nyuuzyou/wikis huggingface dataset:
Product Rule for Counting/Examples/Choices from 2 and 3

== Example of Use of Product Rule for Counting ==
The canonical example concerns choices from the menu at a restaurant:
You may select exactly one dish from each category:
;Starters
:$(1): \quad$ Crottled Greeps
:$(2): \quad$ Stone Soup
:$(3): \quad$ Petty-Dwarf Roots
;Main Course
:$(1): \quad$ Hufu Salad
:$(2): \quad$ Braised Trake in Funistrada
The diner then has $2 \times 3 = 6$ possible different meals:
:$(1): \quad$ Crottled Greeps with Hufu Salad
:$(2): \quad$ Crottled Greeps with Braised Trake in Funistrada
:$(3): \quad$ Stone Soup with Hufu Salad
:$(4): \quad$ Stone Soup with Braised Trake in Funistrada
:$(5): \quad$ Petty-Dwarf Roots with Hufu Salad
:$(6): \quad$ Petty-Dwarf Roots with Braised Trake in Funistrada
== Sources ==
* : $\text {3-1}$ Permutations and Combinations: Example $\text {3-1}$
Category:Product Rule for Counting
Product Rule for Counting/Examples/Choices from 4, 3 and 2

== Example of Use of Product Rule for Counting ==
Let $N$ be the number of ways you can choose at least $1$ item of fruit from:
:$4$ (indistinguishable) oranges
:$3$ (indistinguishable) bananas
:$2$ (indistinguishable) apples
Then:
:$N = 59$
== Proof ==
You can choose:
:$0$, $1$, $2$, $3$ or $4$ oranges, giving you $5$ options
:$0$, $1$, $2$ or $3$ bananas, giving you $4$ options
:$0$, $1$, or $2$ apples, giving you $3$ options
Each of these options is independent of each other.
Hence the Product Rule for Counting can be applied.
Hence the total number of ways you can choose a selection of fruit from the above is $5 \times 4 \times 3 = 60$.
This includes the option of choosing no items of fruit at all.
We specifically want to exclude that option.
Hence:
:$N = 5 \times 4 \times 3 - 1 = 59$
== Sources ==
* : $\text I$. Algebra: Permutations and Combinations: Exercises $\text I$: $8$
Category:Product Rule for Counting
Product Rule for Counting/Examples/Choices from 5 and 3

== Example of Use of Product Rule for Counting ==
There are:
:$5$ different ways to travel from $A$ to $B$
:$3$ different ways to travel from $B$ to $C$.
Hence there are $5 \times 3 = 15$ different ways to travel from $A$ to $C$.
== Sources ==
* : Chapter $\text I$ Permutations and Combinations: The $r$, $s$ Principle: Example $1$.
Category:Product Rule for Counting
Product Rule for Counting/General Theorem

== Theorem ==
Suppose a process can be broken into $m$ successive, ordered, stages, with the $i$th stage having $r_i$ possible outcomes (for $i = 1, \ldots, m$).
Let the number of outcomes at each stage be independent of the choices in previous stages
Let the composite outcomes be all distinct.
Then the total procedure has $\ds \prod_{i \mathop = 1}^m r_i$ different composite outcomes.
== Proof ==
== Also known as ==
Some sources give this as the General Combinatorial Principle.
== Also see ==
* Fundamental Principle of Counting
== Sources ==
* : Chapter $\text I$ Permutations and Combinations: The $r$, $s$ Principle: Example $2$.
* : $5.1$ Basic Counting Principles
Category:Product Rule for Counting
Product Rule for Curl

== Theorem ==
Let $R$ be a region of space embedded in Cartesian $3$ space $\R^3$.
Let $\mathbf A$ be a vector field over $\mathbf V$.
Let $U$ be a scalar field over $\mathbf V$.
Then:
where:
:$\curl$ denotes the curl operator
:$\grad$ denotes the gradient operator
:$\times$ denotes vector cross product
== Proof ==
From Curl Operator on Vector Space is Cross Product of Del Operator and definition of the gradient operator:
where $\nabla$ denotes the del operator.
Hence we are to demonstrate that:
Let $\mathbf A$ be expressed as a vector-valued function on $\mathbf V$:
:$\mathbf A := \tuple {\map {A_x} {\mathbf r}, \map {A_y} {\mathbf r}, \map {A_z} {\mathbf r} }$
where $\mathbf r = \tuple {x, y, z}$ is the position vector of an arbitrary point in $R$.
Let $\tuple {\mathbf i, \mathbf j, \mathbf k}$ be the standard ordered basis on $\R^3$.
Then:
== Also presented as ==
This result can also be presented as:
:$\nabla \times \paren {U \mathbf A} = \map U {\nabla \times \mathbf A} + \paren {\nabla U} \times \mathbf A$
or:
:$\nabla \times \paren {U \mathbf A} = \map U {\nabla \times \mathbf A} - \mathbf A \times \paren {\nabla U}$
presupposing the implementations of $\curl$ and $\grad$ as operations using the del operator.
== Sources ==
* : Chapter $\text {IV}$: The Operator $\nabla$ and its Uses: $7$. Divergence and Curl of $S \mathbf A$: $(4.14)$
* : $\S 22$: Miscellaneous Formulas involving $\nabla$: $22.39$
Category:Curl Operator
Product Rule for Derivatives

== Theorem ==
Let $\map f x, \map j x, \map k x$ be real functions defined on the open interval $I$.
Let $\xi \in I$ be a point in $I$ at which both $j$ and $k$ are differentiable.
Let $\map f x = \map j x \map k x$.
Then:
:$\map {f'} \xi = \map j \xi \map {k'} \xi + \map {j'} \xi \map k \xi$
It follows from the definition of derivative that if $j$ and $k$ are both differentiable on the interval $I$, then:
:$\forall x \in I: \map {f'} x = \map j x \map {k'} x + \map {j'} x \map k x$
Using Leibniz's notation for derivatives, this can be written as:
:$\map {\dfrac \d {\d x} } {y \, z} = y \dfrac {\d z} {\d x} + \dfrac {\d y} {\d x} z$
where $y$ and $z$ represent functions of $x$.
=== General Result ===
== Proof ==
== Examples ==
== Also see ==
* Derivative of Product of Real Function and Vector-Valued Function
* Derivative of Vector Cross Product of Vector-Valued Functions
* Derivative of Dot Product of Vector-Valued Functions
* Derivative of Product of Operator-Valued Functions
* Leibniz's Rule in One Variable, of which this is the special case of the first derivative
== Historical Note ==
== Sources ==
* : $\text {II}$. Calculus: Differentiation
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Derivatives: $3.3.3$
* : $\S 13$: General Rules of Differentiation: $13.7$
* : $5$. Differential Calculus: Appendix: Differentiation Rules: $3.$ Product rule
* 
* 
* : Appendix: Table $1$: Derivatives
* 
* 
* : Appendix: Table $1$: Derivatives
Category:Product Rule for Derivatives
Category:Differential Calculus
Product Rule for Derivatives/Examples

== Examples of Use of Product Rule for Derivatives == 
=== Example: $2 a x e^{a x^2}$ ===
=== Example: $x \sin x$ ===
=== Example: $x \cot x$ ===
=== Example: $x^2 \arctan x$ ===
=== Example: $x e^x \sin x$ ===
=== Example: $\cot x e^{-x}$ ===
Category:Examples of Use of Product Rule for Derivatives
Product Rule for Derivatives/Examples/2 a x times Exponential of a x^2

== Examples of Use of Product Rule for Derivatives == 
:$\map {\dfrac \d {\d x} } {2 a x e^{a x^2} } = 2 a e^{a x^2} \paren {1 + 2 a x^2}$
== Proof ==
Let $u = 2 a x$.
Let $v = e^{a x^2}$.
Thus we have:
:$y = u v$
and so:
Category:Examples of Use of Product Rule for Derivatives
Product Rule for Derivatives/Examples/Cotangent of x times Exponential of -x

== Examples of Use of Product Rule for Derivatives == 
:$\map {\dfrac \d {\d x} } {\cot x e^{-x} } = -e^{-x} \paren {\cot x + \cosec^2 x}$
== Proof ==
Let $u = \cot x$.
Let $v = e^{-x}$.
Thus we have:
:$y = u v$
and so:
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Exercises $\text {IX}$: $22$.
Category:Examples of Use of Product Rule for Derivatives
Product Rule for Derivatives/Examples/x squared times Arctangent of x

== Examples of Use of Product Rule for Derivatives == 
:$\map {\dfrac \d {\d x} } {x^2 \arctan x} = 2 x \arctan x + \dfrac {x^2} {1 + x^2}$
== Proof ==
Let $u = x^2$.
Let $v = \arctan x$.
Thus we have:
:$y = u v$
and so:
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Exercises $\text {IX}$: $17$.
Category:Examples of Use of Product Rule for Derivatives
Product Rule for Derivatives/Examples/x times Cotangent of x

== Examples of Use of Product Rule for Derivatives == 
:$\map {\dfrac \d {\d x} } {x \cot x} = \cot x - x \cosec^2 x$
== Proof ==
Let $u = x$.
Let $v = \cot x$.
Thus we have:
:$y = u v$
and so:
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Exercises $\text {IX}$: $16$.
Category:Examples of Use of Product Rule for Derivatives
Product Rule for Derivatives/Examples/x times Exponential of x times Sine of x

== Examples of Use of Product Rule for Derivatives == 
:$\map {\dfrac \d {\d x} } {x e^x \sin x} = e^x \paren {\paren {1 + x} \sin x + x \cos x}$
== Proof ==
Let $u = x$.
Let $v = e^x$.
Let $w = \sin x$.
Thus we have:
:$y = u v w$
and so:
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Exercises $\text {IX}$: $21$.
Category:Examples of Use of Product Rule for Derivatives
Product Rule for Derivatives/Examples/x times Sine of x

== Examples of Use of Product Rule for Derivatives == 
:$\map {\dfrac \d {\d x} } {x \sin x} = x \cos x + \sin x$
== Proof ==
Let $u = x$.
Let $v = \sin x$.
Thus we have:
:$y = u v$
and so:
== Sources ==
* 
* 
Category:Examples of Use of Product Rule for Derivatives
Product Rule for Derivatives/General Result

== Theorem ==
Let $\map {f_1} x, \map {f_2} x, \ldots, \map {f_n} x$ be real functions differentiable on the open interval $I$.
then:
:$\forall x \in I: \ds \map {D_x} {\prod_{i \mathop = 1}^n \map {f_i} x} = \sum_{i \mathop = 1}^n \paren {\map {D_x} {\map {f_i} x} \prod_{j \mathop \ne i} \map {f_j} x}$
=== $3$ Factors ===
== Proof ==
Proof by Principle of Mathematical Induction:
For all $n \in \N_{\ge 1}$, let $\map P n$ be the proposition:
:$\ds \map {D_x} {\prod_{i \mathop = 1}^n \map {f_i} x} = \sum_{i \mathop = 1}^n \paren {\map {D_x} {\map {f_i} x} \prod_{j \mathop \ne i} \map {f_j} x}$
$\map P 1$ is true, as this just says:
:$\map {D_x} {\map {f_1} x} = \map {D_x} {\map {f_1} x}$
=== Basis for the Induction ===
$\map P 2$ is the case:
:$\map {D_x} {\map {f_1} x \map {f_2} x} = \map {D_x} {\map {f_1} x} \map {f_2} x + \map {f_1} x \map {D_x} {\map {f_2} x}$
which has been proved in Product Rule for Derivatives.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show 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 our induction hypothesis:
:$\ds \map {D_x} {\prod_{i \mathop = 1}^k \map {f_i} x} = \sum_{i \mathop = 1}^k \paren {\map {D_x} {\map {f_i} x} \prod_{j \mathop \ne i} \map {f_j} x}$
Then we need to show:
:$\ds \map {D_x} {\prod_{i \mathop = 1}^{k + 1} \map {f_i} x} = \sum_{i \mathop = 1}^{k + 1} \paren {\map {D_x} {\map {f_i} x} \prod_{j \mathop \ne i} \map {f_j} x}$
=== Induction Step ===
This is our induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \map {D_x} {\prod_{i \mathop = 1}^n \map {f_i} x} = \sum_{i \mathop = 1}^n \paren {\map {D_x} {\map {f_i} x} \prod_{j \mathop \ne i} \map {f_j} x}$ for all $n \in \N$
== Mnemonic device ==
:$\paren {f g}' = f g' + f' g$
:$\paren {f g h}' = f' g h + f g' h + f g h'$
and in general, making sure to exhaust all possible combinations, making sure that there are as many summands as there are functions being multiplied.
== Also see ==
* Derivative of Product of Real Function and Vector-Valued Function
* Derivative of Vector Cross Product of Vector-Valued Functions
* Derivative of Dot Product of Vector-Valued Functions
Category:Product Rule for Derivatives
Product Rule for Derivatives/General Result/3 Factors

== Theorem ==
Let $\map u x$, $\map v x$ and $\map w x$ be real functions differentiable on the open interval $I$.
Then:
:$\forall x \in I: \map {\dfrac \d {\d x} } {u v w} = u v \dfrac {\d w} {\d x} + u w \dfrac {\d v} {\d x} + v w \dfrac {\d u} {\d x}$
== Proof ==
Let $y = u v w$.
Then:
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Extension to triple product
* : $\S 13$: General Rules of Differentiation: $13.8$
Category:Product Rule for Derivatives
Product Rule for Derivatives/Proof

== Theorem ==
== Proof ==
First we note that from Differentiable Function is Continuous‎, $j$ is continuous at $\xi$.
Hence:
:$(1): \quad \map j {\xi + h} \to \map j \xi$ as $h \to 0$
So:
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Product
* : $\S 10.9 \ \text{(ii)}$
Category:Product Rule for Derivatives
Product Rule for Distributional Derivatives of Distributions multiplied by Smooth Functions

== Theorem ==
Let $\alpha \in \map {C^\infty} \R$ be a smooth real function.
Let $T \in \map {\DD'} \R$ be a distribution.
Then in the distributional sense it holds that:
:$\paren {\alpha T}' = \alpha' T + \alpha T'$
== Proof ==
Let $\phi \in \map \DD \R$ be a test function.
By the Product Rule for Derivatives:
:$\paren {\alpha \phi}' = \alpha' \phi + \alpha \phi'$
Hence:
== Sources ==
* : Chapter $\S 6.4$: A glimpse of distribution theory. Multiplication by $C^\infty$ functions
Category:Distributional Derivatives
Product Rule for Divergence

== Theorem ==
Let $\map {\mathbf V} {x_1, x_2, \ldots, x_n}$ be a vector space of $n$ dimensions.
Let $\mathbf A$ be a vector field over $\mathbf V$.
Let $U$ be a scalar field over $\mathbf V$.
Then:
:$\map {\operatorname {div} } {U \mathbf A} = \map U {\operatorname {div} \mathbf A} + \mathbf A \cdot \grad U$
where
:$\operatorname {div}$ denotes the divergence operator
:$\grad$ denotes the gradient operator
:$\cdot$ denotes dot product.
== Proof ==
From Divergence Operator on Vector Space is Dot Product of Del Operator and definition of the gradient operator:
where $\nabla$ denotes the del operator.
Hence we are to demonstrate that:
:$\nabla \cdot \paren {U \, \mathbf A} = \map U {\nabla \cdot \mathbf A} + \paren {\nabla U} \cdot \mathbf A$
Let $\mathbf A$ be expressed as a vector-valued function on $\mathbf V$:
:$\mathbf A := \tuple {\map {A_1} {\mathbf r}, \map {A_2} {\mathbf r}, \ldots, \map {A_n} {\mathbf r} }$
where $\mathbf r = \tuple {x_1, x_2, \ldots, x_n}$ is an arbitrary element of $\mathbf V$.
Let $\tuple {\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n}$ be the standard ordered basis of $\mathbf V$.
Then:
== Also presented as ==
This result can also be presented as:
:$\nabla \cdot \paren {U \, \mathbf A} = \map U {\nabla \cdot \mathbf A} + \paren {\nabla U} \cdot \mathbf A$
presupposing the implementations of $\operatorname {div}$ and $\grad$ as operations using the del operator.
== Sources ==
* : Chapter $\text {IV}$: The Operator $\nabla$ and its Uses: $7$. Divergence and Curl of $S \mathbf A$: $(4.13)$
* : $\S 22$: Miscellaneous Formulas involving $\nabla$: $22.38$
Category:Divergence Operator
Product Rule for Sequence in Normed Algebra

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {A, \norm {\, \cdot \,} }$ be a normed algebra over $\GF$. 
Let $\sequence {a_n}_{n \in \N}$ and $\sequence {b_n}_{n \in \N}$ be sequences in $A$ converging to $a$ and $b$ respectively. 
Then: 
:$a_n b_n \to a b$ 
== Proof ==
From Convergent Sequence in Normed Vector Space is Bounded, there exists $M > 0$ such that: 
:$\norm {a_n} \le M$ for each $n \in \N$.
We have for $n \in \N$: 
So from Sequence in Normed Vector Space Convergent to Limit iff Norm of Sequence minus Limit is Null Sequence, we have: 
:$a_n b_n \to a b$ 
Category:Normed Algebras
Category:Convergent Sequences (Normed Vector Spaces)
Product Sigma-Algebra Generated by Projections

== Theorem ==
Let $\struct {X, \Sigma_1}$ and $\struct {Y, \Sigma_2}$ be measurable spaces.
Let $\Sigma_1 \otimes \Sigma_2$ be the product $\sigma$-algebra on $X \times Y$.
Let $\pr_1: X \times Y \to X$ and $\pr_2: X \times Y \to Y$ be the first and second projections, respectively.
Then:
:$\Sigma_1 \otimes \Sigma_2 = \map \sigma {\pr_1, \pr_2}$
where $\sigma$ denotes generated $\sigma$-algebra.
== Proof ==
Let $R = \{A \times B: A \in \Sigma_1, B \in \Sigma_2\}$ be a generator for the product $\sigma$-algebra $\Sigma_1 \otimes \Sigma_2$.
Note that by Preimage of Element under Projection:
:$\pr_1^{-1}(A) = A \times Y \in R$ 
And:
:$\pr_2^{-1}(B) = X \times B \in R$ 
for any $A \in \Sigma_1$ and $B \in \Sigma_2$.
Let:
:$G = \{\pr_i^{-1}(C_i): C_i \in \Sigma_i, i = 1, 2\}$
Thus:
:$G \subseteq R$.
However, for any $A \times B \in R$:
:$A \times B = \pr_1^{-1}(A) \cap \pr_2^{-1}(B) \in \sigma(G)$
So we have:
:$R \subseteq \sigma(G)$.
But:
:$G \subseteq R$ 
and:
:$R \subseteq \sigma(G)$
we use Generated Sigma-Algebra Preserves Subset and ($\sigma(G) = \sigma(\sigma(G))$).
Thus:
: $\sigma(R) = \sigma(G)$.
It follows from the definition:
:$\Sigma_1 \otimes \Sigma_2 = \sigma(R) = \sigma(G) = \sigma (\pr_1, \pr_2)$
== Sources ==
* : $13.10 \ \text{(i)}$
Category:Product Sigma-Algebras
Product Space Basis Induced from Factor Space Bases

== Theorem ==
Let $\family {\struct{S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be an indexed family of topological spaces for $\alpha$ in some indexing set $I$.
Let $\BB_\alpha$ be a basis for the topology $\tau_\alpha$ for each $\alpha \in I$.
Let $\struct {S, \tau} = \ds \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$.
Let $\BB$ be the set of all cartesian products of the form $\ds \prod_{\alpha \mathop \in I} U_\alpha$ where:
:for all but finitely many indices $\alpha : U_\alpha = S_\alpha$
:for all $\alpha \in I : U_\alpha \ne S_\alpha \implies U_\alpha \in \BB_\alpha$
Then $\BB$ is a basis for the topology on the product space $\struct{S, \tau}$.
== Proof ==
Let $x = \family {x_\alpha}_{\alpha \mathop \in I} \in S$.
Let $W \in \tau$ such that $x \in W$.
From Natural Basis of Product Topology, the set $\BB'$ of cartesian products of the form $\ds \prod_{\alpha \mathop \in I} V_\alpha$ where:
:for all $\alpha \in I : V_\alpha \in \tau_\alpha$
:for all but finitely many indices $\alpha : V_\alpha = X_\alpha$
is a basis for $\tau$.
Therefore, there exists $V = \ds \prod_{\alpha \mathop \in I} V_\alpha \in \BB'$:
:$x \in V \subseteq W$
Let $J = \set{\alpha \in I : V_\alpha \ne S_\alpha}$.
By definition of $\BB'$, $J$ is finite.
Now for all $\alpha \in J$, $x_\alpha \in V_\alpha$ and $V_\alpha \in \tau_\alpha$.
Since $\BB_\alpha$ is a basis for $\tau_\alpha$ then for all $\alpha \in J$ there exists $N_\alpha \in \BB_\alpha$:
:$x_\alpha \in N_\alpha \subseteq V_\alpha$
For all $\alpha \in I$, let:
:$U_\alpha = \begin {cases} N_\alpha & : \alpha \in J \\ S_\alpha & : \alpha \notin J \end {cases}$
Let $U = \ds \prod_{\alpha \mathop \in I} U_\alpha$.
For $\alpha \in J$:
:$U_\alpha = N_\alpha \subseteq V_\alpha$
:$x_\alpha \in N_\alpha = U_\alpha$
:$U_\alpha = N_\alpha \in \BB_\alpha$
For $\alpha \notin J$:
:$U_\alpha = S_\alpha = V_\alpha$
:$x_\alpha \in S_\alpha = U_\alpha$
Thus:
:$x \in U$
:$U \subseteq V \subseteq W$
By the definition of $\BB$ and because $J$ is finite:
:$U \in \BB$.
The result follows.
Category:Topological Bases
Category:Product Topology
Product Space Local Basis Induced from Factor Spaces Local Bases

== Theorem ==
Let $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be an indexed family of topological spaces for $\alpha$ in some indexing set $I$.
Let $\struct {S, \tau} = \ds\prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$.
Let $x = \family {x_\alpha} \in S$.
Let $\BB_\alpha$ be a local basis for $x_\alpha$ in the topological space $\struct {S_\alpha, \tau_\alpha}$ for each $\alpha \in I$.
Let $\BB_x$ be the set of all cartesian products of the form $\ds \prod_{\alpha \mathop \in I} U_\alpha$ where:
:for all but finitely many indices $\alpha : U_\alpha = S_\alpha$
:for all $\alpha \in I : U_\alpha \ne S_\alpha \implies U_\alpha \in \BB_\alpha$
Then $\BB_x$ is a local basis for $x$ in the product space $\struct{S, \tau}$.
== Proof ==
Let $W \in \tau$ such that $x \in W$.
From Natural Basis of Product Topology, the set $\BB$ of cartesian products of the form $\ds \prod_{\alpha \mathop \in I} V_\alpha$ where:
:for all $\alpha \in I : V_\alpha \in \tau_\alpha$
:for all but finitely many indices $\alpha : V_\alpha = X_\alpha$
is a basis for $\tau$.
Therefore, there exists $V = \ds \prod_{\alpha \mathop \in I} V_\alpha \in \BB$:
:$x \in V \subseteq W$
Let $J = \set {\alpha \in I : V_\alpha \ne S_\alpha}$.
By definition of $\BB$, $J$ is finite.
Now for all $\alpha \in J$, $x_\alpha \in V_\alpha$ and $V_\alpha \in \tau_\alpha$.
Since $\BB_\alpha$ is a local basis for $\tau_\alpha$ then for all $\alpha \in J$ there exists $N_\alpha \in \BB_\alpha$:
:$x_\alpha \in N_\alpha \subseteq V_\alpha$
For all $\alpha \in I$, let:
:$U_\alpha = \begin {cases} N_\alpha & : \alpha \in J \\ S_\alpha & : \alpha \notin J \end {cases}$
Let $U = \ds \prod_{\alpha \mathop \in I} U_\alpha$.
For $\alpha \in J$:
:$U_\alpha = N_\alpha \subseteq V_\alpha$
:$x_\alpha \in N_\alpha = U_\alpha$
:$U_\alpha = N_\alpha \in \BB_\alpha$
For $\alpha \notin J$:
:$U_\alpha = S_\alpha = V_\alpha$
:$x_\alpha \in S_\alpha = U_\alpha$
Thus:
:$x \in U$
:$U \subseteq V \subseteq W$
By definition of $\BB_x$ and since $J$ is finite:
:$U \in \BB_x$.
The result follows.
Category:Topological Bases
Category:Product Spaces
Category:Local Bases
Product Space is Completely Hausdorff iff Factor Spaces are Completely Hausdorff

== 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 $T_{2 \frac 1 2}$ (completely Hausdorff) space each of $\struct{S_\alpha, \tau_\alpha}$ is a $T_{2 \frac 1 2}$ (completely Hausdorff) space.
== Proof ==
=== Necessary Condition ===
=== Sufficient Condition ===
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces
Category:Completely Hausdorff Spaces
Category:Product Topology
Category:Product Space is Completely Hausdorff iff Factor Spaces are Completely Hausdorff
Product Space is Completely Hausdorff iff Factor Spaces are Completely Hausdorff/Necessary Condition

== 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 completely Hausdorff space.
Then for each $\alpha \in I$, $\struct {S_\alpha, \tau_\alpha}$ is a completely Hausdorff space.
== Proof ==
Suppose $T$ is a completely Hausdorff 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 completely Hausdorff property is hereditary, $T_\alpha$ is completely Hausdorff.
From Completely Hausdorff Space is Preserved under Homeomorphism, $\struct {S_\alpha, \tau_\alpha}$ is completely Hausdorff.
As $\alpha \in I$ was arbitrary, the result follows.
Category:Product Space is Completely Hausdorff iff Factor Spaces are Completely Hausdorff