nyuuzyou/wikis huggingface dataset:
Product of Functions of Bounded Variation is of Bounded Variation

== Theorem ==
Let $a, b$ be real numbers with $a < b$.
Let $f, g : \closedint a b \to \R$ be functions of bounded variation.
Let $\map {V_f} {\closedint a b}$ and $\map {V_g} {\closedint a b}$ be the total variations of $f$ and $g$ respectively. 
Then the pointwise product $f \cdot g$ is of bounded variation with:
:$\map {V_{f \cdot g} } {\closedint a b} \le A \map {V_f} {\closedint a b} + B \map {V_g} {\closedint a b}$
where:
:$\map {V_{f \cdot g} } {\closedint a b}$ denotes the total variation of $f \cdot g$ on $\closedint a b$ 
:$A, B$ are non-negative real numbers.
== Proof ==
For each finite subdivision $P$ of $\closedint a b$, write: 
:$P = \set {x_0, x_1, \ldots, x_n }$
with:
:$a = x_0 < x_1 < x_2 < \cdots < x_{n - 1} < x_n = b$
By Function of Bounded Variation is Bounded: 
:$f$ and $g$ are bounded. 
So, there exists $A, B \in \R$ such that: 
:$\size {\map f x} \le B$
:$\size {\map g x} \le A$
for all $x \in \closedint a b$.
Then:
Since $f$ and $g$ are of bounded variation, there exists $M, K \in \R$ such that: 
:$\map {V_f} {P ; \closedint a b} \le M$
:$\map {V_g} {P ; \closedint a b} \le K$
for all finite subdivisions $P$.
We therefore have: 
:$\map {V_{f \cdot g} } {P ; \closedint a b} \le A M + B K$
so $f \cdot g$ is of bounded variation.
Further, we have: 
== Sources ==
* : $\S 6.4$: Total Variation: Theorem $6.9$
Category:Total Variation of Real Function
Category:Bounded Variation
Product of Functions of Exponential Order

== Theorem ==
Let $f, g: \R \to \F$ be functions, where $\F \in \set {\R, \C}$.
Let $f$ be of exponential order $a$ and $g$ be of exponential order $b$.
Then $f g: t \mapsto \map f t \map g t$ is of exponential order $a+b$.
== Proof ==
Let $t$ be sufficiently large so that both $f$ and $g$ are of exponential order on some shared unbounded closed interval.
By the definition of exponential order:
Category:Exponential Order
Product of GCD and LCM

== Theorem ==
:$\lcm \set {a, b} \times \gcd \set {a, b} = \size {a b}$
where:
:$\lcm \set {a, b}$ denotes the lowest common multiple of $a$ and $b$
:$\gcd \set {a, b}$ denotes the greatest common divisor of $a$ and $b$.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Proof 5 ==
== Sources ==
* : Introduction $\S 6$: The division process in $I$
* : $\text {2-2}$ Divisibility: Exercise $4$
* : Chapter $1$: Properties of the Natural Numbers: $\S 23 \gamma$
Category:Greatest Common Divisor
Category:Lowest Common Multiple
Category:Product of GCD and LCM
Product of GCD and LCM/Proof 1

== Theorem ==
== Proof ==
It is sufficient to prove that $\lcm \set {a, b} \times \gcd \set {a, b} = a b$, where $a, b \in \Z_{>0}$.
Now we have:
:$a \divides m \land b \divides m \implies m = a r = b s$
Also, by Bézout's Identity we have:
:$d = a x + b y$
So:
So:
:$m = n \paren {s x + r y}$
Thus:
:$n \divides m \implies n \le \size m$
while:
:$a b = d n = \gcd \set {a, b} \times \lcm \set {a, b}$
as required.
Category:Product of GCD and LCM
Product of GCD and LCM/Proof 2

== Theorem ==
== Proof ==
Let $a = g m$ and $b = g n$, where $g = \gcd \set {a, b}$ and $m$ and $n$ are coprime.
The existence of $m$ and $n$ are proved by Integers Divided by GCD are Coprime.
It follows that:
:$a = g m \divides g m n$
and:
:$b = g n \divides g m n$
So $g m n$ is a common multiple of $a$ and $b$. 
Hence there exists an integer $g k \le g m n$ that is divisible by both $a$ and $b$.
Then:
As $g k \le g m n$, it follows that:
:$k \le m n$
But $m, n$ are coprime.
So:
Category:Product of GCD and LCM
Product of GCD and LCM/Proof 3

== Theorem ==
== Proof ==
Let $d := \gcd \set {a, b}$. 
Then by definition of the GCD, there exist $j_1, j_2 \in \Z$ such that $a = d j_1$ and $b = d j_2$.
Because $d$ divides both $a$ and $b$, it must divide their product:
:$\exists l \in \Z$ such that $a b = d l$
Then we have: 
showing that $a \divides l$ and $b \divides l$.
That is, $l$ is a common multiple of $a$ and $b$.
Now it must be shown that $l$ is the least such number.
Let $m$ be any common multiple of $a$ and $b$.
Then there exist $k_1, k_2 \in \Z$ such that $m = a k_1 = b k_2$.
By Bézout's Identity:
:$\exists x, y \in \Z: d = a x + b y$
So:
Thus:
:$m = l \paren {b k_2 + a k_1}$
that is, $l \divides m$.
Hence by definition of the LCM:
:$\lcm \set {a, b} = l$
In conclusion:
:$a b = d l = \gcd \set {a, b} \cdot \lcm \set {a, b}$
Category:Product of GCD and LCM
Product of GCD and LCM/Proof 4

== Theorem ==
== Proof ==
From Fundamental Theorem of Arithmetic, let:
From LCM from Prime Decomposition:
:$\lcm \set {m, n} = p_1^{\max \set {k_1, l_1} } p_2^{\max \set {k_2, l_2} } \dotsm p_r^{\max \set {k_r, l_r} }$
From GCD from Prime Decomposition:
:$\gcd \set {m, n} = p_1^{\min \set {k_1, l_1} } p_2^{\min \set {k_2, l_2} } \dotsm p_r^{\min \set {k_r, l_r} }$
From Sum of Maximum and Minimum, for all $i \in \set {1, 2, \ldots, r}$:
:$\min \set {k_i, l_i} + \max \set {k_i, l_i} = k_i + l_i$
Hence:
== Sources ==
* : $\text {2-4}$ The Fundamental Theorem of Arithmetic: Exercise $8$
Category:Product of GCD and LCM
Product of GCD and LCM/Proof 5

== Theorem ==
== Proof ==
Let $d := \gcd \set {a, b}$. 
Then by definition of the GCD, there exist $r, s\in \Z$ such that $a = d r$ and $b = d s$.
Let $m = \dfrac {a b} d$.
Then:
:$m = a s = r b$
which makes $m$ a common multiple of $a$ and $b$.
Let $c \in \Z_{>0}$ be a common multiple of $a$ and $b$.
Let us say that:
:$c = a u = b v$
From Bézout's Identity:
:$\exists x, y \in \Z: d = a x + b y$
Then:
That is:
:$m \divides c$
where $\divides$ denotes divisibility.
So by Absolute Value of Integer is not less than Divisors:
:$m \le c$
Hence by definition of the LCM:
:$\lcm \set {a, b} = m$
In conclusion:
:$\lcm \set {a, b} = \dfrac {a b} d = \dfrac {a b} {\gcd \set {a, b} }$
and the result follows.
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.3$ The Euclidean Algorithm
Category:Product of GCD and LCM
Product of Generating Elements of Dihedral Group

== Theorem ==
Let $D_n$ be the dihedral group of order $2 n$.
Let $D_n$ be defined by its group presentation:
:$D_n = \gen {\alpha, \beta: \alpha^n = \beta^2 = e, \beta \alpha \beta = \alpha^{−1} }$
Then for all $k \in \Z_{\ge 0}$:
:$\beta \alpha^k = \alpha^{n - k} \beta$
== Proof ==
The proof proceeds by induction.
For all $k \in \Z_{\ge 0}$, let $\map P k$ be the proposition:
:$\beta \alpha^k = \alpha^{n - k} \beta$
$\map P 0$ is the case:
Thus $\map P 0$ is seen to hold.
=== Basis for the Induction ===
We have:
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true.
So this is the induction hypothesis:
:$\beta \alpha^r = \alpha^{n - r} \beta$
from which it is to be shown that:
:$\beta \alpha^{r + 1} = \alpha^{n - r - 1} \beta$
=== Induction Step ===
This is the induction step:
So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall k \in \Z_{\ge 0}: \beta \alpha^k = \alpha^{n - k} \beta$
Category:Dihedral Groups
Product of Generating Elements of Quaternion Group

== Theorem ==
Let $Q = \Dic 2$ be the quaternion group:
:$\Dic 2 = \gen {a, b: a^4 = e, b^2 = a^2, a b a = b}$
Then for all $k \in \Z_{\ge 0}$:
:$b a^k = a^{-k} b$
== Proof ==
The proof proceeds by induction.
For all $k \in \Z_{\ge 0}$, let $\map P k$ be the proposition:
:$b a^k = a^{-k} b$
$\map P 0$ is the case:
Thus $\map P 0$ is seen to hold.
=== Basis for the Induction ===
We have:
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true.
So this is the induction hypothesis:
:$b a^r = a^{-r} b$
from which it is to be shown that:
:$b a^{r + 1} = a^{-r - 1} b$
=== Induction Step ===
This is the induction step:
So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall k \in \Z_{\ge 0}: b a^k = a^{-k} b$
Category:Quaternion Group
Product of Generating Functions

== Theorem ==
Let $\map G z$ be the generating function for the sequence $\sequence {a_n}$.
Let $\map H z$ be the generating function for the sequence $\sequence {b_n}$.
Then $\map G z \map H z$ is the generating function for the sequence $\sequence {c_n}$, where:
:$\forall n \in \Z_{\ge 0}: c_n = \ds \sum_{k \mathop = 0}^n a_k b_{n - k}$
=== General Rule ===
== Proof ==
By definition of generating function:
:$\map G z = \ds \sum_{n \mathop \ge 0} a_n z^n$
:$\map H z = \ds \sum_{n \mathop \ge 0} b_n z^n$
Then:
$\quad \begin {array}{c|ccccc}
\times & b_0 & b_1 z & b_2 z^2 & b_3 z^3 & \cdots \\
\hline
a_0 & a_0 b_0 & a_0 b_1 z & a_0 b_2 z^2 & a_0 b_3 z^3 & \cdots \\
a_1 z & a_1 b_0 z & a_1 b_1 z^2 & a_1 b_2 z^3 & a_1 b_3 z^4 & \cdots \\
a_2 z^2 & a_2 b_0 z^2 & a_2 b_1 z^3 & a_2 b_2 z^4 & a_2 b_3 z^5 & \cdots \\
a_3 z^3 & a_3 b_0 z^3 & a_3 b_1 z^4 & a_3 b_2 z^5 & a_3 b_3 z^6 & \cdots \\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots 
\end {array}$
The result follows by definition of generating function.
== Sources ==
* : $\S 1.2.9$: Generating Functions: $(6)$
Category:Generating Functions
Product of Generating Functions/General Rule

== Theorem ==
Let $\map {G_0} z, \map {G_1} z, \map {G_2} z, \ldots$ be any number of generating functions (up to countably infinite) for the sequences $\sequence {a_0 n}, \sequence {a_1 n}, \sequence {a_2 n}, \ldots$
Then:
== Proof ==
== Sources ==
* : $\S 1.2.9$: Generating Functions: $(9)$
Category:Generating Functions
Product of Geometric Sequences from One

== Theorem ==
Let $Q_1 = \sequence {a_j}_{0 \mathop \le j \mathop \le n}$ and $Q_2 = \sequence {b_j}_{0 \mathop \le j \mathop \le n}$ be geometric sequences of integers of length $n + 1$.
Let $a_0 = b_0 = 1$.
Then the sequence $P = \sequence {p_j}_{0 \mathop \le j \mathop \le n}$ defined as:
:$\forall j \in \set {0, \ldots, n}: p_j = b^j a^{n - j}$
is a geometric sequence.
== Proof ==
By Form of Geometric Sequence of Integers with Coprime Extremes, the $j$th term of $Q_1$ is given by:
:$a_j = a^j$
and of $Q_1$ is given by:
:$b_j = b^j$
Let the terms of $P$ be defined as:
:$\forall j \in \left\{{0, 1, \ldots, n}\right\}: p_j = b^j a^{n - j}$
Then from Form of Geometric Sequence of Integers it follows that $P$ is a geometric sequence.
== Sources ==
* : Book $\text{VIII}$. Propositions
Category:Geometric Sequences of Integers
Product of Hausdorff Factor Spaces is Hausdorff

== Theorem ==
Let $T_\alpha = \struct {S_\alpha, \tau_\alpha}$ and $T_\beta = \struct {S_\beta, \tau_\beta}$ be topological spaces.
Let $T = T_\alpha \times T_\beta$ be the product space of $T_\alpha$ and $T_\beta$
Let $T_\alpha$ and $T_\beta$ both be $T_2$ (Hausdorff) spaces.
Then $T$ is also a $T_2$ (Hausdorff) space.
=== General Result ===
== Proof ==
Let $T_\alpha$ and $T_\beta$ be Hausdorff spaces.
Let $\tuple {a, b}$ and $\tuple {c, d}$ be two distinct points of the product space $T$.
Let $a = c$.
Then as $\tuple {a, b} \ne \tuple {c, d}$ it follows that $b \ne d$.
As $T_\beta$ is Hausdorff, there exists two disjoint open sets $U, V \subseteq T_\beta$ such that $b \in U, d \in V$.
Then $T_\alpha \times U$ and $T_\alpha \times V$ are two open disjoint sets in the product space $T$ containing $\tuple {a, b}$ and $\tuple {c, d}$.
Let $a \ne c$.
As $T_\alpha$ is Hausdorff, there exists two disjoint open sets $U, V \subseteq T_\alpha$ such that $a \in U, c \in V$.
Then $U \times T_\beta$ and $V \times T_\beta$ are two open disjoint sets in the product space $T$ containing $\tuple {a, b}$ and $\tuple {c, d}$.
== Sources ==
* : $4$: The Hausdorff condition: $4.2$: Separation axioms: Proposition $4.2.4 \ \text{(b)}$
* : $4$: The Hausdorff condition: Exercise $4.3: 2$
Category:Hausdorff Spaces
Category:Product Spaces
Product of Hausdorff Factor Spaces is Hausdorff/General Result

== 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 $T_2$ (Hausdorff) spaces.
Then $T$ is a $T_2$ (Hausdorff) space.
== Proof ==
Let $x, y \in S : x \ne y$.
Then $x_\alpha \ne y_\alpha$ for some $\alpha \in I$.
Since $\struct {S_\alpha, \tau_\alpha}$ is Hausdorff 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:
 
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 $\tau$:
:$\map {\pr_\alpha^\gets} U, \map {\pr_\alpha^\gets} V \in \tau$
Since $x, y \in S$ were arbitrary, it follows that $T$ is a $T_2$ (Hausdorff) space.
Category:Hausdorff Spaces
Category:Product Topology
Product of Imaginary Unit with Itself

== Theorem ==
Let $\tuple {0, 1}$ denote the imaginary unit.
Then:
:$\tuple {0, 1} \times \tuple {0, 1} = \tuple {-1, 0}$
where $\times$ denotes complex multiplication.
== Proof ==
== Sources ==
* : Chapter $\text I$: Functions of a Complex Variable: $\S 1$. Complex Numbers: The abbreviated notation
Category:Imaginary Parts
Category:Complex Multiplication
Product of Incidence Matrix of BIBD with its Transpose

== Theorem ==
Let $A$ be the block incidence matrix for a BIBD with parameters $v, k, \lambda$.
Then:
:$A^\intercal \cdot A = \sqbrk {a_{ij} } = \paren {r - \lambda} I_v + \lambda J_v$
where:
:$A$ is $v \times b$
:$A^\intercal$ is the transpose of $A$
:$J_v$ is the all $v \times v$ $1$'s matrix
:$I_v$ is the $v \times v$ identity matrix.
That is:
:$A^\intercal \cdot A = \begin{bmatrix}
r & \lambda & \cdots & \lambda \\
\lambda & r & \cdots & \lambda \\
\vdots & \vdots & \ddots & \vdots \\
\lambda & \lambda & \cdots & r \\
\end{bmatrix}$
== Proof ==
Let row $i$ of $A$ be multiplied by column $i$ of $A^\intercal$.
This is the same as multiplying row $i$ of $A$ by row $i$ of $A$.
Each row of $A$ has $r$ entries (since any point must be in $r$ blocks).
Then:
:$\sqbrk {a_{ii} } = r = \sum $ of the all the $1$'s in row $i$
This completes the main diagonal.
Let row $i$ of $A$ be multiplied by column $j$ of $A^\intercal$.
This is the same as multiplying row $i$ of $A$ by row $j$ of $A$.
This will give the number of times point $i$ is the same block as point $j$.
Therefore:
:$i \ne j \implies \sqbrk {a_{ij} } = \lambda$
So:
:$A^\intercal \cdot A = \sqbrk {a_{ij} } = \paren {r - \lambda} I_v + \lambda J_v$
Category:Balanced Incomplete Block Designs
Product of Indexed Suprema of Non-Negative Numbers

== Theorem ==
Let $\family {a_i}_{i \mathop \in I}$ be a family of elements of the non-negative real numbers $\R_{\ge 0}$ indexed by $I$.
Let $\family {b_j}_{j \mathop \in J}$ be a family of elements of the non-negative real numbers $\R_{\ge 0}$ indexed by $J$.
Let $\map R i$ and $\map S j$ be propositional functions of $i \in I$, $j \in J$.
Let $\ds \sup_{\map R i} a_i$ and $\ds \sup_{\map S j} b_j$ be the indexed suprema on $\family {a_i}$ and $\family {b_j}$ respectively.
Then:
:$\ds \paren {\sup_{\map R i} a_i} \paren {\sup_{\map S j} b_j} = \sup_{\map R i} \paren {\sup_{\map S j} \paren {a_i b_j} }$
== Proof ==
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $35$
Category:Suprema
Product of Indices of Real Number/Integers

== Theorem ==
Let $r \in \R_{> 0}$ be a positive real number.
Let $n, m \in \Z$ be positive integers.
Let $r^n$ be defined as $r$ to the power of $n$.
Then:
:$\paren {r^n}^m = r^{n m}$
== Proof ==
From Product of Indices of Real Number: Positive Integers, we have that:
:$m \in \Z_{\ge 0}: \forall n \in \Z: \paren {r^n}^m = r^{n m}$
It remains to be shown that:
:$\forall m \in \Z_{<0}: \forall n \in \Z: \paren {r^n}^m = r^{n m}$
As $m < 0$ we have that $m = -p$ for some $p \in \Z_{> 0}$.
Thus:
Hence the result, by replacing $-p$ with $m$.
== Sources ==
* : $\S 1.2.2$: Numbers, Powers, and Logarithms: $(5)$
* : $\S 1.2.2$: Numbers, Powers, and Logarithms: Exercise $7$
Category:Product of Indices of Real Number
Product of Indices of Real Number/Positive Integers

== Theorem ==
Let $r \in \R_{> 0}$ be a positive real number.
Let $n, m \in \Z_{\ge 0}$ be positive integers.
Let $r^n$ be defined as $r$ to the power of $n$.
Then:
:$\paren {r^n}^m = r^{n m}$
== Proof ==
Proof by induction on $m$:
For all $m \in \Z_{\ge 0}$, let $\map P m$ be the proposition:
:$\forall n \in \Z_{\ge 0}: \paren {r^n}^m = r^{n m}$
$\map P 0$ is true, as this just says:
:$\paren {r^n}^0 = 1 = r^0 = r^{n \times 0}$
=== Basis for the Induction ===
$\map P 1$ is true, by definition of power to an integer:
:$\paren {r^n}^1 = r^n = r^{n \times 1}$
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\forall n \in \Z_{\ge 0}: \paren {r^n}^k = r^{n k}$
Then we need to show:
:$\forall n \in \Z_{\ge 0}: \paren {r^n}^{\paren {k + 1} } = r^{n \paren {k + 1} }$
=== 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:
:$\forall n, m \in \Z_{\ge 0}: \paren {r^n}^m = r^{n m}$
== Sources ==
* : $\S 0.1$. Arithmetic: Example $1: \ \text{II}$
Category:Product of Indices of Real Number
Product of Indices of Real Number/Rational Numbers

== Theorem ==
Let $r \in \R_{> 0}$ be a (strictly) positive real number.
Let $x, y \in \Q$ be rational numbers.
Let $r^x$ be defined as $r$ to the power of $x$.
Then:
:$\paren {r^x}^y = r^{x y}$
== Proof ==
Let $x = \dfrac p q, y = \dfrac u v$.
Consider $\paren {\paren {r^x}^y}^{q v}$.
Then:
== Sources ==
* : $\S 1$: Real Numbers: Exercise $\S 1.12 \ (3) \, \text{(ii)}$
* : $\S 1.2.2$: Numbers, Powers, and Logarithms: Exercise $9$
Category:Product of Indices of Real Number
Product of Injective Spaces is Injective

== Theorem ==
Let $I$ be a non-empty indexing set.
Let $\family {\struct {S_i, \tau_i} }_{i \mathop \in I}$ be an indexed family of injective topological spaces.
Then $\ds \prod_{i \mathop \in I} \struct {S_i, \tau_i}$ is injective space.
== Proof ==
== Sources ==
* 
Category:Topology
Product of Integers of form 4n + 1

== Theorem ==
Let $m, n \in \Z$ such that both $m$ and $n$ are of the form $4 k + 1$ where $k \in \Z$.
Then $m n$ is also of the form $4 k + 1$.
== Proof ==
Let $m = 4 k_1 + 1, n = 4 k_2 + 1$.
Then:
== Sources ==
* : $\S 2.2$: Divisibility and factorization in $\mathbf Z$: Exercise $8$
Category:Arithmetic
Category:Number Theory
Product of Integral Multiples

== Theorem ==
Let $\struct {F, +, \times}$ be a field.
Let $a, b \in F$ and $m, n \in \Z$.
Then:
:$\paren {m \cdot a} \times \paren {n \cdot b} = \paren {m n} \cdot \paren {a \times b}$
where $m \cdot a$ is as defined in Integral Multiple.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 2$. Elementary Properties
Category:Field Theory
Category:Product of Integral Multiples
Product of Integral Multiples/Proof 1

== Theorem ==
== Proof ==
Let the zero of $F$ be $0_F$.
=== Base Result ===
First we need to show that:
:$\paren {m \cdot a} \times b = m \cdot \paren {a \times b}$
This will be done by induction:
For all $m \in \N$, let $\map P n$ be the proposition:
:$\paren {m \cdot a} \times b = m \cdot \paren {a \times b}$
First we verify $\map P 0$.
When $m = 0$, we have:
So $\map P 0$ holds.
==== Basis for the Induction ====
Now we verify $\map P 1$:
So $\map P 1$ holds.
This is our basis for the induction.
==== Induction Hypothesis ====
Now we need to show 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 our induction hypothesis:
:$\paren {k \cdot a} \times b = k \cdot \paren {a \times b}$
Then we need to show:
:$\paren {\paren {k + 1} \cdot a} \times b = \paren {k + 1} \cdot \paren {a \times b}$
==== 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:
:$\forall m \in \N: \paren {m \cdot a} \times b = m \cdot \paren {a \times b}$
The result for $m < 0$ follows directly from Powers of Group Elements.
=== Full Result ===
Proof by induction:
For all $n \in \N$, let $\map P n$ be the proposition:
:$\forall m \in \Z: \paren {m \cdot a} \times \paren {n \cdot b} = \paren {m n} \cdot \paren {a \times b}$
First we verify $\map P 0$.
When $n = 0$, we have:
So $\map P 0$ holds.
==== Full Result - Basis for the Induction ====
Next we verify $\map P 1$.
When $n = 1$, we have:
So $\map P 1$ holds.
This is our basis for the induction.
==== Full Result - Induction Hypothesis ====
Now we need to show 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 our induction hypothesis:
:$\paren {m \cdot a} \times \paren {k \cdot b} = \paren {m k} \cdot \paren {a \times b}$
Then we need to show:
:$\paren {m \cdot a} \times \paren {\paren {k + 1} \cdot b} = \paren {m \paren {k + 1} } \cdot \paren {a \times b}$
==== Full Result - 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:
:$\forall m \in \Z: \forall n \in \N: \paren {m \cdot a} \times \paren {n \cdot b} = \paren {m n} \cdot \paren {a \times b}$
The result for $n < 0$ follows directly from Powers of Group Elements.
Category:Product of Integral Multiples
Product of Integral Multiples/Proof 2

== Theorem ==
== Proof ==
First let $m = 0$ or $n = 0$.
, let $m = 0$.
The case where $n = 0$ follows the same lines.
We have:
Next let $m > 0$ and $n > 0$.
The results for $m < 0$ and $n < 0$ follow directly from Powers of Group Elements.
Category:Product of Integral Multiples
Product of Limits of Real Sequences (1 + x over n)^n and (1 - x over n)^n equals 1

== Theorem ==
Let $\sequence {a_n}$ be the sequence defined as:
:$a_n = \paren {1 + \dfrac x n}^n$
Let $\sequence {b_n}$ be the sequence defined as:
:$b_n = \paren {1 - \dfrac x n}^n$
Then the product of the limits of $\sequence {a_n}$ and $\sequence {b_n}$ equals $1$
== Proof ==
From Real Sequence $\paren {1 + \dfrac x n}^n$ is Convergent, $\sequence {a_n}$ is convergent.
Setting $x \to -x$, it follows that $\sequence {\paren {1 + \dfrac {\paren {-x} } n}^n} = \sequence {b_n}$ is also convergent.
Then:
From Real Sequence $\paren {1 + \dfrac x n}^n$ is Convergent, $\sequence {\paren {1 - \dfrac {x^2} {n^2} }^{n^2} }$ is convergent to a positive limit.
It follows that:
:$\exists a, b \in \R_{>0}: a < \paren {1 - \dfrac {x^2} {n^2} }^{n^2} < b$
for sufficiently large $n$.
Hence:
:$a^{1/n} < \paren {1 - \dfrac x n}^n \paren {1 - \dfrac x n}^n < b^{1/n}$
The result follows from Limit of Root of Positive Real Number and the Squeeze Theorem for Real Sequences.
== Sources ==
* : $\S 5$: Subsequences: Exercise $\S 5.7 \ (6)$
Category:Examples of Convergent Real Sequences
Product of Lindelöf Spaces is not always Lindelöf

== 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 a Lindelöf space.
Then it is not necessarily the case that $\struct {S, \tau}$ is also Lindelöf space.
== Proof ==
Let $T$ be the Sorgenfrey line.
Let $T' = T \times T$ be Sorgenfrey's half-open square topology.
From Sorgenfrey Line is Lindelöf, $T$ is a Lindelöf space.
From Sorgenfrey's Half-Open Square Topology is Not Lindelöf, $T'$ is not a Lindelöf space.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Invariance Properties
Category:Lindelöf Spaces
Category:Product Spaces