nyuuzyou/wikis huggingface dataset:
Generalized Sum over Union of Disjoint Index Sets

== Theorem ==
Let $\struct {G, +}$ be a commutative topological semigroup.
Let $I$ and $J$ be disjoint indexing sets.
Let $K = I \cup J$.
Let $\family{g_k}_{k \mathop \in K}$ be an indexed family of elements of $G$.
Let the generalized sums $\ds \paren{\sum_{i \mathop \in I} g_i}$ and $\ds \paren{\sum_{j \mathop \in J} g_j}$ converge.
Then:
:the generalized sum $\ds \sum_{k \mathop \in K} g_k$ converges
and:
:$\ds \sum_{k \mathop \in K} g_k = \paren{\sum_{i \mathop \in I} g_i} + \paren{\sum_{j \mathop \in J} g_j}$
== Proof ==
Let $0_G$ be the identity of the semigroup $\struct {G, +}$.
Let $\family{f_k}_{k \mathop \in K}$ be an indexed family of elements of $G$ defined by:
:$\forall k \in K : f_k = \begin{cases}
g_k & : k \in I \\
0_G & : k \in J 
\end{cases}$
Let $\family{h_k}_{k \mathop \in K}$ be an indexed family of elements of $G$ defined by:
:$\forall k \in K : h_k = \begin{cases}
0_G & : k \in I \\
g_k & : k \in J 
\end{cases}$
From Generalized Sum Restricted to Non-zero Summands:
:$\ds \sum_{k \mathop \in K} f_k = \sum_{i \mathop \in I} g_i$
and:
:$\ds \sum_{k \mathop \in K} h_k = \sum_{j \mathop \in J} g_j$
From Sum Rule for Convergent Generalized Sums:
:$\ds \sum_{k \mathop \in K} \paren{f_k + h_k} = \sum_{k \mathop \in K} f_k + \sum_{k \mathop \in K} h_k = \sum_{i \mathop \in I} g_i + \sum_{j \mathop \in J} g_j$
We have:
Hence:
:$\ds \sum_{k \mathop \in K} g_k = \paren{\sum_{i \mathop \in I} g_i} + \paren{\sum_{j \mathop \in J} g_j}$
Category:Generalized Sums
Generalized Sum with Countable Non-zero Summands

== Theorem ==
Let $V$ be a Banach space.
Let $\norm {\, \cdot \,}$ denote the norm on $V$.
Let $\family{v }_{i \in I}$ be an indexed family of elements of $V$.
Let $J$ be a countably infinite subset of $I$ such that $\set{i \in I : v_i \ne 0} \subseteq J$.
Let $\set{j_0, j_1, j_2, \ldots}$ be a countably infinite enumeration of $J$.
Let $r \in \R_{\mathop > 0}$.
Then:
:the generalized sum $\ds \sum_{i \mathop \in I} v_i$ converges absolutely to $r$ 
:
:the series $\ds \sum_{n \mathop = 1}^\infty v_{j_n}$ converges absolutely to $r$
=== Corollary ===
== Proof ==
From Corollary of Generalized Sum Restricted to Non-zero Summands:
:the generalized sum $\ds \sum_{i \mathop \in I} v_i$ converges absolutely to $r$ 
:the generalized sum $\ds \sum_{j \mathop \in J} v_j$ converges absolutely to $r$ 
From Absolute Net Convergence Equivalent to Absolute Convergence: 
:the generalized sum $\ds \sum_{j \mathop \in J} v_j$ converges absolutely to $r$ 
:
:the series $\ds \sum_{n \mathop = 1}^\infty v_{j_n}$ converges absolutely to $r$
Category:Generalized Sums
Category:Banach Spaces
Category:Generalized Sum with Countable Non-zero Summands
Generalized Sum with Countable Non-zero Summands/Corollary

== Theorem ==
Let $V$ be a Banach space.
Let $\norm {\, \cdot \,}$ denote the norm on $V$.
Let $\family{v }_{i \in I}$ be an indexed family of elements of $V$.
Let $J, K$ be countably infinite subsets of $I$ such that $\set{i \in I : v_i \ne 0} \subseteq J \cap K$.
Let $\set{j_0, j_1, j_2, \ldots}$ and $\set{k_0, k_1, k_2, \ldots}$ be countably infinite enumerations of $J$ and $K$ respectively.
Let $r \in \R_{\mathop > 0}$.
Then:
:the series $\ds \sum_{n \mathop = 1}^\infty v_{j_n}$ converges absolutely to $r$
:
:the series $\ds \sum_{n \mathop = 1}^\infty v_{k_n}$ converges absolutely to $r$
== Proof ==
From Generalized Sum with Countable Non-zero Summands:
:the series $\ds \sum_{n \mathop = 1}^\infty v_{j_n}$ converges absolutely to $r$
:the generalized sum $\ds \sum_{i \mathop \in I} v_i$ converges absolutely to $r$
:
:the series $\ds \sum_{n \mathop = 1}^\infty v_{k_n}$ converges absolutely to $r$
Category:Generalized Sum with Countable Non-zero Summands
Generalized Sum with Finite Non-zero Summands

== Theorem ==
Let $G$ be a commutative topological semigroup with identity $0_G$.
Let $\family{g }_{i \in I}$ be an indexed family of elements of $G$.
Let $\set{i \in I : g_i \ne 0_G}$ be finite.
Let $\set{i_1, i_2, \cdots, i_n}$ be a finite enumeration of $\set{i \in I : g_i \ne 0_G}$.
Then the generalized sum $\ds \sum_{i \mathop \in I} g_i$ converges to the summation $\ds \sum_{k \mathop = 1}^n g_{i_k}$
== Proof ==
Let $J = \set{i \in I : g_i \ne 0_G}$.
From Finite Generalized Sum Converges to Summation:
:the generalized sum $\ds \sum_{j \mathop \in J} g_j$ converges to the summation $\ds \sum_{k \mathop = 1}^n g_{i_k}$
From Generalized Sum Restricted to Non-zero Summands:
:the generalized sum $\ds \sum_{i \mathop \in I} g_i$ converges to the summation $\ds \sum_{k \mathop = 1}^n g_{i_k}$
Category:Generalized Sums
Generated Finite Partition of Generated Finite Sub-Sigma-Algebra is Itself

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $\eta$ be a finite partition of $\Omega$.
Then:
:$\map \xi {\map \sigma \eta} = \eta$
where:
:$\map \sigma \cdot$ denotes the generated $\sigma$-algebra
:$\map \xi \cdot$ denotes the generated finite partition.
== Proof ==
== Sources ==
* $4.1$: Partitions and Subalgebras
Category:Probability Theory
Category:Ergodic Theory
Generated Finite Sub-Sigma-Algebra of Generated Finite Partition is Itself

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $\BB\subseteq \Sigma$ be a finite sub-$\sigma$-algebra.
Then:
:$\map \sigma {\map \xi \BB} = \BB$
where:
:$\map \xi \cdot$ denotes the generated finite partition
:$\map \sigma \cdot$ denotes the generated $\sigma$-algebra.
== Proof ==
== Sources ==
* $4.1$: Partitions and Subalgebras
Category:Probability Theory
Category:Ergodic Theory
Generated Sigma-Algebra Contains Generated Dynkin System

== Theorem ==
Let $X$ be a set.
Let $\GG \subseteq \powerset X$ be a collection of subsets of $X$.
Then $\map \delta \GG \subseteq \map \sigma \GG$.
Here $\delta$ denotes generated Dynkin system, and $\sigma$ denotes generated $\sigma$-algebra.
== Proof ==
By Sigma-Algebra is Dynkin System, $\map \sigma \GG$ is a Dynkin system.
The definition of $\map \delta \GG$ now ensures that $\map \delta \GG \subseteq \map \sigma \GG$.
== Sources ==
* : $5.3$
Category:Dynkin Systems
Category:Sigma-Algebras Generated by Collection of Subsets
Generated Sigma-Algebra Contains Generated Sigma-Algebra of Subset

== Theorem ==
Let $\map \sigma \FF$ be the $\sigma$-algebra generated by $\EE$.
Let $\map \sigma \FF$ contain a set of sets $\EE$.
Let $\map \sigma \EE$ be the $\sigma$-algebra generated by $\EE$.
Then $\map \sigma \EE \subseteq \map \sigma \FF$.
== Proof ==
Apply Sigma-Algebra Contains Generated Sigma-Algebra of Subset to $\map \sigma \FF$.
== Sources ==
	
* : $\S 1.2$
Category:Sigma-Algebras Generated by Collection of Subsets
Generated Sigma-Algebra Preserves Finiteness

== Theorem ==
Let $X$ be a set, and let $A_1, \ldots, A_n \subseteq X$ be subsets of $X$.
Then $\map \sigma {\set {A_1, \ldots, A_n} }$ is a finite set, where $\sigma$ denotes generated $\sigma$-algebra.
== Proof ==
Proceed by induction on $n$, that is, on the number of generators.
=== Basis for the Induction ===
The case $n = 0$ is verified as follows:
:$\map \sigma \O = \set {\O, X}$
from Sigma-Algebra Generated by Empty Set.
Thus $\map \sigma \O$ is finite.
This is the basis for the induction.
=== Induction Hypothesis ===
Fix $n \in \N$.
Assume that $\map \sigma {\set {A_1, \ldots, A_n} }$ is finite.
This is the induction hypothesis.
=== Induction Step ===
This is the induction step:
For brevity, write $\Sigma_n$ for $\map \sigma {A_1, \ldots, A_n}$.
Observe that:
:$\set {A_1, \ldots, A_{n + 1} } \subseteq \Sigma_n \cup \set {A_{n + 1} } \subseteq \Sigma_{n + 1}$
Hence Condition on Equality of Generated Sigma-Algebras applies to yield:
:$\Sigma_{n + 1} = \map \sigma {\Sigma_n \cup \set {A_{n + 1} } }$
From Sigma-Algebra Extended by Single Set, it follows that:
:$\map \sigma {\Sigma_n \cup \set {A_{n + 1} } } = \set {\paren {E_1 \cap A_{n + 1} } \cup \paren {E_2 \cap A_{n + 1}^c}: E_1, E_2 \in \Sigma_n}$
Thus there is a surjection $f: \Sigma_n \times \Sigma_n \to \Sigma_{n + 1}$, defined by:
:$\map f {E_1, E_2} := \paren {E_1 \cap A_{n + 1} } \cup \paren {E_2 \cap A_{n + 1}^c}$
By Cardinality of Codomain of Surjection:
:$\card {\Sigma_{n + 1} } \le \card {\Sigma_n \times \Sigma_n}$
and the latter equals $\card {\Sigma_n}^2$ by Cardinality of Cartesian Product of Finite Sets.
By the induction hypothesis, this is necessarily finite.
Therefore, $\Sigma_{n + 1}$ is also finite.
The result follows by the Principle of Mathematical Induction.
== Sources ==
* : $\S 3$: Problem $5 \text{(ii)}$
Category:Sigma-Algebras Generated by Collection of Subsets
Generated Sigma-Algebra Preserves Subset

== Theorem ==
Let $X$ be a set.
Let $\FF, \GG \subseteq \powerset X$ be collections of subsets of $X$.
Suppose that:
:$\FF \subseteq \GG$
Then:
:$\map \sigma \FF \subseteq \map \sigma \GG$
where $\map \sigma \GG$ denotes the $\sigma$-algebra generated by $\GG$
== Proof ==
By definition of $\sigma$-algebra generated by $\GG$:
:$\GG \subseteq \map \sigma \GG$
It follows that also:
:$\FF \subseteq \map \sigma \GG$
By definition of $\sigma$-algebra generated by $\FF$:
:$\FF \subseteq \map \sigma \FF$
We are given that:
:$\FF \subseteq \GG$.
Hence:
:$\map \sigma \FF \subseteq \map \sigma \GG$
== Sources ==
* : $3.5 \ \text{(iii)}$
Category:Sigma-Algebras Generated by Collection of Subsets
Generated Sigma-Algebra by Generated Monotone Class

== Theorem ==
Let $X$ be a set, and let $\GG \subseteq \powerset X$ be a nonempty collection of subsets of $X$.
Suppose that $\GG$ satisfies the following condition:
:$(1):\quad A \in \GG \implies \relcomp X A \in \GG$
that is, $\GG$ is closed under complement in $X$.
Then:
:$\map {\mathfrak m} \GG = \map \sigma \GG$
where $\mathfrak m$ denotes generated monotone class, and $\sigma$ denotes generated $\sigma$-algebra.
=== Corollary ===
== Proof ==
By Sigma-Algebra is Monotone Class, and the definition of generated monotone class, it follows that:
:$\map {\mathfrak m} \GG \subseteq \map \sigma \GG$
Next, define $\Sigma$ by:
:$\Sigma := \set {M \in \map {\mathfrak m} \GG: X \setminus M \in \map {\mathfrak m} \GG}$
By $(1)$, it follows that $\GG \subseteq \Sigma$.
Next, we will show that $\Sigma$ is a $\sigma$-algebra.
To this end, let $G \in \GG$ be arbitrary.
By $(1)$, also $X \setminus G \in \GG$.
Hence from the definition of monotone class:
:$\O = G \cap \paren {X \setminus G} \in \map {\mathfrak m} \GG$
:$X = G \cup \paren {X \setminus G} \in \map {\mathfrak m} \GG$
by virtue of Set Difference Intersection with Second Set is Empty Set and Set Difference and Intersection form Partition.
Since $\O = X \setminus X$, it follows that:
:$X, \O \in \Sigma$
Further, from Set Difference with Set Difference, it follows that:
:$E \in \Sigma \implies X \setminus E \in \Sigma$
Lastly, for any sequence $\sequence {E_n}_{n \mathop \in \N}$ in $\Sigma$, the definition of monotone class implies that:
:$\ds \bigcup_{n \mathop \in \N} E_n \in \map {\mathfrak m} \GG$
Now to ensure that it is in fact in $\Sigma$, compute:
All of the $X \setminus E_n$ are in $\map {\mathfrak m} \GG$ as each $E_n$ is in $\Sigma$.
Hence, from the definition of monotone class, we conclude:
:$\ds \bigcap_{n \mathop \in \N} \paren {X \setminus E_n} \in \map {\mathfrak m} \GG$
which subsequently implies that:
:$\ds \bigcup_{n \mathop \in \N} E_n \in \Sigma$
Thus, having verified all three axioms, $\Sigma$ is a $\sigma$-algebra.
Since $\GG \subseteq \Sigma$, this means, by definition of generated $\sigma$-algebra, that:
:$\map \sigma \GG \subseteq \Sigma \subseteq \map {\mathfrak m} \GG$
By definition of set equality:
:$\map {\mathfrak m} \GG = \map \sigma \GG$
== Sources ==
* : $\S 3$: Problem $11 \ \text{(ii), (iii)}$
Category:Monotone Classes
Category:Sigma-Algebras Generated by Collection of Subsets
Category:Generated Sigma-Algebra by Generated Monotone Class
Generated Sigma-Algebra by Generated Monotone Class/Corollary

== Corollary to Generated Sigma-Algebra by Generated Monotone Class ==
Let $X$ be a set.
Let $\GG \subseteq \powerset X$ be a non-empty set of subsets of $X$.
Define $\relcomp X \GG$ by:
:$\relcomp X \GG := \set {\relcomp X A: A \in \GG}$
Then:
:$\map \sigma \GG = \map {\mathfrak m} {\GG \cup \relcomp X \GG}$
where:
:$\sigma$ denotes generated $\sigma$-algebra
:$\mathfrak m$ denotes generated monotone class.
== Proof ==
From Set is Subset of Union:
:$\GG \subseteq \GG \cup \relcomp X \GG$
Further, as $\map \sigma \GG$ is a $\sigma$-algebra:
:$S \in \map \sigma \GG \implies \relcomp X s = X \setminus S \in \map \sigma \GG$
from Set Difference as Intersection with Relative Complement.
Since $\GG \subseteq \map \sigma \GG$:
:$\GG \cup \relcomp X \GG \subseteq \map \sigma \GG$
By Condition on Equality of Generated Sigma-Algebras:
:$\map \sigma \GG = \map \sigma {\GG \cup \relcomp X \GG}$
Applying Generated Sigma-Algebra by Generated Monotone Class:
:$\map \sigma \GG = \map {\mathfrak m} {\GG \cup \relcomp X \GG}$
Category:Monotone Classes
Category:Generated Sigma-Algebra by Generated Monotone Class
Category:Sigma-Algebras Generated by Collection of Subsets
Generated Submodule is Linear Combinations

== Theorem ==
Let $G$ be a unitary $R$-module.
Let $S \subseteq G$.
Then the submodule $H$ generated by $S$ is the set of all linear combinations of $S$.
== Proof ==
First the extreme case:
The smallest submodule of $G$ containing $\O$ is $\set {e_G}$.
By definition of linear combination of empty set, $\set {e_G}$ is the set of all linear combinations of $\O$.
Now the general case:
Let $\O \subset S \subseteq G$.
Let $L$ be the set of all linear combinations of $S$.
Since $G$ is a unitary $R$-module, every element $x \in S$ is the linear combination $1_R x$, so $S \subseteq L$.
But $L$ is closed for addition and scalar multiplication, so is a submodule.
Thus $H \subseteq L$.
But as every linear combination of $S$ clearly belongs to any submodule of $G$ which contains $S$, we also have $L \subseteq H$.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Theorem $27.3$
Category:Unitary Modules
Category:Linear Algebra
Generated Submodule may not equal Set of Linear Combinations

== Theorem ==
Let $\struct { R, +_R, \times_R }$ be a ring.
Let $\struct { M, +_M, \circ }_R$ be an $R$-module.
Let $S$ be a subset of $M$.
Let $H_1$ be the submodule generated by $S$.
Let $H_2$ be the set of all linear combinations of elements of $S$. 
Then it is possible to select $\struct { R, +_R, \times_R }$, $\struct { M, +_M, \circ }_R$, and $S$ such that:
:$H_1 \ne H_2$
== Proof ==
Let $R = \set {2 k : k \in \Z}$ be the set of all even integers.
From Integers form Commutative Ring, it follows that $\Z$ is a ring.
From Ideal of Ring/Examples/Set of Even Integers, it follows that $R$ is an ideal of $\Z$.
From Ideal is Subring, it follows that $\struct {R, +, \times}$ is a ring, where $+$ denotes integer addition, and $\times$ denotes integer multiplication.
Set $M := R$.
It follows that $\struct {R, +, \times }_R$ is a module over $R$, where the scalar multiplication $\times$ is defined as the integer multiplication $\times$ in $R$.
That is, $R$ is considered as a module over itself.
Set $S := \set{ 2 }$.
Let $H_1$ be the submodule generated by $S$, which means that:
:$\ds H_1 := \bigcap \set { M' \subseteq R : \set {2} \subseteq M', \textrm {$M'$ is a submodule of $R$} } $
Let $M'$ be a submodule of $R$ such that $2 \in M'$.
We show that $M' = R$.
From Congruence Modulo Integer is Equivalence Relation, it follows that all $x \in \Z$ is of the form $x := 4 k' + r$ for some $k \in \Z, r \in \set{ 0,1,2,3 }$.
Suppose $x \in R$ such that $x = 4 k'$ for some $k' \in \Z$
Then:
:$4 k' = 2 k' \times 2$
As $M'$ is closed for scalar product, it follows that $4 k' \in M'$.
Suppose $x \in R$ such that $x = 4 k' + 2$ for some $k' \in \Z$.
As $M'$ is closed for addition, and $2 \in M'$, it follows that $4 k' + 2 \in M'$.
If instead $x = 4 k' + 1$ or $x = 4 k' + 3$, then $x$ cannot be an even integer, so $x \notin R$.
It follows that $R \subseteq M'$.
As $M'$ is a submodule of $R$, it follows that $M' \subseteq R$.
As we have $M' = R$, it follows that:
:$\bigcap \set { M' \subseteq R : \set {2} \subseteq M', \textrm {$M'$ is a submodule of $R$} } = \bigcap \set {R} = R$
Let $H_2$ be the set of all linear combinations of elements of $\set {2}$, which means that:
:$H_2 = \ds \set {\sum_{i \mathop = 1}^n \lambda_i \times 2 : n \in \N_{\ge 1}, \lambda_1, \ldots, \lambda_n \in R}$
With $\ds \sum_{i \mathop= 1}^n \lambda_i \times 2 \in H_2$, it follows that:
As $\ds \sum_{i \mathop= 1}^n k_i \in \Z$, it follows that:
:$H_2 = \set { 4 k : k \in Z}$
Hence, $H_1 \ne H_2$.
== Also see ==
* Equivalence of Definitions of Generated Submodule over Ring with Unity, which shows that the submodules defined as $H_1$ and $H_2$ are always equal, if $R$ is a ring with unity.
Category:Generators of Modules
Category:Linear Combinations
Generated Submonoid is Intersection of Submonoids containing Generator

== Theorem ==
Let $\struct {M, \circ}$ be a monoid whose identity is $e_M$.
Let $S \subseteq M$.
Let $\struct {H, \circ}$ be the submonoid of $\struct {M, \circ}$ generated by $S$.
Then $\struct {H, \circ}$ is the intersection of all submonoids of $\struct {M, \circ}$ containing $S \cup \set {e_M}$.
== Proof ==
Let $\struct {H, \circ}$ be the submonoid of $\struct {M, \circ}$ generated by $S$.
Then by definition $H$ is the smallest (with respect to set inclusion) submonoid of $\struct {M, \circ}$ containing $S \cup \set {e_M}$.
Let $\mathbb S$ be the set of submonoids of $\struct {M, \circ}$ containing $S \cup \set {e_M}$.
We need to show that $H = \ds \bigcap \mathbb S$.
Because $H$ is a submonoid of $\struct {M, \circ}$ containing $S \cup \set {e_M}$:
:$H \in \mathbb S$
By Intersection is Subset:
:$\ds \bigcap \mathbb S \subseteq H$
By Intersection of Submonoids with Monoid Identity is Submonoid:
:$\ds \bigcap \mathbb S$ is a submonoid of $\struct {M, \circ}$ containing $S \cup \set {e_M}$.
Because $H$ is the smallest (with respect to set inclusion) submonoid of $\struct {M, \circ}$ containing $S \cup \set {e_M}$:
:$H \subseteq \ds \bigcap \mathbb S$
By definition of set equality:
:$H = \ds \bigcap \mathbb S$
Hence the result.
== Sources ==
* : Chapter $1$: Basic Definitions and Results
Category:Submonoids
Generated Submonoid is Set of Words of Generator

== Theorem ==
Let $\struct {M, \circ}$ be a monoid whose identity is $e_M$.
Let $S \subseteq M$.
Let $\gen S$ be the submonoid of $\struct {M, \circ}$ generated by $S$.
Then:
:$\gen S = \set {\ds \sum_{i \mathop = 1}^r n_i \cdot s_i : r \in \N, n_i \in \N, s_i \in S}$
where:
:$n_i \cdot s_i$ denotes the power of $s_i$:
::$n \cdot a = \begin {cases}
e_M & : n = 0 \\
\paren {\paren {n - 1} \cdot a} \circ a & : n > 0
\end {cases}$
:$\ds \sum_{i \mathop = 1}^r n_i \cdot s_i := \paren {n_1 \cdot s_1} \circ \paren {n_2 \cdot s_2} \circ \cdots \circ \paren {n_r \cdot s_r}$
That is, $\gen S$ is the set of words of $S$.
== Proof ==
== Sources ==
* : Chapter $1$: Basic Definitions and Results
Category:Submonoids
Category:Words (Abstract Algebra)
Generated Subsemigroup is not necessarily Same as Generated Group

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $A \subseteq G$ be a subset of $G$.
Let $\struct {S, \circ}$ be the subsemigroup of $\struct {G, \circ}$ generated by $S$.
Let $\struct {H, \circ}$ be the subgroup of $\struct {G, \circ}$ generated by $S$.
Then it is not necessarily the case that $\struct {S, \circ}$ is the same as $\struct {H, \circ}$.
== Proof ==
Proof by Counterexample:
Let $\struct {\Z, +}$ be the additive group of integers.
Let $A$ be the set of positive odd integers.
From Generator of Subsemigroup: Positive Odd Numbers:
:the subsemigroup of $\struct {\Z, +}$ generated by $A$ is the semigroup of strictly positive integers under addition
while from Generator of Subgroup: Positive Odd Numbers
:the subgroup of $\struct {\Z, +}$ generated by $A$ is $\struct {\Z, +}$ itself.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings
Category:Generators of Groups
Category:Subsemigroups
Generated Topology is a Topology

== Theorem ==
Let $X$ be a set.
Let $\SS \subseteq \powerset X$, where $\powerset X$ is the power set of $X$.
Let $\TT_\SS$ be the generated topology for $\SS$.
Then $\TT_\SS$ is a topology on $X$.
== Proof ==
We show that $\SS^* = \set {\bigcap S: S \subseteq \SS \text{ finite} }$ (cf. the definition of the generated topology) is a basis.
By definition of synthetic basis, we need to prove:
:$(1): \quad X = \bigcup \SS^*$
:$(2): \quad$ For any $U_1, U_2 \in \SS^*$ and $x \in U_1 \cap U_2$ there is a $U \in \SS^*$ such that $x \in U \subseteq U_1 \cap U_2$
First note that $X = \bigcap \O \in \SS^*$ and therefore $\bigcup \SS^* = X$.
Additionally, if $U_1, U_2 \in \SS^*$, then there exist finite sets $S_1, S_2 \subseteq \SS$ such that $U_1 = \bigcap S_1$ and $U_2 = \bigcap S_2$ by the definition of $\SS^*$.
Thus we have:
:$U_1 \cap U_2 = \bigcap \paren {S_1 \cup S_2}$
Because $S_1 \cup S_2$ is again a finite set it follows that $U_1 \cap U_2 \in \SS^*$.
This implies $(2)$.
Category:Topology
Generating Finite Partition Preserves Order

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $\BB, \CC \subseteq \Sigma$ be finite sub-$\sigma$-algebras.
Then:
:$\BB \subseteq \CC \iff \map \xi \BB \le \map \xi \CC$
where:
:$\map \xi \cdot$ denotes the generated finite partition
:$\le$ denotes the order by refinement of partition.
== Proof ==
== Sources ==
* { $4.1$: Partitions and Subalgebras
Category:Probability Theory
Category:Ergodic Theory
Generating Finite Sub-Sigma-Algebra Preserves Order

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $\eta, \gamma$ be finite partitions of $\Omega$.
Then:
:$\eta \le \gamma \iff \map \sigma \eta \subseteq \map \sigma \gamma$
where:
:$\map \sigma \cdot$ denotes the generated $\sigma$-algebra
:$\le$ denotes the order by refinement of partition.
== Proof ==
== Sources ==
* $4.1$: Partitions and Subalgebras
Category:Probability Theory
Category:Ergodic Theory
Generating Fraction for Lucas Numbers

== Theorem ==
The fraction:
:$\dfrac {199} {9899}$
has a decimal expansion which contains within it the start of the Lucas sequence:
:$0 \cdotp 02010 \, 30407 \, 11 \ldots$
=== Corollary ===
== Proof ==
By long division:
 0.0201030407111829
 ----------------------
9899 ) 199.0000000000000000
 197 98
 --- --
 1 0200
 9899
 - ----
 30100
 29697
 -----
 40300
 39596
 -----
 70400
 69293
 -----
 11070
 9899
 -----
 11710
 9899
 -----
 18110
 9899
 -----
 82110
 79192
 -----
 29180
 19798
 -----
 93820
 89091
 -----
 4729
By Generating Function for Lucas Numbers:
The first few terms are contained in the decimal expansion, as long as $L_{k + 1} < 100$, where there is no carry.
== Sources ==
* : $0 \cdotp 02010 \, 30407 \, 11 \ldots$
Category:Lucas Numbers
Generating Fraction for Lucas Numbers/Corollary

== Corollary to Generating Fraction for Lucas Numbers ==
The fraction:
:$\dfrac {1999} {998 \, 999}$
has a decimal expansion which contains within it the start of the Lucas sequence:
:$0 \cdotp 00200 \, 10030 \, 04007 \, 011 \ldots$
and in general, the fraction:
:$\dfrac {2 \times 10^n - 1} {10^{2 n} - 10^n - 1}$
contains the Lucas sequence spread out with $n$ digits between each term.
== Proof ==
By Generating Function for Lucas Numbers:
The first few terms are contained in the decimal expansion, as long as $L_{k + 1} < 10^n$, where there is no carry.
For $n = 3$:
== Sources ==
* : $0 \cdotp 02010 \, 30407 \, 11 \ldots$
Category:Lucas Numbers
Generating Function Divided by Power of Parameter

== Theorem ==
Let $\map G z$ be the generating function for the sequence $\sequence {a_n}$.
Let $m \in \Z_{\ge 0}$ be a non-negative integer.
Then $\dfrac 1 {z^m} \paren {\map G z - \ds \sum_{k \mathop = 0}^{m - 1} a_k z^k}$ is the generating function for the sequence $\sequence {a_{n + m} }$.
== Proof ==
Hence the result.
== Sources ==
* : $\S 1.2.9$: Generating Functions: $(4)$
Category:Generating Functions