.2$
Category:Order of Groups
Finite Orbit under Group of Automorphisms of Field implies Separable over Fixed Field

== Theorem ==
Let $E$ be a field.
Let $G \le \Aut E$ be a subgroup of its automorphism group.
Let $F = \map {\operatorname {Fix}_E} G$ be its fixed field.
Let $\alpha \in E$ have a finite orbit under $G$.
Then $\alpha$ is separable over $F$.
== Proof ==
Let $\Lambda$ be the orbit of $\alpha$ under $G$.
By:
:Minimal Polynomial of Element with Finite Orbit under Group of Automorphisms over Fixed Field in terms of Orbit
the product:
:$\map p x = \ds \prod_{\beta \in \Lambda} \paren {x - \beta}$
is the minimal polynomial of $\alpha$ over $F$.
By Product of Distinct Monic Linear Polynomials is Separable, $p$ is separable.
Thus $\alpha$ is separable over $F$.
Category:Field Extensions
Finite Order Elements of Infinite Abelian Group form Normal Subgroup

== Theorem ==
Let $G$ be an infinite abelian group.
Let $H \subseteq G$ be the subset of $G$ defined as:
:$H := \set {x \in G: x \text { is of finite order in } G}$
Then $H$ forms a normal subgroup of $G$.
=== Corollary ===
== Proof ==
Let $e$ be the identity element of $G$.
From Identity is Only Group Element of Order 1, $\order e = 1$ and so $H \ne \O$.
Let $a \in H$.
Then by Order of Group Element equals Order of Inverse:
:$\order a = \order {a^{-1} }$
and so $a \in H$.
Let $a, b \in H$.
From Order of Product of Abelian Group Elements Divides LCM of Orders of Elements:
:$\order {a b} \divides \lcm \set {\order a, \order b}$
where:
:$\order a$ denotes the order of $a$
:$\divides$ denotes divisibility
:$\lcm$ denotes the lowest common multiple.
Thus $a b$ is also of finite order.
Thus by definition:
:$a b \in H$
By the Two-Step Subgroup Test it follows that $H$ is a subgroup of $G$.
By Subgroup of Abelian Group is Normal, $H$ is normal in $G$.
== Sources ==
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $14$
Category:Normal Subgroups
Category:Infinite Groups
Category:Abelian Groups
Finite Order Elements of Infinite Abelian Group form Normal Subgroup/Corollary

== Corollary to Finite Order Elements of Infinite Abelian Group form Normal Subgroup ==
Let $G$ be an infinite abelian group.
Let $H \subseteq G$ be the subset of $G$ defined as:
:$H := \set {x \in G: x \text { is of finite order in } G}$
All the elements of the quotient group $G / H$ are of infinite order except the identity.
== Proof ==
Let $e$ be the identity element of $G$.
From Finite Order Elements of Infinite Abelian Group form Normal Subgroup, $H$ forms a normal subgroup of $G$.
Hence $G / H$ is defined as the quotient group.
From Quotient Group is Group, $\order {e H} = H$ is the identity element of $G / H$.
From Identity is Only Group Element of Order 1, $\order {e H} = 1$.
Let $x H$ be of finite order in $G / H$, where $x \in G$.
Let $\order {x H} = m$.
Then $\paren {x H}^m = H$ by definition of finite order element.
Thus:
:$\exists n \in \Z_{>0}: \paren {x^m}^n = e$
and so $x \in H$.
That is:
:$x H = H$
and so $x H$ is the identity element of $G / H$.
Thus if $x H$ is of finite order in $G / H$, it has to be the identity element.
== Sources ==
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $14$
Category:Normal Subgroups
Category:Infinite Groups
Category:Abelian Groups
Finite Ordinal Plus Transfinite Ordinal

== Theorem ==
Let $n$ be a finite ordinal.
Let $x$ be a transfinite ordinal.
Then:
: $n + x = x$
== Proof ==
By Transfinite Induction on $x$.
The proof will use $<$, $\in$, and $\subset$ interchangeably. This is justified by Transitive Set is Proper Subset of Ordinal iff Element of Ordinal.
=== Base Case ===
By our hypothesis, $\omega \le x$, so $x \nless \omega$, so we may begin our induction at $\omega$.
From these conclusions, we may deduce that:
:$\ds \omega = \bigcup_{y \mathop \in \omega} \paren {n + y}$
=== Inductive Case ===
=== Limit Case ===
== Sources ==
* : $\S 8.10$
Category:Ordinal Arithmetic
Category:Transfinite Arithmetic
Category:Finite Ordinals
Finite Ordinal Times Ordinal

== Theorem ==
Let $m$ and $n$ be finite ordinals.
Let $m \ne 0$, where $0$ is the zero ordinal.
Let $x$ be a limit ordinal.
Then:
:$m \times \paren {x + n} = x + \paren {m \times n}$
== Proof ==
=== Lemma ===
By Ordinal Multiplication is Left Distributive:
:$m \times \paren {x + n} = \paren {m \times x} + \paren {m \times n}$
It remains to prove that $x = m \times x$.
Since $x$ is a limit ordinal, it follows that:
== Sources ==
* : $\S 8.29$
Category:Ordinal Arithmetic
Category:Transfinite Arithmetic
Category:Finite Ordinals
Finite Ordinal Times Ordinal/Lemma

== Lemma ==
Let $m$ be a finite ordinal.
Let $m \ne 0$, where $0$ is the zero ordinal.
Then:
:$m \times \omega = \omega$
where $\omega$ denotes the minimally inductive set.
== Proof ==
Also, by Subset is Right Compatible with Ordinal Multiplication:
:$\omega \le \paren {m \times \omega}$
The lemma follows from the definition of equality.
Category:Ordinal Arithmetic
Category:Transfinite Arithmetic
Category:Finite Ordinals
Category:Minimally Inductive Set
Finite Ordinal is equal to Natural Number

== Theorem ==
Let $n$ be an element of the minimally inductive set.
Let $x$ be an ordinal.
Then:
:$n \sim x \implies n = x$
== Proof ==
Let $n \ne x$.
Then either $n < x$ or $x < n$ by Ordinal Membership is Trichotomy.
If $x < n$, then by Subset of Finite Set is Finite both $x$ and $n$ are finite.
Therefore by No Bijection between Finite Set and Proper Subset:
:$x \not \sim n$
Suppose $n < x$.
Then $x \sim n$ implies that $x$ is finite by definition.
By No Bijection between Finite Set and Proper Subset:
:$x \not \sim n$
Therefore by Rule of Transposition:
:$n \sim x \implies n = x$
== Sources ==
* : $\S 10.20$
Category:Cardinals
Category:Minimally Inductive Set
Finite Ordinal is not Subset of one of its Elements

== Theorem ==
Let $n$ be a finite ordinal.
Then:
:$\nexists x \in n: n \subseteq x$
that is, $n$ is not a subset of one of its elements.
== Proof ==
Let $S$ be the set of all those finite ordinals $n$ which are not a subset of any of its elements.
That is:
:$n \in S \iff n \in \omega \land \forall x \in n: n \nsubseteq x$
We know that $0 = \O$ is not a subset of any of its elements, as $\O$ by definition has no elements.
So $0 \in S$.
Now suppose $n \in S$.
From Set is Subset of Itself:
:$n \subseteq n$
But as $n \in S$ it follows by definition of $S$ that:
:$n \notin n$
By definition of the successor of $n$, it follows that:
:$n^+ \nsubseteq n$
Now from Subset Relation is Transitive:
:$n^+ \subseteq x \implies n \subseteq x$
But since $n \in S$ it follows that:
:$x \notin n$
So:
:$n^+ \nsubseteq n$
and:
:$\forall x \in n: n^+ \nsubseteq x$
So $n^+$ is not a subset of any of its elements.
That is:
:$n^+ \in S$
So by the Principle of Mathematical Induction:
:$S = \omega$
Hence the result.
== Sources ==
* : $\S 12$: The Peano Axioms
Category:Finite Ordinals
Finite Product Space is Connected iff Factors are Connected

== Theorem ==
Let $T_1 = \struct {S_1, \tau_1}, T_2 = \struct {S_2, \tau_2}, \dotsc, T_n = \struct {S_n, \tau_n}$ be topological spaces.
Let $T = \ds \prod_{i \mathop = 1}^n T_i$ be the product space of $T_1, T_2, \ldots, T_n$.
Then $T$ is connected each of $T_1, T_2, \ldots, T_n$ are connected.
=== General Case ===
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 2}$, let $\map P n$ be the proposition:
:$T$ is connected each of $T_1, T_2, \ldots, T_n$ are connected.
=== Basis for the Induction ===
$\map P 2$ is the case:
:The product space $T_1 \times T_2$ is connected $T_1$ and $T_2$ are connected.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\ds \prod_{i \mathop = 1}^k T_i$ is connected each of $T_1, T_2, \ldots, T_k$ are connected.
from which it is to be shown that:
:$\ds \prod_{i \mathop = 1}^{k + 1} T_i$ is connected each of $T_1, T_2, \ldots, T_{k + 1}$ are connected.
=== Induction Step ===
This is the induction step:
By definition of product space:
:$\ds \prod_{i \mathop = 1}^{k + 1} T_i = \paren {\prod_{i \mathop = 1}^k T_i} \times T_{k + 1}$
But from the basis for the induction:
:$\ds \prod_{i \mathop = 1}^{k + 1} T_i$ is connected $\ds \prod_{i \mathop = 1}^k T_i$ is connected and $T_{k + 1}$ is connected
and from the induction hypothesis:
:$\ds \prod_{i \mathop = 1}^k T_i$ is connected each of $T_1, T_2, \ldots, T_k$ are connected.
Hence:
:$\ds \prod_{i \mathop = 1}^{k + 1} T_i$ is connected each of $T_1, T_2, \ldots, T_{k + 1}$ are connected.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction:
:For all $n \in \Z_{\ge 2}$, $T$ is connected each of $T_1, T_2, \ldots, T_n$ are connected.
== Also see ==
* Product Space is Path-connected iff Factor Spaces are Path-connected
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness: Functions and Products
Category:Product Spaces
Category:Connected Spaces
Finite Product Space is Connected iff Factors are Connected/Basis for the Induction

== Theorem ==
Let $T_1$ and $T_2$ be topological spaces.
Then the product space $T_1 \times T_2$ is connected $T_1$ and $T_2$ are connected.
== Proof ==
==== Necessary Condition ====
Let $T_1 \times T_2$ be connected.
By Projection from Product Topology is Continuous, $T_1$ and $T_2$ are continuous images under the projections $\pr_1$ and $\pr_2$.
Hence by Continuous Image of Connected Space is Connected, $T_1$ and $T_2$ are connected.
==== Sufficient Condition ====
Suppose that $T_1$ and $T_2$ are connected.
Define:
:$C_y = T_1 \times \set y$ for each $y \in T_2$
:$B = \set {x_0} \times T_2$ for some fixed $x_0 \in T_1$.
Each $C_y$ is homeomorphic to $T_1$ by Topological Product with Singleton.
By Connectedness is a Topological Property, each $C_y$ is therefore connected.
By the same argument, $B$ is also connected.
Also:
:$C_y \cap B = \set {\tuple {x_0, y} }$ and hence is non-empty
:$\ds T_1 \times T_2 = B \cup \bigcup_{y \mathop \in T_2} C_y$.
So by the corollary to Union of Connected Sets with Non-Empty Intersections is Connected, it follows that $T_1 \times T_2$ is connected.
== Also see ==
* Product Space is Path-connected iff Factor Spaces are Path-connected
== Sources ==
* : $6.2$: Connectedness: Proposition $6.2.17$
Category:Product Spaces
Category:Connected Spaces
Finite Product Space is Connected iff Factors are Connected/General Case

== Theorem ==
Let $I$ be an indexing set.
Let $\family {T_\alpha}_{\alpha \mathop \in I}$ be an indexed family of topological spaces.
Let $T = \ds \prod_{\alpha \mathop \in I} T_\alpha$ be the Cartesian space of $\family {T_\alpha}_{\alpha \mathop \in I}$.
Let $T = \ds \overline {\bigcup_{\alpha \mathop \in I} S_\alpha}$.
Let $\tau$ be a topology on $T$ such that the subsets ${S'}_\alpha \subseteq \ds \prod T_\alpha$ where ${S'}_\alpha = \set {\family {y_\beta} \in T: y_\beta = x \beta \text { for all } \beta \ge \alpha}$ is homeomorphic to $S_{\alpha - 1} \times T_\alpha$.
Then $T$ is connected each of $T_\alpha: \alpha \in I$ are connected.
== Proof ==
Let the Axiom of Choice be assumed.
Let $I$ be well-ordered.
Let $x = \family {x_\alpha} \in T$ be some arbitrary fixed element of $T$.
Let $S_\alpha = \set {\family {y_\beta} \in T: y_\beta = x \beta \text { for all } \beta \ge \alpha}$.
We have that $S_\alpha$ is homeomorphic to $S_{\alpha - 1} \times T_\alpha$.
Then from Finite Product Space is Connected iff Factors are Connected, $S_\alpha$ is connected $S_{\alpha - 1}$ is.
Let $\alpha$ be a limit ordinal.
Then:
:$S_\alpha = \ds \paren {\bigcup_{\beta \mathop < \alpha} S_\beta}^-$
where $X^-$ denotes the closure of $X$.
So if each $S_\beta$ is connected for $\beta < \alpha$, it follows that $S_\alpha$ must likewise be connected, as $\family \gamma$ is nested.
Thus:
:$T = \ds \overline {\bigcup_{\alpha \mathop \in I} S_\alpha}$ is connected.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness: Functions and Products
Category:Product Spaces
Category:Connected Spaces