nyuuzyou/wikis huggingface dataset:
Product of Countably Compact Spaces is not always Countably Compact

== Theorem ==
Let $I$ be an indexing set.
Let $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be a family of topological spaces indexed by $I$.
Let $\ds \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$.
Let each of $\struct {S_\alpha, \tau_\alpha}$ be countably compact.
Then it is not necessarily the case that $\struct {S, \tau}$ is also countably compact.
== Proof ==
Let $T$ denote the Novak space.
Let $T \times T$ denote the Cartesian product of the Novak space with itself under the product topology.
From Novak Space is Countably Compact, $T$ is a countably compact space.
But from Cartesian Product of Novak Spaces is not Countably Compact, $T \times T$ is not a countably compact space.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Invariance Properties
Category:Countably Compact Spaces
Category:Product Spaces
Product of Cut with Zero Cut equals Zero Cut

== Theorem ==
Let $\alpha$ be a cut.
Let $0^*$ denote the rational cut associated with the (rational) number $0$.
Then:
:$\alpha 0^* = 0^*$
where $\alpha 0^*$ denotes the product of $\alpha$ and $0^*$.
== Proof ==
By definition, we have that:
:$\alpha \beta := \begin {cases}
\size \alpha \, \size \beta & : \alpha \ge 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha < 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha \ge 0^*, \beta < 0^* \\
\size \alpha \, \size \beta & : \alpha < 0^*, \beta < 0^*
\end {cases}$
where:
:$\size \alpha$ denotes the absolute value of $\alpha$
:$\size \alpha \, \size \beta$ is defined as in Multiplication of Positive Cuts
:$\ge$ denotes the ordering on cuts.
From Absolute Value of Cut is Zero iff Cut is Zero:
:$\size {0^*} = 0^*$
Thus:
:$\alpha 0^* := \begin {cases}
\size \alpha \, 0^* & : \alpha \ge 0^* \\
-\paren {\size \alpha \, 0^*} & : \alpha < 0^*
\end {cases}$
From Absolute Value of Cut is Greater Than or Equal To Zero Cut, we have that:
:$\size \alpha = \beta$
where $\beta \ge 0^*$.
By definition of Multiplication of Positive Cuts:
$\beta 0^*$ is the set of all rational numbers $r$ such that either:
:$r < 0$
or
:$\exists p \in \beta, q \in 0^*: r = p q$
where $p \ge 0$ and $q \ge 0$
By definition of $0^*$, it is not possible for $q \ge 0$.
Thus:
:$\beta 0^* = \set {r \in \Q: r < 0}$
Hence by definition of a cut:
:$\beta 0^* = 0^*$
By Identity Element for Addition of Cuts and definition of negative of cut:
:$-0^* = 0^*$
and the result follows.
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Dedekind Cuts: $1.26$. Theorem $\text {(d)}$
Category:Cuts
Category:Multiplication
Product of Cuts is Zero Cut iff Either Factor equals Zero Cut

== Theorem ==
Let $\alpha$ and $\beta$ be cuts.
Let $0^*$ denote the rational cut associated with the (rational) number $0$.
Then:
:$\alpha \beta = 0^*$
:
:$\alpha = 0^*$ or $\beta = 0^*$
where $\alpha \beta$ denotes the product of $\alpha$ and $\beta$.
== Proof ==
By definition, we have that:
:$\alpha \beta := \begin {cases}
\size \alpha \, \size \beta & : \alpha \ge 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha < 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha \ge 0^*, \beta < 0^* \\
\size \alpha \, \size \beta & : \alpha < 0^*, \beta < 0^*
\end {cases}$
where:
:$\size \alpha$ denotes the absolute value of $\alpha$
:$\size \alpha \, \size \beta$ is defined as in Multiplication of Positive Cuts
:$\ge$ denotes the ordering on cuts.
First we note that from Absolute Value of Cut is Zero iff Cut is Zero:
:$\size {0^*} = 0^*$
=== Necessary Condition ===
Let $\alpha = 0^*$ or $\beta = 0^*$.
Then from Product of Cut with Zero Cut equals Zero Cut it follows directly that:
:$\alpha \beta = 0^*$
=== Sufficient Condition ===
Let $\alpha \beta = 0^*$.
Suppose $\alpha > 0^*$ and $\beta > 0^*$.
By definition of multiplication of positive cuts:
$\alpha \beta$ is the set of all rational numbers $r$ such that either:
:$r < 0$
or
:$\exists p \in \alpha, q \in \beta: r = p q$
where $p \ge 0$ and $q \ge 0$.
We have that $\alpha \beta = 0^*$.
 $r = p q$ where $p \ge 0$ and $q \ge 0$.
Then $r \ge 0$.
But then $r \notin 0^*$ by definition of $0^*$.
Thus there is no $p \in \alpha, q \in \beta$ such that $r = p q$ where $p \ge 0$ and $q \ge 0$.
It follows that $\alpha > 0^*$ and $\beta > 0^*$ cannot both be satisfied.
It follows further that for $\size \alpha \, \size \beta = 0^*$ it is not possible for $\size \alpha > 0^*$ and $\size \beta > 0^*$.
The result follows.
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Dedekind Cuts: $1.26$. Theorem $\text {(e)}$
Category:Cuts
Category:Multiplication
Product of Cyclotomic Polynomials

== Theorem ==
Let $n > 0$ be a (strictly) positive integer.
Then:
:$\ds \prod_{d \mathop \divides n} \map {\Phi_d} x = x^n - 1$
where:
:$\map {\Phi_d} x$ denotes the $d$th cyclotomic polynomial
:the product runs over all divisors of $n$.
== Proof ==
From the Polynomial Factor Theorem and Complex Roots of Unity in Exponential Form:
:$\ds x^n - 1 = \prod_\zeta \paren {x - \zeta}$
where the product runs over all complex $n$th roots of unity.
In the , each factor $x - \zeta$ appears exactly once, in the factorization of $\map {\Phi_d} x$ where $d$ is the order of $\zeta$.
Thus the polynomials are equal.
== Also see ==
* Möbius Inversion Formula for Cyclotomic Polynomials
Category:Cyclotomic Polynomials
Product of Diagonal Matrices is Diagonal

== Theorem ==
Let $A$ and $B$ be $n \times n$ diagonal matrices. 
Then the matrix product $A B$ is an $n \times n$ diagonal matrix.
Further: 
:$\paren {A B}_{i j} = \begin {cases} \paren A_{i i} \paren B_{i i} & i = j \\ 0 & i \ne j \end {cases}$
== Proof ==
We have: 
:$\ds \paren {A B}_{ij} = \sum_{k \mathop = 1}^n \paren A_{i k} \paren B_{k j}$
Since $A$ and $B$ are diagonal:
:$\paren A_{i k} = 0$ for $i \ne k$,
and:
:$\paren B_{k j} = 0$ for $k \ne j$.
If $i \ne j$, for each $k$ we either have $i \ne k$ or $k \ne j$, so:
:$\paren A_{i k} \paren B_{k j} = 0$ for each $1 \le k \le n$.
Hence if $i \ne j$:
:$\ds \paren {A B}_{i j} = \sum_{k \mathop = 1}^n \paren A_{i k} \paren B_{k j} = 0$
so $A B$ is diagonal.
Let $i = j$.
Then, if $k \ne i$ we have: 
:$\paren A_{i k} \paren B_{k j} = \paren A_{i k} \paren B_{k i} = 0$
so:
:$\ds \sum_{k \mathop = 1}^n \paren A_{i k} \paren B_{k j} = \paren A_{i i} \paren B_{i i}$
We can conclude:
:$\paren {A B}_{i j} = \begin {cases} \paren A_{i i} \paren B_{i i} & i = j \\ 0 & i \ne j \end {cases}$
Category:Diagonal Matrices
Category:Conventional Matrix Multiplication
Product of Diagonals from Point of Regular Polygon

== Theorem ==
Let $A_0, A_1, \ldots, A_{n - 1}$ be the vertices of a regular $n$-gon $P = A_0 A_1 \cdots A_{n - 1}$ which is circumscribed by a unit circle.
Then:
:$\ds \prod_{k \mathop = 2}^{n - 2} A_0 A_k = \frac {n \csc^2 \frac \pi n} 4$
where $A_0 A_k$ is the length of the line joining $A_0$ to $A_k$.
== Proof ==
First it is worth examining the degenerate case $n = 3$, where there are no such lines.
In this case:
The continued product $\ds \prod_{k \mathop = 2}^{n - 2} A_0 A_k$ is vacuous here.
By convention, such a vacuous product is defined as being equal to $1$.
So the result holds for $n = 3$.
:
Let $n > 3$.
The illustration above is a diagram of the case where $n = 7$.
From Complex Roots of Unity are Vertices of Regular Polygon Inscribed in Circle, the vertices of $P$ can be identified with the complex $n$th roots of unity.
Thus:
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Miscellaneous Problems: $160$
Category:Polygons
Product of Differences between 1 and Complex Roots of Unity

== Theorem ==
Let $\alpha$ be a primitive complex $n$th root of unity.
Then:
:$\ds \prod_{k \mathop = 1}^{n - 1} \paren {1 - \alpha^k} = n$
== Proof ==
From Power of Complex Number minus 1: Corollary:
:$\ds \sum_{k \mathop = 0}^{n - 1} z^k = \prod_{k \mathop = 1}^{n - 1} \paren {z - \alpha^k}$
The result follows by setting $z = 1$.
== Sources ==
* : $\S 3$. Roots of Unity: Example $5$.
Category:Complex Roots of Unity
Product of Directed Sets is Directed Set

== Theorem ==
Let $\struct {\Lambda_1, \preceq_1}$ and $\struct {\Lambda_2, \preceq_2}$ be directed sets. 
Define a relation $\sqsubseteq$ on $\Lambda_1 \times \Lambda_2$ by: 
:$\tuple {\lambda_1, \lambda_2} \sqsubseteq \tuple {\mu_1, \mu_2}$
for $\tuple {\lambda_1, \lambda_2}, \tuple {\mu_1, \mu_2} \in \Lambda_1 \times \Lambda_2$ :
:$\lambda_1 \preceq_1 \mu_1$ and $\lambda_2 \preceq_2 \mu_2$
Then $\struct {\Lambda_1 \times \Lambda_2, \sqsubseteq}$ is a directed set.
== Proof ==
We first show that $\sqsubseteq$ is a preordering on $\Lambda_1 \times \Lambda_2$.
Let $\tuple {\lambda, \mu} \in \Lambda_1 \times \Lambda_2$.
We have $\lambda \preceq_1 \lambda$ and $\mu \preceq_2 \mu$ since $\Lambda_1$ and $\Lambda_2$ are preorderings.
So we have $\tuple {\lambda, \mu} \sqsubseteq \tuple {\lambda, \mu}$, and so $\preceq$ is reflexive.
Now suppose that $\tuple {\lambda_1, \mu_1}, \tuple {\lambda_2, \mu_2}, \tuple {\lambda_3, \mu_3} \in \Lambda_1 \times \Lambda_2$ are such that:
:$\tuple {\lambda_1, \mu_1} \sqsubseteq \tuple {\lambda_2, \mu_2}$
and:
:$\tuple {\lambda_2, \mu_2} \sqsubseteq \tuple {\lambda_3, \mu_3}$
Then we have: 
:$\lambda_1 \preceq_1 \lambda_2$ and $\lambda_2 \preceq_1 \lambda_3$
and:
:$\mu_1 \preceq_2 \mu_2$ and $\mu_2 \preceq_2 \mu_3$
Since $\preceq_1$ is transitive, we have: 
:$\lambda_1 \preceq_1 \lambda_3$
Since $\preceq_2$ is transitive, we have: 
:$\mu_1 \preceq_2 \mu_3$
So we have: 
:$\tuple {\lambda_1, \mu_1} \sqsubseteq \tuple {\lambda_3, \mu_3}$
So $\sqsubseteq$ is transitive.
We now show that $\sqsubseteq$ is directed. 
Let $\tuple {\lambda_1, \mu_1}, \tuple {\lambda_2, \mu_2} \in \Lambda_1 \times \Lambda_2$.
Then since $\preceq_1$ is directed, there exists $\lambda \in \Lambda_1$ such that: 
:$\lambda_1 \preceq_1 \lambda$ and $\lambda_2 \preceq_2 \lambda$.
Since $\preceq_2$ is directed, there exists $\mu \in \Lambda_2$ such that:
:$\mu_1 \preceq_2 \mu$ and $\mu_2 \preceq_2 \mu$.
So we have: 
:$\tuple {\lambda_1, \mu_1} \sqsubseteq \tuple {\lambda, \mu}$
and:
:$\tuple {\lambda_2, \mu_2} \sqsubseteq \tuple {\lambda, \mu}$
So $\tuple {\Lambda_1 \times \Lambda_2, \sqsubseteq}$ is a directed set.
Category:Directed Preorderings
Product of Discrete Random Variables

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ and $Y$ be discrete random variables on $\struct {\Omega, \Sigma, \Pr}$.
Let $V: \Omega \to \R$ be defined as:
:$\forall \omega \in \Omega: \map V \omega = \map X \omega \map Y \omega$
Then $V$ is also a discrete random variable on $\struct {\Omega, \Sigma, \Pr}$.
== Proof ==
To show that $V$ is a discrete random variable on $\struct {\Omega, \Sigma, \Pr}$, we need to show that:
:$(1): \quad $The image of $U$ and $V$ are countable subsets of $\R$;
:$(2): \quad \forall x \in \R: \set {\omega \in \Omega: \map V \omega = x} \in \Sigma$.
First we consider any $V_V = \set {\omega \in \Omega: \map V \omega = v}$ such that $V_v \ne \O$.
We have that:
: $V_v = \set {\omega \in \Omega: \map X \omega \map Y \omega = v}$
Consider any $\omega \in V_v$.
If $v = 0$ the result follows immediately, so we assume that $v \ne 0$.
Then:
:$\omega \in X_x \cap Y_x$
where:
: $X_x = \set {\omega \in \Omega: \map X \omega = x}, Y_x = \set {\omega \in \Omega: \map Y \omega = \dfrac v x}$
Both $X_x \in \Sigma$ and $Y_x \in \Sigma$ (as $X$ and $Y$ are discrete random variables.
As $\struct {\Omega, \Sigma, \Pr}$ is a probability space then $X_x \cap Y_x \in \Sigma$.
Now note that $\ds V_v = \bigcup_{x \mathop \in \R} \paren {X_x \cap Y_x}$.
That is, it is the union of all such intersections of sets whose discrete random variables multiply together to make $v$.
As $X_x$ is a countable set it follows that $V_v$ is a countable union of countable sets.
From Countable Union of Countable Sets is Countable it follows that $X_x$ is a countable set.
And, by dint of $\struct {\Omega, \Sigma, \Pr}$ being a probability space, $V_v \in \Sigma$.
Thus $V$ is a discrete random variables on $\struct {\Omega, \Sigma, \Pr}$.
== Sources ==
* : $\S 2.1$: Probability mass functions: Exercise $1$
Category:Probability Theory
Product of Distances of Polar and Pole from Center of Circle

== Theorem ==
Let $\CC$ be a circle of radius $r$ whose center is at $O$.
Let $P$ be a point.
Let $\LL_1$ be the polar of $P$ with respect to $\CC$.
Let $\LL_2$ be the line $OP$.
Let $N$ be the point of intersection of $\LL_1$ and $\LL_2$.
Then:
:$ON \cdot OP = r^2$
== Proof ==
Let $U$ and $V$ be the points where $OP$ intersects $\CC$.
:
From Harmonic Property of Pole and Polar wrt Circle, $\tuple {UV, NP}$ form a harmonic range.
That is:
== Sources ==
* : Chapter $\text {III}$. The Circle: $6$.
Category:Polars of Points
Product of Division Products

== Theorem ==
Let $\struct {R, +, \circ}$ be a commutative ring with unity.
Let $\struct {U_R, \circ}$ be the group of units of $\struct {R, +, \circ}$.
Let $a, b \in R, c, d \in U_R$.
Then:
:$\dfrac a c \circ \dfrac b d = \dfrac {a \circ b} {c \circ d}$
where $\dfrac x z$ is defined as $x \circ \paren {z^{-1} }$, that is, $x$ divided by $z$.
== Proof ==
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $23$. The Field of Rational Numbers: Theorem $23.7 \ (4)$
Category:Commutative Algebra
Product of Divisor Sum and Euler Phi Functions

== Theorem ==
Let $n$ be an integer such that $n \ge 2$.
Let the prime decomposition of $n$ be:
:$n = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}$
Let $\map {\sigma_1} n$ be the divisor sum function of $n$.
Let $\map \phi n$ be the Euler phi function of $n$.
Then:
:$\ds \map {\sigma_1} n \map \phi n = n^2 \prod_{1 \mathop \le i \mathop \le r} \paren {1 - \frac 1 {p_i^{k_i + 1} } }$
== Proof ==
From Euler Phi Function of Integer:
:$\ds \map \phi n = \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i - 1} \paren {p_i - 1}$
From Divisor Sum of Integer:
:$\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$
So:
:$\ds \map {\sigma_1} n \map \phi n = \prod_{1 \mathop \le i \mathop \le r} \paren {\frac {p_i^{k_i + 1} - 1} {p_i - 1} } p_i^{k_i - 1} \paren {p_i - 1}$
Taking a general factor of this product:
So:
:$\ds \map {\sigma_1} n \map \phi n = \prod_{1 \mathop \le i \mathop \le r} p_i^{2 k_i} \paren {1 - \frac 1 {p_i^{k_i + 1} } }$
Hence:
:$\ds \prod_{1 \mathop \le i \mathop \le r} p_i^{2 k_i} = \paren {\prod_{1 \mathop \le i \mathop \le r} p_i^{k_i} }^2 = n^2$
and the result follows.
Category:Divisor Sum Function
Category:Euler Phi Function
Product of Divisors

== Theorem ==
Let $n$ be an integer such that $n \ge 1$.
Let $\map D n$ denote the product of the divisors of $n$.
Then:
:$\map D n = n^{\map {\sigma_0} n / 2}$
where $\map {\sigma_0} n$ denotes the divisor count function of $n$.
== Proof 1 ==
== Proof 2 ==
Category:Product of Divisors
Category:Divisor Count Function
Category:Divisors
Product of Divisors/Proof 1

== Theorem ==
== Proof ==
We have by definition that:
:$\map D n = \ds \prod_{d \mathop \divides n} d$
Also by definition, $\map {\sigma_0} n$ is the number of divisors of $n$.
Suppose $n$ is not a square number.
Let $p \divides n$, where $\divides$ denotes divisibility.
Then:
:$\exists q \divides n : p q = n$
Thus the divisors of $n$ come in pairs whose product is $n$.
From Divisor Count Function is Odd Iff Argument is Square, $\map {\sigma_0} n$ is even.
Thus $\dfrac {\map {\sigma_0} n} 2$ is an integer.
Thus there are exactly $\dfrac {\map {\sigma_0} n} 2$ pairs of divisors of $n$ whose product is $n$.
Thus the product of the divisors of $n$ is:
:$\ds \prod_{d \mathop \divides n} d = n^{\map {\sigma_0} n / 2}$
Now suppose $n$ is square such that $n = r^2$.
Then from Divisor Count Function is Odd Iff Argument is Square, $\map {\sigma_0} n$ is odd.
Hence the number of divisors of $n$ not including $r$ is $\map {\sigma_0} n - 1$.
As before, these exist in pairs whose product is $n$.
Thus:
Hence the result.
Category:Product of Divisors
Product of Divisors is Divisor of Product

== Theorem ==
Let $a, b, c, d \in \Z$ be integers such that $a, c \ne 0$.
Let $a \divides b$ and $c \divides d$, where $\divides$ denotes divisibility.
Then:
:$a c \divides b d$
== Proof ==
By definition of divisibility:
Then:
and the result follows by definition of divisibility.
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Theorem $2 \text{-} 2 \ (3)$
Category:Divisors
Product of Element in *-Star Algebra with its Star is Hermitian

== Theorem ==
Let $\struct {A, \ast}$ be a $\ast$-algebra.
Let $a \in A$.
Then $a^\ast a$ and $a a^\ast$ are Hermitian.
== Proof ==
We have:
and:
So $a^\ast a$ and $a a^\ast$ are both Hermitian.
== Sources ==
* : $2.1$: $C^\ast$-Algebras
Category:Hermitian Elements of *-Algebras
Product of Even and Odd Functions

== Theorem ==
Let $\OO$ be an odd real function defined on some symmetric set $S$.
Let $\EE$ be an even real function defined on some symmetric set $S'$.
Let $\OO \EE$ be their pointwise product, defined on the intersection of the domains of $\OO$ and $\EE$.
Then $\OO \EE$ is odd. 
That is:
:$\forall x \in S \cap S': \map {\paren {\OO \EE}} {-x} = - \map {\paren {\OO \EE} } x$.
== Proof ==
The result follows from the definition of an odd function.
== Also see ==
* Product of Odd Functions
* Product of Even Functions
Category:Real Analysis
Product of Exponential Generating Functions

== Theorem ==
Let $\map G z$ be the exponential generating function for the sequence $\sequence {\dfrac {a_n} {n!} }$.
Let $\map H z$ be the exponential generating function for the sequence $\sequence {\dfrac {b_n} {n!} }$.
Then $\map G z \map H z$ is the generating function for the sequence $\sequence {\dfrac {c_n} {n!} }$, where:
:$\forall n \in \Z_{\ge 0}: c_n = \ds \sum_{k \mathop \in \Z} \dbinom n k a_k b_{n - k}$
== Proof ==
== Sources ==
* : $\S 1.2.9$: Generating Functions: $(10)$
Category:Generating Functions
Category:Product of Exponential Generating Functions
Product of Exponential Generating Functions/Proof

== Theorem ==
== Proof ==
Let $\map G z \map H z$ be the generating function for the sequence $\sequence {c'_n}$.
By definition of generating function:
Then:
Hence the result.
== Sources ==
* : $\S 1.2.9$: Generating Functions: Exercise $2$
Category:Product of Exponential Generating Functions
Product of Factors of Perfect Number

== Theorem ==
Let $P$ be the perfect number $2^{n - 1} \paren {2^n - 1}$.
Then:
:$\ds \prod_{d \mathop \divides P} d = P^n$
== Proof ==
The factors of $P$ are:
:$1, 2, 4, \dots, 2^{n - 1}, 2^n - 1, 2 \paren {2^n - 1}, \dots, 2^{n - 1} \paren {2^n - 1}$
Therefore their product is:
== Sources ==
* : $28$
Category:Perfect Numbers
Product of Field Negatives

== Theorem ==
Let $\struct {F, +, \times}$ be a field whose zero is $0_F$.
Let $a, b \in F$.
Then:
:$-\paren {a \times b} = \paren {-a} \times \paren {-b} = a \times b$
== Proof ==
== Sources ==
* : Chapter $1$: Axioms for the Real Numbers: $2$. Fields: Theorem $2 \ \text {(v)}$
Category:Field Theory
Product of Finite Sequence of Matrices

== Theorem ==
Let $\mathbf A_1, \mathbf A_2, \ldots, \mathbf A_n$ be matrices.
Let the order of $\mathbf A_j$ be $d_j \times d_{j + 1}$.
Let $\ds \mathbf C := \prod_{j \mathop = 1}^n \mathbf A_j = \mathbf A_1 \mathbf A_2 \cdots \mathbf A_n$ be the (conventional) matrix product of $\mathbf A_1, \mathbf A_2, \ldots, \mathbf A_n$.
Then:
:$(1): \quad \ds \map c {i_1, i_{n + 1} } = \sum_{i_n \mathop = 1}^{d_n} \dotsm \sum_{i_3 \mathop = 1}^{d_3} \sum_{i_2 \mathop = 1}^{d_2} \map {a_1} {i_1, i_2} \map {a_2} {i_2, i_3} \dotsm \map {a_{n - 1} } {i_{n - 1}, i_n} \map {a_n} {i_n, i_{n + 1} }$
where:
:$\map {a_1} {i_1, i_2}$ (for example) denotes the element of $\mathbf A_1$ whose indices are $i_1$ and $i_2$
:the order of $\mathbf C$ is $d_1 \times d_{n+1}$.
== Proof ==
Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\ds \map c {i_1, i_{n + 1} } = \sum_{i_n \mathop = 1}^{d_n} \dotsm \sum_{i_3 \mathop = 1}^{d_3} \sum_{i_2 \mathop = 1}^{d_2} \map {a_1} {i_1, i_2} \map {a_2} {i_2, i_3} \dotsm \map {a_{n - 1} } {i_{n - 1}, i_n} \map {a_n} {i_n, i_{n + 1} }$
$\map P 1$ is trivially true, as this just says:
:$\map c {i_1, i_2} = \map {a_1} {i_1, i_2}$
=== Basis for the Induction ===
$\map P 2$ is the case:
:$\ds \map c {i_1, i_3} = \sum_{i_2 \mathop = 1}^{d_2} \map {a_1} {i_1, i_2} \map {a_2} {i_2, i_3}$
which is the definition of (conventional) matrix product.
The order of $\mathbf C$ is $d_1 \times d_3$, while the orders of $\mathbf A_1$ and $\mathbf A_2$ are $d_1 \times d_2$ and $d_2 \times d_3$ respectively.
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:
:$\ds \map c {i_1, i_{k + 1} } = \sum_{i_k \mathop = 1}^{d_k} \dotsm \sum_{i_3 \mathop = 1}^{d_3} \sum_{i_2 \mathop = 1}^{d_2} \map {a_1} {i_1, i_2} \map {a_2} {i_2, i_3} \dotsm \map {a_k} {i_k, i_{k + 1} }$
Then we need to show:
:$\ds \map c {i_1, i_{k + 2} } = \sum_{i_{k + 1} \mathop = 1}^{d_{k + 1} } \sum_{i_k \mathop = 1}^{d_k} \dotsm \sum_{i_3 \mathop = 1}^{d_3} \sum_{i_2 \mathop = 1}^{d_2} \map {a_1} {i_1, i_2} \map {a_2} {i_2, i_3} \dotsm \map {a_k} {i_k, i_{k + 1} } \, \map {a_{k + 1} } {i_{k + 1}, i_{k + 2} }$
=== Induction Step ===
This is our induction step:
Let $\ds \mathbf C_1 := \prod_{j \mathop = 1}^k \mathbf A_j$.
Let $\ds \mathbf C := \prod_{j \mathop = 1}^{k + 1} \mathbf A_j = \mathbf C_1 \mathbf A_{k + 1}$.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore for all $n \in \N_{>0}$:
:$\ds \map c {i_1, i_{n + 1} } = \sum_{i_n \mathop = 1}^{d_n} \dotsm \sum_{i_3 \mathop = 1}^{d_3} \sum_{i_2 \mathop = 1}^{d_2} \map {a_1} {i_1, i_2} \map {a_2} {i_2, i_3} \dotsm \map {a_{n - 1} } {i_{n - 1}, i_n} \map {a_n} {i_n, i_{n + 1} }$
Category:Conventional Matrix Multiplication
Product of Finite Sets is Finite/Proof 1

== Theorem ==
== Proof ==
By the definition of Cartesian product:
:$S \times T = \set {\tuple {s, t}: s \in S, t \in T}$
Then by definition of set union:
:$S \times T = \ds \bigcup_{s \mathop \in S} \set s \times T$
Also, the mapping $g_s: \set s \times T \to T$ defined by:
:$\map {g_s} {s, t} = t$
is a bijection.
Therefore, since $T$ is finite, so is $\set s \times T$ for all $s \in S$.
Since $S$ is finite, the result follows from Finite Union of Finite Sets is Finite.
== Sources ==
* : $1$: Set Theory and Logic: $\S 6$: Finite Sets: Corollary $6.8$
Category:Product of Finite Sets is Finite
Product of Finite Sets is Finite/Proof 2

== Theorem ==
== Proof ==
Let $\card S$ denote the cardinal number of $S$.
Let $\cdot$ denote ordinal multiplication.
By Cardinal Product Equinumerous to Ordinal Product, it follows that $S \times T \sim \card S \cdot \card T$.
But then $\card S$ and $\card T$ are members of the minimally inductive set.
Therefore, $\card S \cdot \card T \in \omega$ by Natural Number Multiplication is Closed.
Since $S \times T$ is equinumerous to a member of the minimally inductive set, it follows that $S \times T$ is finite.
== Sources ==
* : $\S 10.29 \ (2)$
Category:Product of Finite Sets is Finite