nyuuzyou/wikis huggingface dataset:
Bolzano-Weierstrass Theorem/Lemma 0

== Lemma for Bolzano-Weierstrass Theorem ==
Let $S' \subseteq S \subseteq \R$.
Then every limit point of $S'$ is a limit point of $S$.
== Proof ==
Consider an arbitrary limit point $l$ of $S'$.
Fix $\epsilon \in \R_{>0}$.
By definition of limit point:
:$\paren {\map {B_\epsilon} l \setminus \set l} \cap S' \ne \O$
where $\map {B_\epsilon} l$ denotes the open $\epsilon$-ball of $l$.
Thus there exists a real $s_\epsilon$ in both $\map {B_\epsilon} l \setminus \set l$ and $S'$.
But since $S' \subseteq S$, $s_\epsilon \in S'$ implies $s_\epsilon \in S$.
So, in other words, $s_\epsilon$ is in both $\map {B_\epsilon} l \setminus \set l$ and $S$.
That is:
:$\paren {\map {B_\epsilon} l \setminus \set l} \cap S \ne \O$
This is exactly what it means for $l$ to be a limit point of $S$.
As $l$ is arbitrary, this result is true for every limit point $l$ of $S'$.
== Also known as ==
Category:Bolzano-Weierstrass Theorem
Bolzano-Weierstrass Theorem/Lemma 1

== Lemma for Bolzano-Weierstrass Theorem ==
Let $S$ be a non-empty subset of the real numbers such that its supremum $\map \sup S$ exists.
Let $\map \sup S \notin S$.
Then $\map \sup S$ is a limit point of $S$.
== Proof ==
 $\map \sup S$ is not a limit point of $S$.
By the negation of the definition of a limit point, there is an $\epsilon \in \R_{>0}$ such that:
:$\paren {\map {B_\epsilon} {\map \sup S} \setminus \set {\map \sup S} } \cap S = \O$
Since $\map \sup S \notin S$, adding back $\map \sup S$ to $\map {B_\epsilon} {\map \sup S} \setminus \set {\map \sup SS}$ still gives an empty intersection with $S$.
That is:
:$\map {B_\epsilon} {\map \sup S} \cap S = \openint {\map \sup S - \epsilon} {\map \sup S + \epsilon} \cap S = \O$
So, since $\openint {\map \sup S - \epsilon} {\map \sup S} \subset \openint {\map \sup S - \epsilon} {\map \sup S + \epsilon}$, we also have:
:$\openint {\map \sup S - \epsilon} {\map \sup S} \cap S = \O$
Now, because $\epsilon > 0$, $\openint {\map \sup S - \epsilon} {\map \sup S}$ is non-empty.
So, there is a real $r$ such that $\map \sup S - \epsilon < r < \map \sup S$.
This $r$ is an upper bound on $S$.
We show this as follows:
Note that for any $s \in S$:
:$s < \map \sup S$
Suppose:
:$\map \sup S - \epsilon < s < \map \sup S$
Then:
:$s \in \openint {\map \sup S - \epsilon} {\map \sup S}$
This contradicts what we established earlier: that $\openint {\map \sup S - \epsilon} {\map \sup S}$ cannot have an element of $S$.
That is, $r$ is an upper bound on $S$.
Hence we finally have that:
:$s \le \map \sup S - \epsilon < r < \map \sup S$
which makes $r$ a lower upper bound on $S$ than $\map \sup S$.
This contradicts the Continuum Property of $\map \sup S$.
Hence by Proof by Contradiction it must be the case that $\map \sup S$ is a limit point of $S$.
== Also known as ==
Category:Bolzano-Weierstrass Theorem
Bolzano-Weierstrass Theorem/Lemma 2

== Theorem ==
Let $S$ be a non-empty subset of the real numbers such that its infimum $\map \inf s$ exists.
Let $\map \inf s \notin S$.
Then $\map \inf s$ is a limit point of $S$.
== Proof ==
The proof follows exactly the same lines as Lemma $1$.
== Also known as ==
== Sources ==
* : $1$: Complex Numbers: Point Sets: $1$. Weierstrass-Bolzano Theorem
Category:Bolzano-Weierstrass Theorem
Bolzano-Weierstrass Theorem/Proof 1

== Theorem ==
== Proof ==
Let $\sequence {x_n}$ be a bounded sequence in $\R$.
By the Peak Point Lemma, $\sequence {x_n}$ has a monotone subsequence $\sequence {x_{n_r} }$.
Since $\sequence {x_n}$ is bounded, so is $\sequence {x_{n_r} }$.
Hence, by the Monotone Convergence Theorem (Real Analysis), the result follows.
== Also see ==
* Heine-Borel Theorem
== Sources ==
* : $\S 5$: Subsequences: Bolzano-Weierstrass Theorem: $\S 5.10$
* : $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces
Category:Bolzano-Weierstrass Theorem
Bolzano-Weierstrass Theorem/Proof 2

== Theorem ==
== Proof ==
Let $\sequence {x_n}_{n \mathop \in \N}$ be a bounded sequence in $\R$.
By definition there are real numbers $a, b \in \R$ such that $ x_n \in \openint{a}{b}$ for all $n \in \N$.
We will construct a subsequence $\sequence {x_{n_i}}_{i \mathop \in \N}$ and two sequences of real numbers $\sequence {b_i}_{i \mathop \in \N}$, $\sequence {c_i}_{i \mathop \in \N}$ as follows:
Set $a_0 = a$, and $b_0 = b$.
For $i \in \N$, at least one of the sets:
:$\set {x_n: n > n_{i-1} , a_i < x_n < \dfrac {a_i + b_i} 2 }, \set {x_n : n > n_{i-1} , \dfrac {a_i + b_i} 2 < x_n < b_i }, \set {x_n : n > n_{i-1} , x_n = \dfrac {a_i + b_i } 2 }$
contains infinitely many elements.
If the set $\set {x_n : n > n_{i-1} , x_n = \dfrac {a_i + b_i} 2 }$ is infinite, the elements of this set form a subsequence that converges to $\dfrac {a_i + b_i} 2$, and the proof is done.
If instead the set $\set {x_n: n > n_{i-1} , a_i < x_n < \dfrac {a_i + b_i} 2 }$ is infinite, define $x_{n_i}$ as the element of that set with the smallest index.
In this case, set $a_{i+1} = a_i$, and $b_{i+1} = \dfrac { a_i + b_i } 2 $.
If instead the set $\set {x_n : n > n_{i-1} , \dfrac {a_i + b_i} 2 < x_n < b_i }$ is infinite, define $x_{n_i}$ as the element of that set with the smallest index.
In this case, set $a_{i+1} = \dfrac { a_i + b_i } 2 $, and $b_{i+1} = b_i$.
In both cases, we have $x_{n_i} \in \openint{ a_j }{ b_j }$ for all $j \in \N$ with $j < i$, so
:$\size { x_{n_j} - x_{n_i} } < b_j - a_j = \dfrac{ a + b } {2^j}$
Repeat this process so we obtain a subsequence $\sequence {x_{n_i} }_{i \mathop \in \N}$.
Given $\epsilon > 0$, we can choose $j \in \N$ so $\size { x_{n_{i}} - x_{n_{i'}} } < \dfrac{ a + b } {2^j} < \epsilon$ for all $i , i' \geq j$.
By , the subsequence is a Cauchy sequence.
By Cauchy's Convergence Criterion on Real Numbers, the subsequence converges.
== Notes ==
This proof uses Cauchy's Convergence Criterion.
To avoid a circular proof, we must create a proof for the criterion that does not depend on the Bolzano-Weierstrass Theorem, such as this.
== Also see ==
* Heine-Borel Theorem
Category:Bolzano-Weierstrass Theorem
Bonferroni Inequalities

== Theorem ==
The Bonferroni inequalities are a generalization of Boole's Inequality:
Let $E = \set {E_1, E_2, \ldots, E_n}$ be a set of $n$ events.
Let $\overline E = \set {\overline {E_1}, \overline {E_2}, \ldots, \overline {E_n} }$ be the set of complementary events to each of $\set {E_1, E_2, \ldots, E_n}$ respectively.
Let:
where:
:$\map \Pr {E_i}$ denotes the probability of $E_i$
:$\bigcap E$ denotes the intersection of $E$.
When $n$ is odd:
and:
When $n$ is even:
and:
In particular:
:$\map \Pr {\bigcap E} > 1 - \ds \sum_{i \mathop = 1}^n \map \Pr {\overline E_i}$
which is a statement of Boole's Inequality.
== Proof ==
== Also see ==
* Bonferroni Correction
== Sources ==
* 
* 
Category:Bonferroni Inequalities
Category:Probability Theory
Bonnet's Recursion Formula

== Theorem ==
Let $\map {P_n} x$ denote the Legendre polynomial of order $n$.
Bonnet's Recursion Formula states:
:$\paren {n + 1} \map {P_{n + 1} } x = \paren {2 n + 1} x \map {P_n} x - n \map {P_{n - 1} } x$
== Proof ==
From Generating Function for Legendre Polynomials, the generating function for $P_n$ is:
:$(1): \quad \ds \frac 1 {\sqrt {1 - 2 x t + t^2} } = \sum_{n \mathop = 0}^\infty \map {P_n} x t^n$
Differentiating both sides of $(1)$ $t$:
Equating coefficients of $t^n$:
Hence the result.
== Also presented as ==
== Sources ==
* : $\S 25$: Legendre Functions: Recurrence Formulas for Legendre Polynomials: $25.20$
* : $\S 28$: Legendre and Associated Legendre Functions: Recurrence Formulas for Legendre Polynomials: $28.20.$
Category:Bonnet's Recursion Formula
Category:Legendre Polynomials
Bonnet's Recursion Formula/Also presented as

== Bonnet's Recursion Formula: Also presented as == 
Bonnet's recursion formula can also be presented in the form:
:$\paren {n + 1} \map {P_{n + 1} } x - \paren {2 n + 1} x \map {P_n} x + n \map {P_{n - 1} } x = 0$
== Sources ==
* : $\S 25$: Legendre Functions: Recurrence Formulas for Legendre Polynomials: $25.20$
* : $\S 28$: Legendre and Associated Legendre Functions: Recurrence Formulas for Legendre Polynomials: $28.20.$
Category:Bonnet's Recursion Formula
Bonnet-Myers Theorem

== Theorem ==
Let $M$ be a complete connected Riemannian manifold.
Suppose all the sectional curvatures of $M$ are bounded below by a positive constant.
Then $M$ is compact and has a finite fundamental group.
== Proof ==
== Sources ==
* : $\S 1$: What Is Curvature? Curvature in Higher Dimensions
Category:Riemannian Manifolds
Bookworm Riddle

== Riddle ==
A $3$-volume set of books is on the shelf of a library, volumes $1$, $2$ and $3$ going as normal from left to right.
A bookworm starts eating from the front cover of volume $1$ and works its way through the the back cover of volume $3$.
Then it eats its way through all $3$ volumes, right?
== Solution ==
Wrong.
Just through volume $2$.
:
As can be seen, when the volumes are arranged on the shelf in the specified order:
:the front cover of volume $1$ abuts the back cover of volume $2$
:the front cover of volume $2$ abuts the back cover of volume $3$.
So the bookworm eats through volume $2$, from the back to the front.
== Sources ==
* : Chapter One: What is a Paradox?
Category:Classic Riddles
Category:Veridical Paradoxes
Boole's Inequality

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $A_1, A_2, \ldots, A_n$ be events in $\Sigma$.
Then:
:$\ds \map \Pr {\bigcup_{i \mathop = 1}^n A_i} \le \sum_{i \mathop = 1}^n \map \Pr {A_i}$
== Proof ==
A direct consequence of the facts that:
:a Probability Measure is Subadditive
:the result Finite Union of Sets in Subadditive Function which gives:
::$\ds \map f {\bigcup_{i \mathop = 1}^n A_i} \le \sum_{i \mathop = 1}^n \map f {A_i}$
:for a subadditive function $f$.
== Also known as ==
This inequality is also known as union bound.
== Sources ==
* : $\S 1.11$: Problems: $3$
Category:Unions of Events
Boolean Algebra as Boolean Lattice

== Theorem ==
Let $\struct {S, \vee, \wedge, \neg}$ be a Boolean algebra.
Let $\preceq$ be the ordering on $S$ defined as:
:$a \preceq b \iff a \vee b = b$
for all $a, b \in S$.
Then $\struct {S, \vee, \wedge, \preceq}$ is a Boolean lattice.
== Proof ==
Hence it suffices to show that:
:$(1): \quad \preceq$ is in fact an ordering
:$(2): \quad \preceq$ is compatible with $\vee$ and $\wedge$
:$(3): \quad$ For all $a, b \in S: a \wedge b \preceq a \vee b$
=== $(1)$: Ordering of $\preceq$ ===
By Operations of Boolean Algebra are Idempotent:
:$a \vee a = a$
for all $a \in S$.
Therefore, $\forall a \in S: a \preceq a$.
Thus, $\preceq$ is reflexive.
Let $a, b, c \in S$ be arbitrary, and suppose $a \preceq b$ and $b \preceq c$.
By definition, that is:
:$\paren 1 \quad a \vee b = b$
:$\paren 2 \quad b \vee c = c$
By substitution of $\paren 1$ into $\paren 2$ we have:
:$\paren {a \vee b} \vee c = c$
By Operations of Boolean Algebra are Associative:
:$a \vee \paren {b \vee c} = c$
By substitution of $\paren 2$:
:$a \vee c = c$
Therefore, $a \preceq c$ by definition.
Since $a, b, c \in S$ were arbitrary, it follows that $\preceq$ is transitive.
Let $a, b \in S$ be arbitrary, and suppose $a \preceq b$ and $b \preceq a$.
By definition, that is:
:$a \vee b = b$
:$b \vee a = a$
By , it follows that $a \vee b = b \vee a$.
It immediately follows that $a = b$ from the assumptions.
Since $a, b \in S$ were arbitrary, it follows that $\preceq$ is antisymmetric.
From the above, $\preceq$ is an ordering by definition.
=== $(2.1)$: Compatibility of $\preceq$ with $\wedge$ ===
By , it suffices to show that, for any $a, b, c \in S$:
:$a \preceq b \implies \paren {c \wedge a} \preceq \paren {c \wedge b}$
Let $a, b, c \in S$ be arbitrary, and suppose $a \preceq b$.
By definition of $\preceq$:
:$a \vee b = b$
Then:
But, by definition of $\preceq$:
:$\paren {c \wedge a} \preceq \paren {c \wedge b}$
Since $a, b, c \in S$ were arbitrary, it follows that $\preceq$ is compatible with $\wedge$.
=== $(2.2)$: Compatibility of $\preceq$ with $\vee$ ===
By , it suffices to show that, for any $a, b, c \in S$:
:$a \preceq b \implies \paren {c \vee a} \preceq \paren {c \vee b}$
Let $a, b, c \in S$ be arbitrary, and suppose $a \preceq b$.
By definition of $\preceq$:
:$a \vee b = b$
Then:
Therefore, $c \vee a \preceq c \vee b$.
Since $a, b, c \in S$ were arbitrary, it follows that $\preceq$ is compatible with $\vee$.
=== $(3)$: Ordering of $\wedge$ and $\vee$ ===
Let $a, b \in S$ be arbitrary.
Then:
Therefore, by definition:
:$a \wedge b \preceq a \vee b$
All of the conditions listed above have been satisfied.
By inspecting the relevant definitions, it follows that $\struct {S, \vee, \wedge, \preceq}$ is a Boolean lattice.
Category:Boolean Algebra is Equivalent to Boolean Lattice
Category:Boolean Algebras
Category:Boolean Lattices
Boolean Group is Abelian

== Theorem ==
Let $G$ be a Boolean group.
Then $G$ is abelian.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 1.1$: Exercise $13$
* : $0$: Some Conventions and some Basic Facts: Exercise $3$
* : Chapter $6$: An Introduction to Groups: Exercise $9$
* : $\S 44.4$ Some consequences of Lagrange's Theorem
* : Chapter $3$: Elementary consequences of the definitions: Exercise $1$
* : $\S 1$: Exercise $8$
Category:Abelian Groups
Category:Boolean Groups
Category:Boolean Group is Abelian
Boolean Group is Abelian/Proof 1

== Theorem ==
== Proof ==
By definition of Boolean group, all elements of $G$, other than the identity, have order $2$.
By Group Element is Self-Inverse iff Order 2 and Identity is Self-Inverse, all elements of $G$ are self-inverse.
The result follows directly from All Elements Self-Inverse then Abelian.
Category:Boolean Group is Abelian
Boolean Group is Abelian/Proof 2

== Theorem ==
== Proof ==
Let $ a, b \in G$.
By definition of Boolean group:
:$\forall x \in G: x^2 = e$
where $e$ is the identity of $G$.
Then:
Thus $a b = b a$ and therefore $G$ is abelian. 
Category:Boolean Group is Abelian
Boolean Interpretation is Well-Defined/Proof 1

== Theorem ==
== Proof ==
By Language of Propositional Logic has Unique Parsability, $\LL_0$ is uniquely parsable.
Therefore, the Principle of Definition by Structural Induction can be applied to $\LL_0$.
By inspection, we see that the definition of the boolean interpretation $v$ follows the bottom-up specification of propositional logic.
Hence the Principle of Definition by Structural Induction implies that $v$ is well-defined.
Category:Boolean Interpretations
Boolean Interpretation is Well-Defined/Proof 2

== Theorem ==
== Proof ==
This is to be done by strong induction on the length of WFFs.
By definition of $v$ being a boolean interpretation, $\map v p$ is well-defined for all $p \in \PP_0$, the vocabulary of $\LL_0$.
A WFF of length $1$ has (trivially) a unique parsing sequence.
Consequently, only a single defining rule for $v$ as a boolean interpretation applies.
So the result holds for all WFFs of length $1$.
Now, suppose the result is true for all WFFs of length $k$ or less.
Let $\mathbf A$ be a WFF of length $k+1$.
There are two possibilities:
Suppose $\mathbf A = \neg \mathbf B$ for some WFF $\mathbf B$.
Then $\mathbf B$ is of length $k$, so by the induction hypothesis has a unique value $v (\mathbf B)$ no matter what parsing sequence is used.
So as $\map v {\mathbf A} = \map {f^\neg} {\map v {\mathbf B} }$, it follows that $\mathbf A$ likewise has a unique value.
Hence the result holds for $k + 1$ in this situation.
Suppose $\mathbf A = \paren {\mathbf B * \mathbf C}$ for some WFFs $\mathbf B$ and $\mathbf C$ and some connective $*$.
By Language of Propositional Logic has Unique Parsability, $*$ must be the unique main connective.
So $\mathbf B$ and $\mathbf C$ are both WFFs shorter than $k + 1$ and therefore by the induction hypothesis have unique values $\map v {\mathbf B}$ and $\map v {\mathbf C}$.
Since $\mathbf A = \paren {\mathbf B * \mathbf C}$ for a unique main connective $*$, it follows that:
:$\map v {\mathbf A} = \map {f^*} {\map v {\mathbf B}, \map v {\mathbf C} }$
is well-defined.
Hence the result holds for $k + 1$ in this situation.
So the result follows by the Second Principle of Mathematical Induction.
== Sources ==
* : $\S 1.5$: Semantics of Propositional Logic: Theorem $1.5.1$
Category:Propositional Logic
Boolean Lattice is Heyting Lattice

== Theorem ==
Let $\struct{B, \vee, \wedge, \neg, \preceq}$ be a Boolean lattice.
Then:
:$\struct{B, \vee, \wedge, \preceq}$ is a Heyting lattice where:
::$\forall x, y \in B : x \to y = \neg x \vee y$
:and:
::$x \to y$ denotes the relative pseudocomplement of $x$ with respect to $y$
== Proof ==
By definition of Boolean lattice and Heyting lattice it remains to show that for all $x, y \in B$ the relative pseudocomplement of $x$ with respect to $y$ exists.
It will be shown that:
:$\forall x, y \in B : \neg x \vee y$ is the relative pseudocomplement of $x$ with respect to $y$
By definition of relative pseudocomplement, it will be shown that:
:$\forall x, y \in B : z \preceq \neg x \vee y$ $z \wedge x \preceq y$
Let:
:$z \in B : z \preceq \neg x \vee y$.
We have:
Let:
:$z \in B : z \wedge x \preceq y$.
We have:
The result follows.
Category:Boolean Lattices
Category:Heyting Algebras
Boolean Prime Ideal Extension

== Theorem ==
Let $B = \tuple {S, \preceq}$ be a Boolean lattice.
Let $I$ be a proper ideal in $B$.
Then, there exists a prime ideal $P$ in $B$ such that $I \subseteq P$.
== Proof from Ultrafilter Lemma ==
Category:Boolean Lattices
Category:Prime Ideals (Order Theory)
Boolean Prime Ideal Theorem

== Theorem ==
Let $\struct {S, \le}$ be a Boolean lattice.
Let $I$ be an ideal in $S$.
Let $F$ be a filter on $S$.
Let $I \cap F = \O$.
Then there exists a prime ideal $P$ in $S$ such that:
:$I \subseteq P$
and:
:$P \cap F = \O$
=== Element Extension Lemma ===
== Proof from the Axiom of Choice ==
== Proof from the Ultrafilter Lemma ==
== Proof from Stone's Representation Theorem for Boolean Algebras ==
== Motivation ==
The Boolean Prime Ideal Theorem is weaker than the Axiom of Choice, but is similarly independent of ZF theory.
It is sufficient to prove a number of important theorems, although such proofs are often more involved than ones relying on the Axiom of Choice.
== Sources ==
Category:Boolean Prime Ideal Theorem
Category:Boolean Algebras
Boolean Prime Ideal Theorem/Element Extension Lemma

== Theorem ==
Let $\struct {B, \vee, \wedge, \le}$ be a Boolean lattice.
Let $F \subseteq B$ be a filter on $B$.
Let $a, x \in B$ such that:
:$a \notin F$
Then either:
:$a \vee x \notin F$
or:
:$a \vee \neg x \notin F$
== Proof ==
 that both:
:$a \vee x \in F$
and:
:$a \vee \neg x \in F$
Then:
which contradicts the hypothesis.
Category:Boolean Lattices
Boolean Prime Ideal Theorem/Extension Lemma

== Theorem ==
Let $\struct {B, \vee, \wedge, \le}$ be a Boolean lattice.
Let $F \subseteq B$ be a filter.
Let $J \subseteq B$ have the finite join property with respect to $F$.
Let $z \in B$.
Then either $J \cup \set z$ or $J \cup \set {\neg z}$ also has the finite join property with respect to $F$.
== Proof ==
 that neither $J \vee z$ nor $J \vee \neg z$ has the finite join property with respect to $F$.
Then there are $x_1, \dots, x_n, y_1, \dots, y_m \in J$ such that $x_1 \vee \dots \vee x_n \vee z \in F$ and $y_1 \vee \dots \vee y_m \vee \neg z \in F$.
Let $q = x_1 \vee \dots \vee x_n \vee y_1 \vee \dots \vee y_m$.
Then, by Join Succeeds Operands:
:$x_1 \vee \dots \vee x_n \le q$
:$y_1 \vee \dots \vee y_m \le q$
and so:
:$q \vee z \in F$
:$q \vee \neg z \in F$
by filter axiom $\paren 3$.
Thus:
This contradicts the fact that $J$ has the finite join property.
Category:Lattice Theory
Boolean Prime Ideal Theorem/Proof 1

== Theorem ==
== Proof ==
Let $T$ be the set of ideals in $S$ that contain $I$ and are disjoint from $F$, ordered by inclusion.
Let $N$ be a chain in $T$.
Then $\ds U = \bigcup N$ is clearly disjoint from $F$ and contains $I$.
Let $x \in U$ and $y \le x$.
Then:
:$\exists A \in N: x \in A$
By the definition of union:
:$x \in U$
Let $x \in U$ and $y \in U$.
Then:
:$\exists A, B \in N: x \in A, y \in B$
By the definition of a chain:
:$A \subseteq B$
or:
:$B \subseteq A$
Hence $U$ is also an ideal.
, suppose $A \subseteq B$.
Then:
:$x \in B$
Since $y$ is also in $B$, and $B$ is an ideal:
:$x \vee y \in U$.
By Zorn's Lemma, $T$ has a maximal element, $M$.
It remains to show that $M$ is a prime ideal:
Every Boolean algebra is a distributive lattice.
So by Maximal Ideal WRT Filter Complement is Prime in Distributive Lattice, $M$ is a prime ideal.
Category:Boolean Prime Ideal Theorem
Boolean Ring has Proper Zero Divisor

== Theorem ==
Let $\left({R, +, \circ}\right)$ be a Boolean ring whose zero is $0_R$.
Suppose that $R$ has more than two elements.
Then $R$ has a proper zero divisor.
== Proof ==
Since $R$ has more than two elements, there exist distinct non-zero elements $x, y \in R$.
Note that $x + y \ne 0_R$ since $x$ and $y$ are distinct (by Idempotent Ring has Characteristic Two).
If $x \circ y = 0$, $x$ is a proper zero divisor.
If $x \circ y \ne 0$, then:
Hence $x \circ y$ is a proper zero divisor.
The result follows, from Proof by Cases.
== Sources ==
* : $\S 1$: Exercise $9$
Category:Boolean Rings
Boolean Satisfiability Problem/Examples/Arbitrary Example 1

== Example of Boolean Satisfiability Problem ==
Let $\LL$ be the set of propositional formulas given by:
:$\neg x_1 \lor \neg x_2$
:$x_3$
:$x_3 \implies x_2$
then a solution to the boolean satisfiability problem for $\LL$ is:
:$x_1 = \operatorname{False}$
:$x_2 = \operatorname{True}$
:$x_3 = \operatorname{True}$
Category:Examples of Boolean Satisfiability Problems
Bootstrapping/Example/Arbitrary Example 1

== Example of Bootstrapping ==
Let $B$ bootstrap samples be taken of a population.
Let $m$ be the median of the population as a whole.
Let ${m_b}^*$ be the median of the $b$th bootstrap sample, where $b = 1, 2, \ldots, B$.
Then the bootstrap estimated standard error of $m$ is given by:
:$\map {\operatorname {se} } m = \sqrt {\dfrac 1 {B - 1} \ds \sum_b \paren { {m_b}^* - {\overline m}^*}^2}$
where ${\overline m}^*$ is the mean of the ${m_b}^*$.
Approximating $95 \%$ confidence limits for the population median are given by the $0 \cdotp 025$ and $0 \cdotp 975$ quantiles of the set of $b$ values ${m_b}^*$.
In practice, good values of $\map {\operatorname {se} } m$ can be obtained with $B = 50$, but values of $B = 1000$ or $B = 2000$ are needed for reliable estimates of confidence intervals.
== Sources ==
* 
* 
Category:Examples of Bootstrapping
Borel-Cantelli Lemma

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\sequence {E_n}_{n \mathop \in \N}$ be a sequence of $\Sigma$-measurable sets.
If:
:$\ds \sum_{n \mathop = 1}^\infty \map \mu {E_n} < \infty$
then:
:$\ds \map \mu {\limsup_{n \mathop \to \infty} E_n} = 0$
where $\limsup$ denotes limit superior of sets.
== Proof ==
By definition of limit superior:
:$\ds \limsup_{n \mathop \to \infty} E_n = \bigcap_{i \mathop = 1}^\infty \bigcup_{j \mathop = i}^\infty E_j$
Thus, by Measure is Monotone and Intersection is Subset:
:$(1): \quad \ds \map \mu {\limsup_{n \mathop \to \infty} E_n} = \map \mu {\bigcap_{i \mathop = 1}^\infty \bigcup_{j \mathop = i}^\infty E_j} \le \map \mu {\bigcup_{j \mathop = i}^\infty E_j}$
for all $i \in \N$.
By Measure is Subadditive:
:$\ds \map \mu {\bigcup_{j \mathop = i}^\infty E_j} \le \sum_{j \mathop = i}^\infty \map \mu {E_j}$
However, by assumption $\ds \sum_{n \mathop = 1}^\infty \map \mu {E_n}$ converges.
By Tail of Convergent Series tends to Zero this implies:
:$\ds \lim_{i \mathop \to \infty} \sum_{n \mathop = i}^\infty \map \mu {E_n} = 0$
Now $(1)$ implies, together with Lower and Upper Bounds for Sequences, that:
:$\ds \map \mu {\limsup_{n \mathop \to \infty} E_n} \le 0$
But as $\mu$ is a measure, the converse inequality also holds.
Hence:
:$\ds \map \mu {\limsup_{n \mathop \to \infty} E_n} = 0$
== Borel-Cantelli Lemma in Probability ==
As each probability space $\struct {X, \Sigma, \Pr}$ is a measure space, the result carries over to probability theory.
Hence, given any countable sequence of events $E_n$, the sum of whose probabilities is finite, the probability that infinitely many of the events occur is zero.
== Also see ==
* Second Borel-Cantelli Lemma
== Sources ==
* 
* : $\S 6$: Problem $9$
* 
Category:Measure Theory
Category:Probability Theory