nyuuzyou/wikis huggingface dataset:
Borel-Cantelli Lemma 10 to Kochen-Stone Theorem

== Theorem ==
Let $A_n$ be a sequence of events with $\ds \sum \map \Pr {A_n} = \infty$ and:
:$\ds \liminf_{k \mathop \to \infty} \frac {\ds \sum_{1 \mathop \le m, n \mathop \le k} \map \Pr {A_n \cap A_m} } {\ds \paren {\sum_{n \mathop = 1}^k \map \Pr {A_n} }^2} < \infty$
Then there is a positive probability that $A_n$ occur infinitely often.
== Proof ==
Fix $\ell < k$.
Let $\ds X = \sum_{n \mathop = \ell}^k 1_{A_n}$.
It follows that:
:$\ds \expect X = \sum_{n \mathop = \ell}^k \Pr(A_n)$
and:
:$\ds \expect {X^2} = \sum_{\ell \mathop \le m, n \mathop \le k} \map \Pr {A_n \cap A_m}$
Using the Paley-Zygmund Inequality for $\theta = 0$, we obtain:
Now, it holds that:
:$\ds \map \Pr {A_n \text { occurs i.o.} } = \lim_{\ell \mathop \to \infty} \lim_{k \mathop \to \infty} \map \Pr {\bigcup_{n \mathop = \ell}^k A_n}$
We have:
:$\ds \frac {\ds \paren {\sum_{n \mathop = 1}^k \map \Pr {A_n} - \sum_{n \mathop = 1}^{\ell - 1} \map \Pr {A_n} }^2} {\ds \sum_{1 \mathop \le m, n \mathop \le k} \map \Pr {A_n \cap A_m} - \sum_{1 \mathop \le m, n \mathop < \ell} \map \Pr {A_n \cap A_m} } \ge \frac {\ds \left(\sum_{n \mathop = 1}^k \map \Pr {A_n} - \sum_{n \mathop = 1}^{\ell - 1} \Pr (A_n)\right)^2} {\ds \sum_{1 \mathop \le m, n \mathop \le k} \map \Pr {A_n \cap A_m} }$
Fix $l \in \N_1$.
Since $\ds \lim_{k \mathop \to \infty}\sum_{n \mathop = 1}^k \map \Pr {A_n} = \infty$, by assumption, if $k$ is sufficiently large:
:$\ds \paren {\sum_{n \mathop = 1}^ k \map \Pr {A_n} - \sum_{n \mathop = 1}^{\ell - 1} \map \Pr {A_n} }^2 \approx \paren {\sum_{n \mathop = 1}^k \map \Pr {A_n} }^2$
So if $k$ is sufficiently large:
:$\ds \frac {\ds \paren {\sum_{n \mathop = 1}^k \map \Pr {A_n} - \sum_{n \mathop = 1}^{\ell - 1} \map \Pr {A_n} }^2} {\ds \sum_{1 \mathop \le m, n \mathop \le k} \map \Pr {A_n \cap A_m} } \approx \frac {\ds \paren {\sum_{n \mathop = 1}^k \map \Pr {A_n} }^2} {\ds \sum_{1 \mathop \le m, n \mathop \le k} \map \Pr {A_n \cap A_m} } = \frac 1 {\frac {\ds \sum_{1 \mathop \le m, n \mathop \le k} \map \Pr {A_n \cap A_m} } {\ds \paren {\sum_{n \mathop = 1}^k \map \Pr {A_n} }^2} }$
Suppose:
:$\ds \liminf_{k \mathop \to \infty} \frac {\ds \sum_{1 \mathop \le m, n \mathop \le k} \map \Pr {A_n \cap A_m} } {\ds \paren {\sum_{n \mathop = 1}^k \map \Pr {A_n} }^2} = c < \infty$
This means that no matter how large $k$ is, there is always some $k' \ge k$, such that:
:$\ds \frac {\ds \sum_{1 \mathop \le m, n \mathop \le k'} \map \Pr {A_n \cap A_m} } {\ds \paren {\sum_{n \mathop = 1}^{k'} \map \Pr {A_n} }^2} \approx c$
So for infinitely many $k$s:
:$\ds \map \Pr {\bigcup_{n \mathop = \ell}^k A_n} \ge \frac 1 {\frac {\ds \sum_{1 \mathop \le m, n \mathop \le k} \map \Pr {A_n \cap A_m} } {\paren {\ds \sum_{n \mathop = 1}^k \map \Pr {A_n} }^2} } \approx \frac 1 c$
Since $\ds \map \Pr {\bigcup_{n \mathop = \ell}^k A_n}$ is decreasing in $k$, we have, approximately:
:$\ds \lim_{k \mathop \to \infty} \map \Pr {\bigcup_{n \mathop = \ell}^k A_n} \ge \frac 1 c$
Since $l$ was arbitrary:
:$\ds \lim_{l \mathop \to \infty} \lim_{k \mathop \to \infty} \map \Pr {\bigcup_{n \mathop = \ell}^k A_n} \ge \frac 1 c > 0$
Category:Kochen-Stone Theorem
Borel-Carathéodory Lemma

== Theorem ==
Let $D \subset \C$ be an open set with $0 \in D$.
Let $R > 0$ be such that the open disk $\map B {0, R} \subset D$.
Let $f: D \to \C$ be analytic with $\map f 0 = 0$.
Let $\map \Re {\map f z} \le M$ for $\cmod z \le R$.
Let $0 < r < R$.
Then for $\cmod z \le r$:
:$(1): \quad \cmod {\map f z} \le \dfrac {2 M r} {R - r}$
:$(2): \quad \cmod {\map {f^{\paren k} } z} \le \dfrac {2 M R k!} {\paren {R - r}^{k + 1} }$ for all $k \ge 1$
== Proof ==
===Lemma===
== Sources ==
* : Chapter $\text {III}$: Further Theory of $\map \zeta s$. Applications: $\S 6$: Theorem $\text E$
* $6$: The Prime Number Theorem: $\S6.1$: A zero-free region: Lemma $6.2$
Category:Complex Analysis
Borel-Carathéodory Lemma/Lemma

== Theorem ==
Let $D \subset \C$ be an open set with $0 \in D$.
Let $R > 0$ be such that the open disk $\map B {0, R} \subset D$.
Let $f: D \to \C$ be analytic with $\map f 0 = 0$.
Let $\map \Re {\map f z} \le M$ for $\cmod z \le R$.
Let $0 < r < R$.
Then:
:$\ds \forall n \in \Z_{\ge 1} : \quad \frac {\cmod {\map {f^{\paren n} } 0} }{ n! } \le \frac {2 M} {R^n}$
== Proof ==
By Cauchy Integral Theorem:
:$\ds \forall k \in \Z_{\ge 0} : \quad \oint_{\partial D} z^{k-1} \map f z \rd z = 0$
Parametrizing $\partial D$ by $R e^{2 \pi i t}$:
:$\ds \forall k \in \Z_{\ge 0} : \quad \int _0^1 e^{2\pi i k t} \map f {R e^{2 \pi ikt} } \rd t = 0$
On the other hand:
Let $\theta_n := \map \arg {\map {f^{\paren n} } 0}$.
Then:
Since:
:$\ds \int_0^1 \map f {R e^{2 \pi i t} } \paren {2 + e^{\paren {2 \pi n t + \theta_n} i } } \rd t = 0$
we have:
Category:Complex Analysis
Borel Sigma-Algebra Generated by Closed Sets

== Theorem ==
Let $\map \BB {S, \tau}$ be a Borel $\sigma$-algebra generated by the set of open sets in $S$.
Then $\map \BB {S, \tau}$ is equivalently generated by the set of closed sets in $S$.
== Proof ==
By definition, a closed set is the relative complement of an open set.
The result follows from Sigma-Algebra Generated by Complements of Generators.
== Also see ==
* Characterization of Euclidean Borel $\sigma$-Algebra:Open equals Closed
== Sources ==
* : $\S 1.2$
Category:Borel Sigma-Algebras
Borel Sigma-Algebra of Subset is Trace Sigma-Algebra

== Theorem ==
Let $\struct {X, \tau}$ be a topological space, and let $A \subseteq X$ be a subset of $X$.
Let $\tau_A$ be the subspace topology on $A$.
Then the following equality of $\sigma$-algebras on $A$ holds:
:$\map \BB {A, \tau_A} = \map \BB {X, \tau}_A$
where $\BB$ signifies Borel $\sigma$-algebra, and $\map \BB {X, \tau}_A$ signifies trace $\sigma$-algebra.
== Proof ==
By definition of Borel $\sigma$-algebra, it holds that:
:$\map \BB {X, \tau} = \map \sigma \tau$
and also, by definition of subspace topology:
:$\tau_A = A \cap \tau = \set {A \cap U: U \in \tau}$
Thus, it follows that:
:$\map \BB {A, \tau_A} = \map \sigma {A \cap \tau}$
Thereby, the desired equality:
:$\map \BB {A, \tau_A} = \map \BB {X, \tau}_A$
follows directly from applying Trace Sigma-Algebra of Generated Sigma-Algebra with $\GG = \tau$.
== Sources ==
* : $\S 3$: Problem $10 \ \text{(ii)}$
Category:Borel Sigma-Algebras
Category:Trace Sigma-Algebras
Borel Sigma-Algebra on Euclidean Space by Monotone Class

== Theorem ==
Let $\sqbrk {\R^n, \tau}$ be the $n$-dimensional Euclidean space.
Then:
:$\map \BB {\R^n, \tau} = \map {\mathfrak m} \tau$
where $\BB$ denotes Borel $\sigma$-algebra, and $\mathfrak m$ denotes generated monotone class.
== Proof ==
Let $U \in \tau$ be an open set, and define $C$ by:
:$C := X \setminus U$
hence $C$ is a closed set.
Further, define, for all $n \in \N$:
:$C_n := \ds \bigcup_{c \mathop \in C} \map B {c; \frac 1 n}$
where $B$ denotes open ball.
The $C_n$ are open sets, being the union of open balls.
It is clear that $C \subseteq C_n$ for all $n \in \N$.
Conversely, as $U$ is open, for any $u \in U$ (that is, $u \notin C$), find $n \in \N$ such that:
:$\map B {u; \dfrac 1 n} \subseteq U$
as is possible from the definition of open set in a metric space.
Thus, for all $c \in C = X \setminus U$, this means:
:$\map d {u, c} \ge \dfrac 1 n$
whence $u \notin C_n$.
That is, we have established that:
:$c \in C \iff \forall n \in \N: c \in C_n$
Phrased in terms of intersection, this means:
:$C = \ds \bigcap_{n \mathop \in \N} C_n$
Thus, since $C_n \in \tau \subseteq \map {\mathfrak m} \tau$:
:$C \in \map {\mathfrak m} \tau$
Now define:
:$\relcomp X \tau := \set {X \setminus U: U \in \tau}$
Then we have shown:
:$\map {\mathfrak m} {\tau \cup \relcomp X \tau} \subseteq \map {\mathfrak m} \tau$
and the reverse inclusion (and hence equality) follows from Generated Monotone Class Preserves Subset.
Now applying Generated Sigma-Algebra by Generated Monotone Class: Corollary:
:$\map \sigma \tau = \map {\mathfrak m} {\tau \cup \relcomp X \tau}$
Combining these two equalities gives the result, by definition of Borel $\sigma$-algebra.
== Sources ==
* : $\S 3$: Problem $12$
Category:Borel Sigma-Algebras
Category:Monotone Classes
Borell-TIS Inequality

== Theorem ==
Let $T$ be a topological space.
Let $\sequence {f_t}_{t \mathop \in T}$ be a centred (i.e. mean zero) Gaussian process on $T$, such that:
:$\norm f_T := \sup_{t \mathop \in T} \size {f_t}$
is almost surely finite.
Let:
:$\sigma_T^2 := \sup_{t \mathop \in T} \operatorname E \size {f_t}^2$
Then $\map {\operatorname E} {\norm f_T}$ and $\sigma_T$ are both finite, and, for each $u > 0$:
:$\map {\operatorname P} {\norm f_T > \map {\operatorname E} {\norm f_T} + u} \le \map \exp {\dfrac {-u^2} {2 \sigma_T^2} }$
== Proof ==
Borsuk-Ulam Theorem

== Theorem ==
Let $n$ be a positive integer.
Let $f: \mathbb S^n \to \R^n$ be a continuous mapping from an $n$-sphere to $\R^n$.
Then there exists $x \in \mathbb S^n$ such that $\map f x = \map f {-x}$.
== Proof ==
Category:Algebraic Topology
Borsuk Null-Homotopy Lemma

== Theorem ==
Let $\Bbb S^2$ denote the unit sphere in $\R^3$.
Let $a,b \in \Bbb S^2$.
Let $\struct {A, \tau_A}$ be a compact topological space.
Let $f : A \to \Bbb S^2 \setminus \set {a,b}$ be a continuous injective mapping.
Let $f$ be null-homotopic.
Then $a$ and $b$ lie in the same component of $\Bbb S^2 \setminus \Img f$.
Here, $\Img f$ denotes the image of $f$.
=== Corollary ===
== Proof ==
== Also known as ==
This lemma is usually referred to as the Borsuk Lemma.
The name Borsuk Null-Homotopy Lemma is used by to distinguish it from other "Borsuk Lemmas" in litterature.
== Sources ==
* : $10$: Separation Theorems in the Plane: $\S 62$: Invariance of Domain
Category:Homotopy Theory
Borsuk Null-Homotopy Lemma/Corollary

== Corollary ==
Let $a,b \in \R^2$.
Let $\struct {A, \tau_A}$ be a compact topological space.
Let $f : A \to \R^2 \setminus \set {a,b}$ be a continuous injective mapping.
Let $f$ be null-homotopic.
Then $a$ and $b$ lie in the same component of $\R^2 \setminus \Img f$.
Here, $\Img f$ denotes the image of $f$.
== Proof ==
We are given that $f$ is null-homotopic.
So there exists a constant mapping:
:$g: A \to \R^2 \setminus \set {a, b}$ 
such that $f$ and $g$ are homotopic.
By Mapping is Constant iff Image is Singleton, it follows that the image of $g$ is equal to $\set y$ for some $y \in \R^2 \setminus \set {a, b}$.
Since $f$ and $g$ are homotopic, there exists a continuous mapping $H : A \times \closedint 0 1 \to \R^2 \setminus \Img f$, such that:
:$\forall x \in A : \map H {x, 0} = \map f x $
:$\forall x \in A : \map H {x, 1} = \map g x = y $
Then $H$ is by definition a null-homotopy.
We are given that $f : A \to \R^2 \setminus \set {a, b}$ is an injective mapping.
So:
:$\forall x_1, x_2 \in A: \map H {x_1, 0} = \map f {x_1} = \map f {x_2} = \map H {x_2, 0} \implies x_1 = x_2$
We are given that $\struct {A, \tau_A}$ is a compact topological space.
So every open cover for $\struct {A, \tau_A}$ has a finite subcover.
So we have:
:$\ds \map {\operatorname{Comp}_x} {\R^2 \setminus \Img f} = \bigcup \leftset{B \subseteq A: x \in B \land B}$ is connected $\rightset{}$
and:
:$a, b \in \ds \map {\operatorname{Comp}_x} {\R^2 \setminus \Img f}$
Hence $a$ and $b$ lie in the same component of $\R^2 \setminus \Img f$.
== Sources ==
* : $10$: Separation Theorems in the Plane: $\S 62$: Invariance of Domain
Category:Homotopy Theory
Bott-Milnor-Kervaire 1,2,4,8 Theorem

== Theorem ==
Let $A$ be a division algebra with real scalars.
Then the dimension of $A$ is either:
:$1$: the real numbers $\R$
:$2$: the complex numbers $\C$
:$4$: the quaternions $\Bbb H$
or:
:$8$: the octonions $\Bbb O$.
== Proof ==
== Historical Note ==
== Sources ==
* : Chapter $\text {B}.26$: Extensions of the Complex Number System. Algebras, Quaternions, and Lagrange's Four Squares Theorem
Category:Division Algebras
Bott-Milnor-Kervaire 1,2,4,8 Theorem/Historical Note

== Historical Note on Bott-Milnor-Kervaire 1,2,4,8 Theorem ==
Bott-Milnor-Kervaire 1,2,4,8 Theorem was proved by , and in $1958$.
== Sources ==
* : Chapter $\text {B}.26$: Extensions of the Complex Number System. Algebras, Quaternions, and Lagrange's Four Squares Theorem
Category:Historical Notes
Bottom-Up Form of Top-Down Grammar defines same Formal Language

== Theorem ==
Let $\LL$ be a formal language.
Let $\TT$ be a top-down grammar for $\LL$.
Let $\BB$ be the bottom-up form of $\TT$.
Then $\BB$ is also a formal grammar for $\LL$.
== Proof ==
Let $\phi$ be a $\BB$-WFF.
If $\phi$ is a letter, then it is a $\TT$-WFF.
For, it may be formed by replacing the starting metasymbol of $\TT$ by $\phi$.
Suppose that $\phi$ is formed from WFFs $\phi_1, \ldots, \phi_n$ by the rule of formation $\mathbf R_\BB$ of $\BB$.
Suppose also that each of $\phi_1, \ldots, \phi_n$ is also a $\TT$-WFF.
Then by applying the rule of formation $\mathbf R$ of $\TT$, we obtain a collation with metasymbols $\psi_1, \ldots, \psi_n$.
By assumption, we can apply rules of formation of $\TT$ to each $\psi_i$ to yield the corresponding $\TT$-WFF $\phi_i$.
It follows that $\phi$ is also a $\TT$-WFF.
By the Principle of Structural Induction, each $\BB$-WFF is also a $\TT$-WFF.
Conversely, suppose that $\phi$ is a $\TT$-WFF.
Let it be formed by applying the rules of formation $\mathbf R_1, \ldots, \mathbf R_n$, in succession.
We will prove that each metasymbol in the inputs of these rules of formation will be replaced by a $\BB$-WFF.
In particular, then, $\phi$ will be a $\BB$-WFF.
Since $\mathbf R_n$ is the last rule of formation applied, it must replace a metasymbol by a letter.
Since letters are $\BB$-WFFs, $\mathbf R_n$ satisfies the assertion.
Suppose that we have established that all metasymbols in the result of $\mathbf R_i$ will be replaced by $\BB$-WFFs.
Then also the metasymbol $\psi$ that $\mathbf R_i$ replaces by a new collation will be replaced by a WFF.
This is because the collation resulting from replacing $\psi$ according to $\mathbf R_i$ contains metasymbols $\psi_i$, which by assumption will be replaced by corresponding $\BB$-WFFs $\phi_i$.
Now the rule of formation $\paren {\mathbf R_i}_\BB$ ensures that $\psi$ will be replaced by a $\BB$-WFF.
So each metasymbol $\psi$ in the input of $\mathbf R_i$ will be replaced by a $\BB$-WFF.
Hence $\phi$ is a $\BB$-WFF.
== Remark ==
This theorem establishes that any formal language has a bottom-up grammar.
We may therefore assume any formal language to be given by a bottom-up grammar, which provides conceptual simplicity.
Category:Formal Languages
Bottom-Up Specification of Propositional Logic/Examples/Example 1

== Example of Bottom-Up Specification of Propositional Logic ==
The following is a WFF of propositional logic:
:$\paren {\paren {p \land q} \implies \paren {\lnot \paren {q \lor r} } }$
== Proof ==
It is assumed that $p$, $q$ and $r$ are WFFs by $\mathbf W: \PP_0$.
By $\mathbf W: \text {OP}$:
:$\paren {p \land q}$ is a WFF
:$\paren {q \lor r}$ is a WFF
By $\mathbf W: \neg$:
:$\paren {\lnot \paren {q \lor r} }$ is a WFF
By $\mathbf W: \text {OP}$:
:$\paren {\paren {p \land q} \implies \paren {\lnot \paren {q \lor r} } }$ is a WFF
== Sources ==
* : $\S 1$: Informal statement calculus: $\S 1.2$: Truth functions and truth tables: Example $1.3$
Category:Language of Propositional Logic
Bottom Relation is Auxiliary Relation

== Theorem ==
Let $L = \left({S, \vee, \preceq}\right)$ be a bounded below join semilattice.
Let $B = \left\{ {\left({\bot, x}\right): x \in S}\right\}$
where $\bot$ denotes the smallest element in $L$.
Then
:$B$ is auxiliary relation.
== Proof ==
By definition of smallest element:
:$\forall x \in S: \bot \preceq x$
Thus by definition of $B$:
:$\forall x, y \in S: \left({x, y}\right) \in B \implies x \preceq y$
We will prove that
:$\forall x, y, z, u \in S: x \preceq y \land \left({y, z}\right) \in B \land z \preceq u \implies \left({x, u}\right) \in B$
Let $x, y, z, u \in S$ such that
:$x \preceq y \land \left({y, z}\right) \in B \land z \preceq u$
By definition of $B$:
:$y = \bot$
By definition of smallest element:
:$\bot \preceq x$
By definition of antisymmetry:
:$x = \bot$
Thus by definition of $B$:
:$\left({x, u}\right) \in B$
We will prove that
:$\forall x, y, z \in S: \left({x, z}\right) \in B \land \left({y, z}\right) \in B \implies \left({x \vee y, z}\right) \in B$
Let $x, y, z \in S$ such that
: $\left({x, z}\right) \in B \land \left({y, z}\right) \in B$
By definition of $B$:
:$x = y = \bot$
By Join is Idempotent:
:$x \vee y = \bot$
Thus by definition of $B$:
:$\left({x \vee y, z}\right) \in B$
Thus by definition of $B$:
:$\forall x \in S: \left({\bot, x}\right) \in B$
Thus by definition:
:$B$ is auxiliary relation.
== Sources ==
* 
* 
Category:Auxiliary Relations
Bottom Relation is Bottom in Ordered Set of Auxiliary Relations

== Theorem ==
Let $L = \struct {S, \vee, \preceq}$ be a bounded below join semilattice.
Let $\map {\it Aux} L$ be the set of all auxiliary relations on $S$.
Let $P = \struct {\map {\it Aux} L, \precsim}$ be an ordered set where:
:$\precsim \mathop = \subseteq \restriction_{ \map {\it Aux} L \times \map {\it Aux} L}$
Let $B = \set {\struct {\bot_L, x}: x \in S}$
where $\bot_L$ denotes smallest element in $L$.
Then
:$B \mathop = \bot_P$
== Proof ==
By Bottom Relation is Auxiliary Relation:
:$B \in \map {\it Aux} L$
By definition:
:$B$ is an upper bound for $\O$ in $P$
We will prove that:
:$\forall R \in \map {\it Aux} L: R$ is an upper bound for $\O \implies B \mathop \precsim R$
Let $R \in \map {\it Aux} L$.
By condition $(iv)$ of definition of auxiliary relation:
:$B \subseteq R$
Thus by definition of $\precsim$:
:$B \precsim R$
By definition of supremum:
:$B = \sup_P \O$
Thus by Supremum of Empty Set is Smallest Element:
:$B = \bot_P$
== Sources ==
* 
* 
Category:Auxiliary Relations
Bottom in Compact Closure

== Theorem ==
Let $L = \left({S, \preceq}\right)$ be a bounded below ordered set.
Let $x \in S$.
Then $\bot \in x^{\mathrm{compact} }$
where $\bot$ denotes the smallest element in $L$,
:$ x^{\mathrm{compact} }$ denotes the compact closure of $x$.
== Proof ==
By Bottom is Compact:
:$\bot$ is a compact element.
By definition of the smallest element:
:$\bot \preceq x$
Thus by definition of compact closure:
:$\bot \in x^{\mathrm{compact} }$
Category:Way Below Relation
Bottom in Compact Subset

== Theorem ==
Let $L = \left({S, \vee, \preceq}\right)$ be a bounded below join semilattice.
Then $\bot \in K\left({L}\right)$
where $\bot$ denotes the smallest element of $L$,
:$K\left({L}\right)$ denotes the compact subset of $L$.
== Proof ==
By Bottom is Way Below Any Element:
:$\bot \ll \bot$
where $\ll$ is the way below relation.
By definition:
:$\bot$ is compact.
Thus by definition of compact subset:
:$\bot \in K\left({L}\right)$
== Sources ==
* 
* 
Category:Way Below Relation
Bottom in Ideal

== Theorem ==
Let $\struct {S, \preceq}$ be a bounded below ordered set.
Let $I$ be a ideal in $S$.
Then $\bot \in I$
where $\bot$ denotes the smallest element of $S$.
== Proof ==
By definition of ideal in ordered set:
:$I$ is non-empty and lower.
By definition of non-empty set:
:$\exists x: x \in I$
By definition of smallest element:
:$\bot \preceq x$
Thus by definition of lower section:
:$\bot \in I$
== Sources ==
* 
Category:Order Theory
Bottom in Ordered Set of Topology

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $P = \struct {\tau, \subseteq}$ be an inclusion ordered set of $\tau$.
Then $P$ is bounded below and $\bot_P = \O$
== Proof ==
By Empty Set is Element of Topology:
:$\O \in \tau$
By Empty Set is Subset of All Sets:
:$\forall A \in \tau: \O \subseteq A$
Hence $P$ is bounded below.
Thus by definition of the smallest element:
:$\bot_P = \O$
== Sources ==
* 
Category:Topology
Category:Order Theory
Bottom is Compact

== Theorem ==
Let $L$ be a bounded below ordered set.
Then $\bot$ is a compact element
where $\bot$ is the smallest element in $L$.
== Proof ==
By Bottom is Way Below Any Element:
:$\bot \ll \bot$
where $\ll$ denotes the way below relation.
Hence $\bot$ is a compact element.
== Sources ==
* 
Category:Way Below Relation
Bottom is Way Below Any Element

== Theorem ==
Let $\left({S, \preceq}\right)$ be a bounded below ordered set.
Let $x \in S$.
Then
:$\bot \ll x$
where $\bot$ denotes the smallest element in $S$.
== Proof ==
Let $D$ be directed subset of $S$ such that
:$D$ admits a supremum and $x \preceq \sup D$
Because $D$ is non-empty, therefore
:$\exists a: a \in D$
By definition of smallest element:
:$\bot \preceq a$
Thus by definition of way below relation;
:$\bot \ll x$
== Sources ==
* 
* 
Category:Way Below Relation
Bottom not in Proper Filter

== Theorem ==
Let $L = \struct {S, \preceq}$ be a bounded below preordered set.
Let $F$ be a filter on $L$.
Then $F$ is proper filter $\bot \notin F$
where $\bot$ denotes the smallest element of $S$.
== Proof ==
=== Sufficient Condition ===
Suppose:
:$F$ is proper.
By definition of proper subset:
:$F \subseteq S$ and $F \ne S$
By definitions of set equality and subset:
:$\exists x: x \in S \land x \notin F$
By definition of smallest element:
:$\bot \preceq x$
Thus by definition of upper section:
:$\bot \notin F$
=== Necessary Condition ===
Suppose $\bot \notin F$
By definition of set equality:
:$F \ne S$
Hence $F$ is proper filter.
== Sources ==
* 
Category:Preorder Theory
Bottom of Lattice is Unique

== Theorem ==
Let $\struct {S, \vee, \wedge, \preceq}$ be a lattice.
Then $S$ has at most one bottom.
== Proof ==
By definition, a bottom for $S$ is a smallest element.
The result follows from Smallest Element is Unique.
== Also see ==
* Top of Lattice is Unique
Category:Lattice Theory
Boubaker's Theorem/Proof of Uniqueness

== Theorem ==
Let $\struct {R, +, \circ}$ be a commutative ring.
Let $\struct {D, +, \circ}$ be an integral subdomain of $R$ whose zero is $0_D$ and whose unity is $1_D$.
Let $X \in R$ be transcendental over $D$.
Let $D \sqbrk X$ be the ring of polynomial forms in $X$ over $D$.
Consider the following properties:
where, for a given positive integer $n$, $p_n \in D \sqbrk X$ is a non-null polynomial such that $p_n$ has $N$ roots $\alpha_k$ in $F$.
Then the subsequence $\sequence {\map {B_{4 n} } x}$ of the Boubaker polynomials is the unique polynomial sequence of $D \sqbrk X$ which verifies simultaneously the four properties $(1) - (4)$.
== Proof ==
Let:
:$\struct {R, +, \circ}$ be a commutative ring
:$\struct {D, +, \circ}$ be an integral subdomain of $R$ whose zero is $0_D$ and whose unity is $1_D$
:$X \in R$ be transcendental over $D$.
It has been demonstrated that the Boubaker Polynomials sub-sequence $\map {B_{4 n} } x$, defined in $D \sqbrk X$ as:
:$\ds \map {B_{4 n} } x = 4 \sum_{p \mathop = 0}^{2 n} \frac {n - p} {4 n - p} \binom {4 n - p} p \paren {-1}^p x^{2 \paren {2 n - p} }$
satisfies the properties:
with $\valueat {\alpha_k} {k \mathop = 1 \,.\,.\, N}$ roots of $B_{4 n}$.
Suppose there exists another $4 n$-indexed polynomial $\map {q_{4 n} } x$, with $N$ roots $\valueat {\beta_k} {k \mathop = 1 \,.\,.\, N}$ in $F$ and which also satisfies simultaneously properties $(1)$ to $(4)$.
Let:
:$\ds \map {B_{4 n} } x = \sum_{p \mathop = 0}^{2 n} a_{4 n, p} x^{2 \paren {2 n - p} }$
and:
:$\ds \map {q_{4 n} } x = \sum_{p \mathop = 0}^{2 n} b_{4 n, p} x^{2 \paren {2 n - p} }$
and:
:$\d_{4 n, p} = a_{4 n, p} - b_{4 n, p}$ for $p = 0 \,.\,.\, 2 n$
then, simultaneous expressions of conditions $(1)$ and $(3)$ give:
:$\quad \ds \sum_{k \mathop = 1}^N \d_{4 n, 2 n} = 0$
:$\quad \ds \sum_{k \mathop = 1}^N \d_{4 n, 2 n - 2} = 0$
It has also been demonstrated that $B_{4 n}$ has exactly $4 n - 2$ real roots inside the domain $\closedint {-2} 2$.
So application of conditions $(3)$ and $(4)$ give $4n-2$ linear equation with variables $\bigvalueat {\d_{4 n, p} } {p \mathop = 0 \,.\,.\, 2 n - 3}$.
Finally, since $B_{4 n}$ contains $2 n$ monomial terms (see definition), we obtain a Cramer system in variables $\bigvalueat {\d_{4 n, p} } {p \mathop = 0 \,.\,.\, 2 n}$, with evident solution:
:$\bigvalueat {\d_{4 n, p} } {p \mathop = 0 \,.\,.\, 2 n} = 0$
and consequently:
:$\bigvalueat {a_{4 n, p} } {p \mathop = 0 \,.\,.\, 2 n} = \bigvalueat {b_{4 n, p} } {p \mathop = 0 \,.\,.\, 2 n}$
which means:
:$\map {q_{4 n} } x = \map {B_{4 n} } x$
Category:Boubaker Polynomials