nyuuzyou/wikis huggingface dataset:
Product Space is Completely Hausdorff iff Factor Spaces are Completely Hausdorff/Sufficient Condition

== Theorem ==
Let $\SS = \family {\struct {S_\alpha, \tau_\alpha} }$ be an indexed family of topological spaces for $\alpha$ in some indexing set $I$.
Let $\ds T = \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\SS$.
Let each of $\struct {S_\alpha, \tau_\alpha}$ for $\alpha \in I$ be completely Hausdorff spaces.
Then $T$ is a completely Hausdorff spaces.
== Proof ==
Let $x, y \in S : x \ne y$ be arbitrary.
Then $x_\alpha \ne y_\alpha$ for some $\alpha \in I$.
Since $\struct {S_\alpha, \tau_\alpha}$ is completely Hausdorff spaces then:
:$\exists U, V \in \tau_\alpha: x_\alpha \in U, y_\alpha \in V : U^- \cap V^- = \O$
Let $\pr_\alpha: S \to S_\alpha$ be the projection of $S$ to $S_\alpha$.
Then:
 
From Preimage of Subset is Subset of Preimage:
:$\map {\pr_\alpha^\gets} U \subseteq \map {\pr_\alpha^\gets} {U^-}$
and
:$\map {\pr_\alpha^\gets} V \subseteq \map {\pr_\alpha^\gets} {V^-}$
From Projection from Product Topology is Open and Continuous:General Result
:$\pr_\alpha: S \to S_\alpha$ is continuous.
From Continuity Defined from Closed Sets:
:$\map {\pr_\alpha^\gets} {U^-}, \map {\pr_\alpha^\gets} {V^-}$ are closed in $T$
 
From Set Closure is Smallest Closed Set in Topological Space:
:$\paren {\map {\pr_\alpha^\gets} U}^- \subseteq \map {\pr_\alpha^\gets} {U^-}$
and
:$\paren {\map {\pr_\alpha^\gets} V}^- \subseteq \map {\pr_\alpha^\gets} {V^-}$
From Subsets of Disjoint Sets are Disjoint:
:$\paren {\map {\pr_\alpha^\gets} U}^- \cap \paren{\map {\pr_\alpha^\gets} V}^- = \O$
By definition of the projection $\pr_\alpha$:
:$\map {\pr_\alpha} x = x_\alpha \in U$
By definition of the preimage under $\pr_\alpha$:
:$x \in \map {\pr_\alpha^\gets} U$
Similarly:
:$y \in \map {\pr_\alpha^\gets} V$
By definition of the product topology $\tau$:
:$\map {\pr_\alpha^\gets} U, \map {\pr_\alpha^\gets} V \in \tau$
As $x, y \in S$ were arbitrary, it follows that $T$ is a completely Hausdorff space by definition.
Category:Product Space is Completely Hausdorff iff Factor Spaces are Completely Hausdorff
Product Space is Homeomorphic to Product Space with Factors Commuted

== Theorem ==
Let $T_1$ and $T_2$ be topological spaces.
Let $T_1 \times T_2$ denote the product space of $T_1$ and $T_2$.
Let $t: T_1 \times T_2 \to T_2 \times T_1$ be the mapping defined as:
:$\forall \tuple {x, y} \in T_1 \times T_2: \map t {x, y} = \tuple {y, x}$
Then $t$ is a homeomorphism.
== Proof ==
$t$ is trivially a bijection.
Let $U$ be open in $T_2 \times T_1$.
Then by definition of product space:
:$U = U_2 \times U_1$
where:
:$U_2$ is open in $T_2$
:$U_1$ is open in $T_1$.
Hence by definition of product space:
:$t^{-1} \sqbrk {U_2 \times U_1} = U_1 \times U_2$ is open in $T_1 \times T_2$.
Hence it has been shown that $t$ is continuous.
Similarly, let $V$ be open in $T_1 \times T_2$.
Then by definition of product space:
:$V = V_1 \times V_2$
where:
:$V_1$ is open in $T_1$
:$V_2$ is open in $T_2$.
Hence by definition of product space:
:$t \sqbrk {V_1 \times V_2} = V_2 \times V_1$ is open in $T_2 \times T_1$.
Hence it has been shown that $t^{-1}$ is continuous.
The result follows by definition of homeomorphism.
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 18$
Category:Product Topology
Category:Homeomorphisms (Topological Spaces)
Product Space is Path-connected iff Factor Spaces are Path-connected

== Theorem ==
Let $\SS = \family {\struct {S_i, \tau_i} }_{i \mathop \in I}$ be an indexed family of topological spaces for $i$ in some indexing set $I$ such that $\forall i \in I: S_i \ne \O$.
Let $\ds T = \struct {S, \tau} = \prod_{i \mathop \in I} \struct {S_i, \tau_i}$ be the product space of $\SS$.
Then $T$ is a path-connected space each of $\struct {S_i, \tau_i}$ is a path-connected space.
== Proof ==
=== Necessary Condition ===
Suppose $S_i$ is path-connected for each $i \in I$. 
Let $x \ne y \in S$ be arbitrary.
Since each $S_i$ is path-connected, we have a continuous mapping:
:$f_i: \closedint 0 1 \to S_i$
such that $\map {f_i} 0 = x_i$ and $\map {f_i} 1 = y_i$ for all $i \in I$.
We then define:
: $f : \closedint 0 1 \to S$
by setting:
:$\forall t \in \closedint 0 1: \map f t_i = \map {f_i} t$
We have that $f_i = \pr_i \circ f$ are all continuous.
So, by Continuous Mapping to Product Space, it follows that $f$ is continuous.
Also:
: $\map f 0_i = \map {f_i} 0 = x_i$
and:
: $\map f 1_i = \map {f_i} 1 = y_i$
Therefore $\map f 0 = x$ and $\map f 1 = y$.
So there exists a continuous mapping:
: $f: \closedint 0 1 \to S$ 
such that $\map f 0 = x$ and $\map f 1 = y$.
Since $x$ and $y$ were arbitrary, we have shown that $S$ is path-connected.
=== Sufficient Condition ===
Conversely, suppose $T$ is a path-connected space.
Let $j$ be an arbitrary $j \in I$.
Let $x_j$ and $y_j$ be points in $S_j$ such that $x_j \ne y_j$.
We have that $\forall i \in I: S_i \ne \O$.
Using the axiom of choice, a choice function can be set up to allow the choice of some $q_i \in S_i$ for each $i \ne j$.
We then define:
: $\forall s \in T: \map {\pr_i} s = q_i$
From Constant Mapping is Continuous, it follows that each of $\pr_i: T \to S_i$ where $i \ne j$ is continuous:
Thus there exist points in $x, y \in S$ such that: 
: $\map {\pr_i} x = q_i$
: $\map {\pr_i} y = q_i$
for all $i \ne j$, and:
: $\map {\pr_j} x = x_j$
: $\map {\pr_j} y = y_j$
Since $T$ is path-connected, there exists a continuous mapping:
: $f: \closedint {0} {1} \to S$ 
such that $\map f 0 = x$ and $\map f 1 = y$.
By Continuous Mapping to Product Space:
: $\pr_j \circ f: \closedint 0 1 \to S_j$
is also continuous.
Furthermore:
: $\map {\pr_j \circ f} 0 = \map {\pr_j} x = x_j$
and:
: $\map {\pr_j \circ f} 1 = \map {\pr_j} y = y_j$
Thus since $x_j$ and $y_j$ were arbitrary, we have shown that $S_j$ is path-connected.
 
Since $j$ was arbitrary, we have shown that all $S_i$ are path-connected.
== Also see ==
* Finite Product Space is Connected iff Factors are Connected
== Sources ==
* : $6.3$: Path-connectedness (in passing)
Category:Path-Connected Spaces
Category:Product Spaces
Product Space is Product in Category of Topological Spaces

== Theorem ==
Let $\mathbf {Top}$ be the category of topological spaces.
Let $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be an indexed family of topological spaces where $I$ is an arbitrary index set.
Let $\struct {\XX, \tau}$ be the product space of $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$.
Then $\struct {\XX, \tau}$ is the product of $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ in $\mathbf{Top}$.
== Proof ==
Category:Category of Topological Spaces
Category:Product Topology
Product Space is T0 iff Factor Spaces are T0

== Theorem ==
Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $T = T_A \times T_B$ be the product space formed from $T_A$ and $T_B$.
Then $T$ is a $T_0$ (Kolmogorov) space $T_A$ and $T_B$ are themselves both $T_0$ (Kolmogorov) spaces.
=== General Result ===
== Proof ==
=== Necessary Condition === 
Let $T$ be a $T_0$ (Kolmogorov) space.
From:
:Subspace of Product Space is Homeomorphic to Factor Space
:Separation Properties Preserved in Subspace
:Separation Axioms Preserved under Homeomorphism
it follows that $T_A$ and $T_B$ are both $T_0$.
=== Sufficient Condition ===
Now suppose that $T$ is not $T_0$.
Then $\exists p, q \in S_A \times S_B$ such that:
:$p \ne q$
and:
:$\forall U \in \tau: \set {p, q} \subseteq U$ or $\set {p, q} \cap U = \O$
If $p \ne q$, then they are different in at least one coordinate.
, let $p_A \ne q_A$.
Let $V \in \tau_A$ such that $p_A \in V$.
Then from Projection from Product Topology is Continuous:
:$\pr_A^{-1} \sqbrk V \in \tau$.
Since $p \in \pr_A^{-1} \sqbrk V$, we have that:
:$q \in \pr_A^{-1} \sqbrk V$
From Projection is Surjection:
:$\map {\pr_i} q = q_A \in V$
So:
:$\forall V \in \tau_A: \set {p_A, q_A} \subseteq V$
or:
:$\set {p_A, q_A} \cap V = \O$
Thus, by definition, $T_A$ is not $T_0$.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces
Category:T0 Spaces
Category:Product Spaces
Product Space is T0 iff Factor Spaces are T0/General Result

== Theorem ==
Let $\SS = \family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be an indexed family of topological spaces where $I$ is an arbitrary index set.
Let $\ds T = \struct {S, \tau} = \prod \struct {S_\alpha, \tau_\alpha}$ be the product space of $\SS$.
Then $T$ is a $T_0$ (Kolmogorov) space each of $\struct{S_\alpha, \tau_\alpha}$ is a $T_0$ (Kolmogorov) space.
== Proof ==
=== 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.
=== 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.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces
Category:T0 Spaces
Category:Product Spaces
Product Space is T1 iff Factor Spaces are T1

== Theorem ==
Let $\mathbb S = \family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be an indexed family of topological spaces for $\alpha$ in some indexing set $I$.
Let $T = \struct {S, \tau} = \ds \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\mathbb S$.
Then $T$ is a $T_1$ (Fréchet) space each of $\struct {S_\alpha, \tau_\alpha}$ is a $T_1$ (Fréchet) space.
== Proof ==
=== Necessary Condition ===
Suppose that for some $\beta$, $\struct {S_\beta, \tau_\beta}$ is not a $T_1$ space.
Then $\exists a, b \in S_\beta$ such that $\forall U_\beta \in \tau_\beta$, $a \in U_\beta \implies b \in U_\beta$.
Consider the elements $y, z \in S$ defined as:
That is, $y$ and $z$ match on all coordinates except that for $\beta$.
Let $H \subseteq S: y \in H$ be open.
Then $z \in H$ as $\forall U_\beta \in \map {\pr_\beta} H: a \in U_\beta \implies b \in U_\beta$
So $T$ is not a $T_1$ (Fréchet) space.
=== Sufficient Condition ===
Suppose $T$ is not a $T_1$ (Fréchet) space.
Then $\exists a, b \in S, a \ne b$ such that for all $U \in \tau$, $a \in U \implies b \in U$.
Then $a$ and $b$ are different in at least one coordinate.
Suppose, $a_\alpha = p, b_\alpha = q$ for some coordinate $\alpha$.
Then for each $U_\alpha \in \tau_\alpha$:
:$\ds a \in U_\alpha \times \prod_{\beta \mathop \ne \alpha} S_\beta \implies b \in U_\alpha \times \prod_{\beta \mathop \ne \alpha} S_\beta$
But:
:$\ds a \in U_\alpha \times \prod_{\beta \mathop \ne \alpha} S_\beta$ $p \in U_\alpha$
:$\ds b \in U_\alpha \times \prod_{\beta \mathop \ne \alpha} S_\beta$ $q \in U_\alpha$
from which we infer, for each $U_\alpha \in \tau_\alpha$:
:$p \in U_\alpha \implies q \in U_\alpha$
It follows that $\struct {S_\alpha, \tau_\alpha}$ is not a $T_1$ (Fréchet) space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces
Category:T1 Spaces
Category:Product Spaces
Product Space is T2 iff Factor Spaces are T2

== Theorem ==
Let $\mathbb S = \family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be an indexed family of topological spaces for $\alpha$ in some indexing set $I$.
Let $\ds T = \struct{S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\mathbb S$.
Then $T$ is a $T_2$ (Hausdorff) space each of $\struct {S_\alpha, \tau_\alpha}$ is a $T_2$ (Hausdorff) space.
== Proof ==
=== Necessary Condition ===
This is shown in Factor Spaces of Hausdorff Product Space are Hausdorff.
=== Sufficient Condition ===
This is shown in Product of Hausdorff Factor Spaces is Hausdorff:General Result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces
Category:Hausdorff Spaces
Category:Product Spaces
Product Space is T3 1/2 iff Factor Spaces are T3 1/2

== Theorem ==
Let $\mathbb S = \family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be an indexed family of topological spaces for $\alpha$ in some indexing set $I$ with $S_\alpha \ne \O$ for every $\alpha \in I$.
Let $\ds T = \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\mathbb S$.
Then $T$ is a $T_{3 \frac 1 2}$ space each of $\struct {S_\alpha, \tau_\alpha}$ is a $T_{3 \frac 1 2}$ space.
== Proof ==
=== Necessary Condition ===
=== Sufficient Condition ===
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces
Category:T3 1/2 Spaces
Category:Product Spaces
Product Space is T3 1/2 iff Factor Spaces are T3 1/2/Factor Spaces are T3 1/2 implies Product Space is T3 1/2

== Theorem ==
Let $\mathbb S = \family {\struct{S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be an indexed family of topological spaces for $\alpha$ in some indexing set $I$ with $S_\alpha \ne \O$ for every $\alpha \in I$.
Let $\ds T = \struct{S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\mathbb S$.
For each $\alpha \in I$, let $\struct{S_\alpha, \tau_\alpha}$ be a $T_{3 \frac 1 2}$ space.
Then $T$ is a $T_{3 \frac 1 2}$ space.
== Proof ==
Let $\struct{S_\alpha, \tau_\alpha}$ is a $T_{3 \frac 1 2}$ space for each $\alpha \in I$. 
Let $x \in S$.
Let $F$ be a closed subset of $S$ such that $x \notin F$.
By definition of a closed subset:
:$S \setminus F \in \tau$
By definition of the product topology, there exists an open set $B$ of the natural basis containing $x$ which is disjoint from $F$.
By definition of the natural basis, $B$ is of the form:
:$\map {\pr_{\alpha_1}^\gets} {U_1} \cap \dotsb \cap \map {\pr_{\alpha_n}^\gets} {U_n}$
where:
:$\pr_{\alpha_k}$ is the $\alpha_k$-th projection from $S$ 
:$U_k$ is open in $S_{\alpha_k}$ for all $1 \le k \le n$
By definition of a $T_{3 \frac 1 2}$ space, for each $1 \le k \le n$ there exists a continuous mapping:
:$f_k: S_{\alpha_k} \to \closedint 0 1$
such that:
:$\map {f_k} {x_{\alpha_k} } = 1$
and:
:$\map {f_k} {S_{\alpha_k} \setminus U_k} = 0$
Let $g_k = f_k \circ \pr_{\alpha_k}$ be the composite mapping of $f_k$ with $\pr_{\alpha_k}$ for each $1 \le k \le n$.
From Composite of Continuous Mappings is Continuous each $g_k: S \to \closedint 0 1$ is a continuous mapping.
We define $g: S \to \closedint 0 1$ by setting:
:$\map g y = \min \set {\map {g_k} {y_{\alpha_k} }: k = 1, \dotsc, n}$
From Minimum Rule for Continuous Functions:
:$g$ is continuous. 
Now:
Let $y \in F$.
By definition of disjoint sets:
:$\exists j \in \set {1, \dotsc, n} : y \notin \map {\pr_{\alpha_j}^\gets} {U_j}$
By definition of the inverse image mapping of $\pr_{\alpha_j}$:
:$y_{\alpha_j} \notin U_j$
Thus:
:$\map {f_j} {y_{\alpha_j} } = 0$
So:
Therefore $T$ is a $T_{3 \frac 1 2}$ space.
Category:T3 1/2 Spaces
Category:Product Spaces
Product Space is T3 1/2 iff Factor Spaces are T3 1/2/Product Space is T3 1/2 implies Factor Spaces are T3 1/2

== Theorem ==
Let $\mathbb S = \family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be an indexed family of non-empty topological spaces for $\alpha$ in some indexing set $I$.
Let $\ds T = \struct{S, \tau} = \prod_{\alpha \mathop \in I} \struct{S_\alpha, \tau_\alpha}$ be the product space of $\mathbb S$.
Let $T$ be a $T_{3 \frac 1 2}$ space.
Then for each $\alpha \in I$, $\struct {S_\alpha, \tau_\alpha}$ is a $T_{3 \frac 1 2}$ space.
== Proof ==
Suppose $T$ is a $T_{3 \frac 1 2}$ space.
Since $S_\alpha \ne \O$ we also have $S \ne \O$ by the axiom of choice.
Let $\alpha \in I$ be arbitrary.
From Subspace of Product Space is Homeomorphic to Factor Space:
:$\struct {S_\alpha, \tau_\alpha}$ is homeomorphic to a subspace $T_\alpha$ of $T$.
From $T_{3 \frac 1 2}$ Property is Hereditary:
:$T_\alpha$ is $T_{3 \frac 1 2}$.
From $T_{3 \frac 1 2}$ Space is Preserved under Homeomorphism:
:$\struct {S_\alpha, \tau_\alpha}$ is $T_{3 \frac 1 2}$.
Because $\alpha \in I$ was arbitrary then the result follows.
Category:T3 1/2 Spaces
Category:Product Spaces
Product Space is T3 iff Factor Spaces are T3

== Theorem ==
Let $\mathbb S = \family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be an indexed family of non-empty topological spaces for $\alpha$ in some indexing set $I$.
Let $\ds T = \struct{S, \tau} = \prod_{\alpha \mathop \in I} \struct{S_\alpha, \tau_\alpha}$ be the product space of $\mathbb S$.
Then $T$ is a $T_3$ space each of $\struct {S_\alpha, \tau_\alpha}$ is a $T_3$ space.
== Proof ==
=== Necessary Condition ===
=== Sufficient Condition ===
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces
Category:T3 Spaces
Category:Product Spaces
Category:Product Space is T3 iff Factor Spaces are T3
Product Space is T3 iff Factor Spaces are T3/Factor Spaces are T3 implies Product Space is T3

== Theorem ==
Let $\mathbb S = \family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be an indexed family of non-empty topological spaces for $\alpha$ in some indexing set $I$.
Let $\ds T = \struct{S, \tau} = \ds \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\mathbb S$.
For each $\alpha \in I$, let $\struct {S_\alpha, \tau_\alpha}$ be a $T_3$ space.
Then $T$ is a $T_3$ space.
== Proof ==
Let $U$ be open in $T$ and $x \in U$.
From Natural Basis of Product Topology, there exists $\ds U' = \prod_{\alpha \mathop \in I} U'_\alpha$ such that:
:for all $\alpha \in I : U_\alpha \in \tau_\alpha$
:$J = \set {\alpha \in I: U_\alpha \ne X_\alpha}$ is finite
:$x \in U'$
:$U' \subseteq U$
By definition of a $T_3$ space, for all $\alpha \in J$:
:$\exists V_\alpha \in \tau_\alpha : x_\alpha \in V_\alpha$ and $V^-_\alpha \subseteq U'_\alpha$
For $\alpha \notin J$ let:
:$V_\alpha = X_\alpha$
and let:
:$\ds V = \prod_{\alpha \mathop \in I} V_\alpha$
By definition of the Cartesian product V:
:$x \in V$
From Natural Basis of Product Topology:
:$V$ open in $T$.
Now consider the Cartesian product of closed sets:
:$\ds \prod_{\alpha \mathop \in I} V^-_\alpha$
From Product of Closed Sets is Closed, $\ds \prod_{\alpha \mathop \in I} V^-_\alpha$ is closed.
From Underlying Set of Topological Space is Closed, for $\alpha \notin J$:
:$V^-_\alpha = X^-_\alpha = X_\alpha$
From Cartesian Product of Family of Subsets, it follows that:
:$\ds V = \prod_{\alpha \mathop \in I} V_\alpha \subseteq \prod_{\alpha \mathop \in I} V^-_\alpha \subseteq \prod_{\alpha \mathop \in I} U'_\alpha \subseteq U$
It follows that $T$ is a $T_3$ space by definition.
Category:Product Space is T3 iff Factor Spaces are T3
Product Space is T3 iff Factor Spaces are T3/Product Space is T3 implies Factor Spaces are T3

== Theorem ==
Let $\mathbb S = \family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be an indexed family of non-empty topological spaces for $\alpha$ in some indexing set $I$.
Let $\ds T = \struct{S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\mathbb S$.
Let $T$ be a $T_3$ space.
Then for each $\alpha \in I$, $\struct {S_\alpha, \tau_\alpha}$ is a $T_3$ space.
== Proof ==
Suppose $T$ is a $T_3$ space.
As $S_\alpha \ne \O$ we also have $S \ne \O$ by the axiom of choice.
Let $\alpha \in I$.
From Subspace of Product Space is Homeomorphic to Factor Space, $\struct {S_\alpha, \tau_\alpha}$ is homeomorphic to a subspace $T_\alpha$ of $T$.
From $T_3$ property is hereditary, $T_\alpha$ is $T_3$.
From T3 Space is Preserved under Homeomorphism, $\struct {S_\alpha, \tau_\alpha}$ is $T_3$.
As $\alpha \in I$ was arbitrary, the result follows.
Category:Product Space is T3 iff Factor Spaces are T3
Product Space of Subspaces is Subspace of Product Space

== Theorem ==
Let $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be a family of topological spaces where $I$ is an arbitrary index set.
Let $\ds T = \struct {X, \tau} = \prod_{i \mathop \in I} \struct {X_i, \tau_i}$ be the product space of $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$.
Let $\family {\struct {Y_i, \upsilon_i} }_{i \mathop \in I}$ be a family of topological spaces such that:
:$\forall i \in I : \struct {Y_i, \upsilon_i}$ is a topological subspace of $\struct {X_i, \tau_i}$
Let $\ds S = \struct {Y, \upsilon} = \prod_{i \mathop \in I} \struct {Y_i, \upsilon_i}$ be the product space of $\family {\struct {Y_i, \upsilon_i} }_{i \mathop \in I}$.
Let $T_Y = \struct {Y, \tau_Y}$ be the topological subspace of $T$.
Then $S = T_Y$.
== Proof ==
From Cartesian Product of Family of Subsets, $Y \subseteq X$.
Thus the topological subspace $T_Y$ is well-defined.
From Natural Basis of Product Topology, a (synthetic) basis for $T$ is:
:$\ds \BB_T = \set {\prod_{i \mathop \in I} U_i : U_i \in \tau_i, U_i = X_i \text{ for all but finitely many } i \in I}$
From Basis for Topological Subspace a (synthetic) basis for $T_Y$ is:
:$\BB_Y = \set {U \cap Y: U \in \BB_T}$
That is:
By definition of a subspace topology,
:$\forall i \in I: \upsilon_i = \set{U \cap S_i : U \in \tau_i}$
From Natural Basis of Product Topology a (synthetic) basis for $S$ is:
:$\ds \BB_S = \set {\prod_{i \mathop \in I} V_i : V_i \in \upsilon_i, V_i = Y_i \text{ for all but finitely many } i \in I}$
That is:
Thus $\BB_S = \BB_Y$ and therefore $S = T_Y$.
Category:Product Topology
Category:Subsets
Product Topology is Coarsest Topology such that Projections are Continuous

== Theorem ==
Let $\mathbb X = \family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be an indexed family of topological spaces where $I$ is an arbitrary index set.
Let $X$ be the cartesian product of $\mathbb X$:
:$\ds X := \prod_{i \mathop \in I} X_i$
Let $\tau$ be the product topology on $X$.
For each $i \in I$, let $\pr_i : X \to X_i$ be the corresponding projection which maps each ordered tuple in $X$ to the corresponding element in $X_i$:
:$\forall x \in X: \map {\pr_i} x = x_i$
Then $\tau$ is the coarsest topology on $X$ such that all the $\pr_i$ are continuous.
== Proof ==
The result follows from the definition of the product topology and Equivalence of Definitions of Initial Topology.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Functions
Category:Projections
Category:Product Topology
Product is Left Identity Therefore Left Cancellable

== Theorem ==
Let $\struct {S, \circ}$ be a semigroup.
Let $e_L \in S$ be a left identity of $S$.
Let $a \in S$ such that:
:$\exists b \in S: b \circ a = e_L$
Then $a$ is left cancellable in $\struct {S, \circ}$.
== Proof ==
Let $x, y \in S$ be arbitrary.
Then:
The result follows by definition of left cancellable.
== Also see ==
* Product is Right Identity Therefore Right Cancellable
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.11$
Category:Semigroups
Category:Cancellability
Category:Identity Elements