nyuuzyou/wikis huggingface dataset:
Strictly Positive Rational Numbers under Multiplication form Countably Infinite Abelian Group

== Theorem ==
Let $\Q_{> 0}$ be the set of strictly positive rational numbers, i.e. $\Q_{> 0} = \set {x \in \Q: x > 0}$.
The structure $\struct {\Q_{> 0}, \times}$ is a countably infinite abelian group.
== Proof ==
From Strictly Positive Rational Numbers under Multiplication form Subgroup of Non-Zero Rational Numbers we have that $\struct {\Q_{> 0}, \times}$ is a subgroup of $\struct {\Q_{\ne 0}, \times}$, where $\Q_{\ne 0}$ is the set of rational numbers without zero: $\Q_{\ne 0} = \Q \setminus \set 0$.
From Subgroup of Abelian Group is Abelian it follows that $\struct {\Q_{> 0}, \times}$ is an abelian group.
From Positive Rational Numbers are Countably Infinite, it follows that $\struct {\Q_{> 0}, \times}$ is a countably infinite group.
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 3$: Examples of Infinite Groups: $\text{(i)}$
* : $\S 1.3$: Example $2$
* : $\S 1$: Some examples of groups: Example $1.05$
* : Chapter $2$: The Definition of Group Structure: $\S 29 \alpha \ (3)$
* : $\S 1.5$
Category:Rational Multiplication
Category:Examples of Abelian Groups
Category:Examples of Infinite Groups
Strictly Positive Rational Numbers under Multiplication form Subgroup of Non-Zero Rational Numbers

== Theorem ==
Let $\Q_{> 0}$ be the set of strictly positive rational numbers, that is $\Q_{> 0} = \set { x \in \Q: x > 0}$.
The structure $\struct {\Q_{> 0}, \times}$ is a subgroup of $\struct {\Q_{\ne 0}, \times}$, where $\Q_{\ne 0}$ is the set of rational numbers without zero: $\Q_{\ne 0} = \Q \setminus \set 0$.
== Proof ==
From Non-Zero Rational Numbers under Multiplication form Infinite Abelian Group we have that $\struct {\Q_{\ne 0}, \times}$ is a group.
We know that $\Q_{> 0} \ne \O$, as (for example) $1 \in \Q_{> 0}$.
Let $a, b \in \Q_{> 0}$.
Then:
:$a b \in \Q_{\ne 0}$ and $ab > 0$
Hence:
:$a b \in \Q_{> 0}$
Let $a \in \Q_{> 0}$.
Then:
:$a^{-1} = \dfrac 1 a \in \Q_{> 0}$
So, by the Two-Step Subgroup Test, $\struct {\Q_{> 0}, \times}$ is a subgroup of $\struct {\Q_{\ne 0}, \times}$.
Category:Rational Multiplication
Category:Subgroups
Strictly Positive Real Numbers are Closed under Division

== Theorem ==
The set $\R_{>0}$ of strictly positive real numbers is closed under division:
:$\forall a, b \in \R_{>0}: a \div b \in \R_{>0}$
== Proof ==
From the definition of division:
:$a \div b := a \times \paren {\dfrac 1 b}$
where $\dfrac 1 b$ is the inverse for real number multiplication.
From Strictly Positive Real Numbers under Multiplication form Uncountable Abelian Group, the algebraic structure $\struct {\R_{>0}, \times}$ forms a group.
Thus it follows that:
:$\forall a, b \in \R_{>0}: a \times \paren {\dfrac 1 b} \in \R$
Therefore real number division is closed in $\R_{>0}$.
== Also see ==
* Strictly Positive Real Numbers are Closed under Multiplication
== Sources ==
* : Chapter $1$: Integral Domains: $\S 2$. Operations: Example $1$
* : Chapter $1$: Axioms for the Real Numbers: $2$. Fields: Example $1$
Category:Real Division
Category:Algebraic Closure
Strictly Positive Real Numbers are Closed under Multiplication

== Theorem ==
The set $\R_{>0}$ of strictly positive real numbers is closed under multiplication:
:$\forall a, b \in \R_{> 0}: a \times b \in \R_{> 0}$
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Strictly Positive Real Numbers are Closed under Division
== Sources ==
* : $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Exercise $2 \ \text{(b)}$
Category:Real Multiplication
Category:Strictly Positive Real Numbers are Closed under Multiplication
Strictly Positive Real Numbers are Closed under Multiplication/Proof 1

== Theorem ==
== Proof ==
Let $a, b \in \R_{> 0}$
We have that the Real Numbers form Ordered Integral Domain.
It follows from Positive Elements of Ordered Ring that:
:$a \times b \in \R_{> 0}$.
Category:Strictly Positive Real Numbers are Closed under Multiplication
Strictly Positive Real Numbers are Closed under Multiplication/Proof 2

== Theorem ==
== Proof ==
Let $b > 0$.
From :
:$a > c \implies a \times b > c \times b$
Thus setting $c = 0$:
:$a > 0 \implies a \times b > 0 \times b$
But from Real Zero is Zero Element:
:$0 \times b = 0$
Hence the result:
:$a, b > 0 \implies a \times b > 0$
Category:Strictly Positive Real Numbers are Closed under Multiplication
Strictly Positive Real Numbers are not Closed under Subtraction

== Theorem ==
The set $\R_{>0}$ of strictly positive real numbers is not closed under subtraction.
== Proof ==
;Proof by Counterexample
Let $a = 1$ and $b = 2$.
Then:
:$a - b = -1$
but $-1$ is not a (strictly) positive real number.
== Sources ==
* : Chapter $1$: Axioms for the Real Numbers: $2$. Fields: Example $1$
Category:Real Subtraction
Category:Algebraic Closure
Strictly Positive Real Numbers under Multiplication form Subgroup of Non-Zero Real Numbers

== Theorem ==
Let $\R_{>0}$ be the set of strictly positive real numbers, that is:
:$\R_{>0} = \set {x \in \R: x > 0}$
The structure $\struct {\R_{>0}, \times}$ forms a subgroup of $\struct {\R_{\ne 0}, \times}$, where $\R_{\ne 0}$ is the set of real numbers without zero, that is:
:$\R_{\ne 0} = \R \setminus \set 0$
== Proof ==
From Non-Zero Real Numbers under Multiplication form Abelian Group we have that $\struct {\R_{\ne 0}, \times}$ is a group.
We know that $\R_{>0} \ne \O$, as (for example) $1 \in \R_{>0}$.
Now, verify that the conditions for Two-Step Subgroup Test are satisfied:
=== Closure under $\times$ ===
Let $a, b \in \R_{>0}$.
We take on board the fact that the Real Numbers form Ordered Integral Domain.
Then:
:$a b \in \R_{\ne 0}$
From Positive Elements of Ordered Ring :
:$a \times b > 0$
so $a b \in \R_{>0}$.
=== Closure under Inverse ===
Let $a \in \R_{>0}$. 
Then $a^{-1} = \dfrac 1 a \in \R_{>0}$.
Hence, by the Two-Step Subgroup Test, $\struct {\R_{>0}, \times}$ is a subgroup of $\struct {\R_{\ne 0}, \times}$.
== Sources ==
* : $\S 5.2$. Subgroups: Example $91$
* : Chapter $\text{II}$: Groups: Subgroups
Category:Real Multiplication
Category:Subgroups
Strictly Positive Real Numbers under Multiplication form Uncountable Abelian Group

== Theorem ==
Let $\R_{>0}$ be the set of strictly positive real numbers:
:$\R_{>0} = \set {x \in \R: x > 0}$
The structure $\struct {\R_{>0}, \times}$ is an uncountable abelian group.
== Proof ==
From Strictly Positive Real Numbers under Multiplication form Subgroup of Non-Zero Real Numbers we have that $\struct {\R_{>0}, \times}$ is a subgroup of $\struct {\R_{\ne 0}, \times}$, where $\R_{\ne 0}$ is the set of real numbers without zero:
:$\R_{\ne 0} = \R \setminus \set 0$
From Subgroup of Abelian Group is Abelian it also follows that $\struct {\R_{>0}, \times}$ is an abelian group.
Its infinite nature follows from the nature of real numbers.
== Sources ==
* : $\S 1$: Some examples of groups: Example $1.06$
Category:Real Multiplication
Category:Examples of Abelian Groups
Category:Examples of Infinite Groups
Strictly Precede and Step Condition and not Precede implies Joins are equal

== Theorem ==
Let $\struct {S, \vee, \preceq}$ be a join semilattice.
Let $p, q, u \in S$ be such that:
:$p \prec q$ and $\paren {\forall s \in S: p \prec s \implies q \preceq s}$ and $u \npreceq p$
Then:
:$p \vee u = q \vee u$
== Proof ==
From the definition of join, it is required to prove that:
:$\forall s \in S: p \preceq s \land u \preceq s \implies q \vee u \preceq s$
Let $s \in S$ be such that:
:$p \preceq s$ and $u \preceq s$
We have:
:$p \ne s$
By definition of strictly precede:
:$p \prec s$
By assumption:
:$q \preceq s$
Thus by definition of the join $q \vee u$:
:$q \vee u \preceq s$
and hence, as desired:
:$q \vee u = p \vee u$
== Sources ==
* 
Category:Join and Meet Semilattices
Strictly Precedes is Strict Ordering

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $\prec$ be the relation on $S$ defined as:
:$a \prec b \iff \paren {a \ne b} \land \paren {a \preceq b}$
That is, $a \prec b$ $a$ strictly precedes $b$.
Then:
:$a \preceq b \iff \paren {a = b} \lor \paren {a \prec b}$
and $\prec$ is a strict ordering on $S$.
== Proof ==
We are given that $\struct {S, \preceq}$ is an ordered set.
=== Antireflexivity ===
Follows immediately:
:$\forall a \in S: a = a \implies a \nprec a$ 
Hence $\prec$ is antireflexive by definition.
=== Transitivity ===
Let $a, b, c \in S$ such that $a \preceq b, b \preceq c$.
$a \preceq b \land b \preceq c \implies a \preceq c$ from transitivity of $\preceq$.
Now suppose $a \preceq b, b \preceq c, a = c$.
Then:
:$a \preceq b$
but because of the antisymmetry of $\preceq$:
:$b \npreceq c$
So a condition for transitivity to be violated will not arise in this circumstance.
If either $a = b$ or $b = c$, then $a \nprec b$ and $b \nprec c$ by antireflexivity of $\prec$.
So $\prec$ is shown to be transitive.
=== Asymmetry ===
A direct application of Antireflexive and Transitive Relation is Asymmetric.
Hence $\prec$ is shown to be asymmetric.
Let $a \preceq b$.
Either $a = b$ or $a \ne b$ by the Law of Excluded Middle.
If $a = b$ the proof is complete.
Suppose $a \ne b$. Then it follows that $a \prec b$.
Thus $a \preceq b \implies \paren {a = b} \lor \paren {a \prec b}$.
Now let $\paren {a = b} \lor \paren {a \prec b}$.
If $a = b$ then $a \preceq b$ by the reflexivity of $\preceq$.
If $a \prec b$ then $a \preceq b$ by definition.
== Sources ==
* : $\S \text I.1$: Lemma $1$
Category:Strict Orderings
Strictly Stationary Stochastic Process/Examples

== Examples of Strictly Stationary Stochastic Processes ==
Let $S$ be a strictly stationary stochastic process giving rise to a time series $T$.
Then the following conditions hold:
=== Joint Probability Mass Function ===
=== Constant Mean Level ===
=== Autocovariance ===
=== Autocorrelation ===
Category:Stationary Stochastic Processes
Strictly Stationary Stochastic Process/Examples/Autocovariance

== Example of Strictly Stationary Stochastic Process ==
Let $S$ be a strictly stationary stochastic process giving rise to a time series $T$.
It is necessary that:
:The autocovariance between every two observations $z_t, z_{t + k}$ separated by a given lag $k$ 
is the same as:
:the autocovariance between every other two observations $z_{t + m}, z_{t + m + k}$separated by a given lag $k$
== Sources ==
* :
::Part $\text {I}$: Stochastic Models and their Forecasting:
:::$2$: Autocorrelation Function and Spectrum of Stationary Processes:
::::$2.1$ Autocorrelation Properties of Stationary Models:
:::::$2.1.2$ Stationary Stochastic Processes: Autocovariance and autocorrelation coefficients
Category:Autocovariance
Category:Stationary Stochastic Processes
Strictly Stationary Stochastic Process/Examples/Constant Mean Level

== Example of Strictly Stationary Stochastic Process ==
Let $S$ be a strictly stationary stochastic process giving rise to a time series $T$.
Consider the expectation of $S$:
:$\mu = \expect {z_t} = \ds \int_{-\infty}^\infty z \map p z \rd z$
where $\map p z$ is the (constant) probability mass function of $S$.
It is necessary that $T$ has a constant mean level, so that:
:The sample mean over a set of $N$ successive values $\set {z_1, z_2, \dotsb, z_N}$
is the same as:
:the sample mean over any other set of $N$ successive values $\set {z_{1 + k}, z_{2 + k}, \dotsb, z_{N + k} }$
for an arbitrary lag $k$.
== Sources ==
* :
::Part $\text {I}$: Stochastic Models and their Forecasting:
:::$2$: Autocorrelation Function and Spectrum of Stationary Processes:
::::$2.1$ Autocorrelation Properties of Stationary Models:
:::::$2.1.2$ Stationary Stochastic Processes: Mean and variance of a stationary process: $(2.1.1)$
Category:Stationary Stochastic Processes
Strictly Stationary Stochastic Process/Examples/Joint Probability Mass Function

== Example of Strictly Stationary Stochastic Process ==
Let $S$ be a strictly stationary stochastic process giving rise to a time series $T$.
It is necessary that:
:The joint probability mass function of any set of $m$ successive values $\set {z_1, z_2, \dotsb, z_m}$
is the same as:
:the joint probability mass function of any other set of $m$ successive values $\set {z_{1 + k}, z_{2 + k}, \dotsb, z_{m + k} }$
for an arbitrary lag $k$.
== Sources ==
* :
::Part $\text {I}$: Stochastic Models and their Forecasting:
:::$2$: Autocorrelation Function and Spectrum of Stationary Processes:
::::$2.1$ Autocorrelation Properties of Stationary Models:
:::::$2.1.2$ Stationary Stochastic Processes
Category:Stationary Stochastic Processes
Strictly Succeed is Dual to Strictly Precede

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $a, b \in S$.
The following are dual statements:
:$a$ strictly succeeds $b$
:$a$ strictly precedes $b$
== Proof ==
By definition, $a$ strictly succeeds $b$ :
:$b \preceq a$ and $b \ne a$
The dual of this statement is:
:$a \succeq b$ and $b \ne a$
by Dual Pairs (Order Theory).
By definition, this means $a$ strictly precedes $b$.
The converse follows from Dual of Dual Statement (Order Theory).
== Also see ==
* Duality Principle (Order Theory)
* Succeed is Dual to Precede
Category:Dual Pairs (Order Theory)
Strictly Well-Founded Relation determines Strictly Minimal Elements

== Theorem ==
Let $A$ be a class.
Let $\RR$ be a strictly well-founded relation on $A$.
Let $B$ be a nonempty class such that $B \subseteq A$.
Then $B$ has a strictly minimal element under $\RR$.
== Proof ==
First a lemma:
=== Lemma ===
Let $\RR' = \paren {B \times B} \cap \RR$.
By Restriction of Strictly Well-Founded Relation is Strictly Well-Founded, $\RR'$ is a strictly well-founded relation.
By the lemma:
:$B$ has a strictly minimal element $m$ under $\RR'$.
By Minimal WRT Restriction, $m$ is a strictly minimal element under $\RR$ in $B$.
== Also see ==
* Well-Founded Proper Relational Structure Determines Minimal Elements‎
* Proper Well-Ordering Determines Smallest Elements
weaker results that do not require the Axiom of Foundation.
== Sources ==
* : $\S 9.21$
Category:Class Theory
Strictly Well-Founded Relation determines Strictly Minimal Elements/Lemma

== Theorem ==
Let $A$ be a non-empty class.
Let $\RR$ be a strictly well-founded relation on $A$.
Then $A$ has a strictly minimal element under $\RR$.
== Proof ==
The general strategy of the proof is as follows:
We will recursively define a certain subset, $a$, of $A$.
We will use the fact that $\RR$ is a strictly well-founded relation to choose a strictly minimal element $m$ of $a$.
Then we will prove that $m$ is in fact a strictly minimal element of $A$.
For each $x \in A$, let $\map {\RR^{-1} } x$ denote the preimage under $\RR$ of $x$ in $A$.
For each class $C$, let $\map R C$ denote the set of elements of $C$ of minimal rank, and let $\map R \O = \O$.
That is:
For a given class $C$, let $\alpha_C$ be the smallest ordinal such that:
:$C \cap \map V {\alpha_C} \ne \O$
where $V$ is the von Neumann hierarchy.
Then let $\map R C$ denote the set $C \cap \map V {\alpha_C}$.
Let $F$ be a function defined recursively:
:$\map F 0 = \map R A$
:$\ds \map F {n + 1} = \bigcup_{y \mathop \in \map F n} \map R {\map {\RR^{-1} } y}$
=== Lemma ===
$\map F n$ is a set for each $n \in \omega$.
=== Proof ===
Proceed by induction:
$\map R A$ is a set, so $\map F 0$ is a set.
Suppose that $\map R {\map F n}$ is a set.
We know that for each $y \in \map F n$, $\map R {\map {\RR^{-1} } y}$ is a set, so by the Axiom of Unions, $\map F {n + 1}$ is a set.
Let $\ds a = \bigcup_{n \mathop \in \omega} \map F n$.
By the Axiom of Unions, $a$ is a set.
Since $\map F n \subseteq A$ for each $n \in \omega$, $a \subseteq A$.
By Non-Empty Class has Element of Least Rank, $\map F 0 \ne \O$, so $a \ne \O$.
Since $\RR$ is strictly well-founded, $a$ has a strictly minimal element $m$ under $\RR$.
 that $m$ is not a strictly minimal element under $\RR$ of $A$.
Then, by Characterization of Minimal Element, 
:$\map {\RR^{-1} } m \ne \O$
By Non-Empty Class has Element of Least Rank, $\map {\RR^{-1} } m$ has an element $w$ of least rank.
Therefore, $a \mathop \cap \map {\RR^{-1} } m \ne \O$, contradicting the fact that $m$ is a strictly minimal element under $\RR$ in $a$.
Thus we conclude that $m$ is a strictly minimal element under $\RR$ of $A$.
== Also see ==
* Well-Founded Proper Relational Structure Determines Minimal Elements‎
* Proper Well-Ordering Determines Smallest Elements
These are weaker results that do not require the Axiom of Foundation.
== Sources ==
* : $\S 9.21$
Category:Class Theory
Strictly Well-Founded Relation has no Relational Loops

== Theorem ==
Let $\prec$ be a strictly well-founded relation on $A$ and let $x_1, x_2, \ldots, x_n \in A$.
Then:
:$\neg \paren {x_1 \prec x_2 \land x_3 \prec x_4 \cdots \land x_n \prec x_1}$
That is, there are no relational loops within $A$.
== Proof ==
Since $x_1, x_2, \ldots, x_n \in A$, there exists a $y$ such that $y = \set {x_1, x_2, \ldots, x_n}$.
Then $y$ is a non-empty subset of $A$.
So, by the definition of a strictly well-founded relation:
:$\exists w \in y: \forall z \in y: \neg w \prec z$
Now, suppose $x_1 \prec x_2 \land x_2 \prec x_3 \cdots \land x_n \prec x_1$.
But since the elements of $y$ are $x_1, x_2, \ldots, x_n$, then this contradicts the previous statement, since:
:$\forall w \in y: \exists z \in y: w \prec z$
Thus a founded relation has no relational loops.
== Also see ==
* Well-Founded Relation has no Relational Loops
* Definition:Strictly Well-Founded Relation
* Axiom:Axiom of Foundation
* Epsilon Relation is Strictly Well-Founded
== Sources ==
* : $6.23$
Category:Set Theory
Strictly Well-Founded Relation is Antireflexive

== Theorem ==
Let $\RR$ be a strictly well-founded relation on a set or class $A$.
Then $\RR$ is antireflexive.
=== Corollary ===
== Proof ==
Let $p \in A$.
Then $\set p \ne \O$ and $\set p \subseteq A$.
Thus, by the definition of strictly well-founded relation:
:$\exists x \in \set p: \forall y \in \set p: \neg \paren {y \mathrel \RR x}$
Since $x \in \set p$, it must be that $x = p$.
It follows that $p \not \mathrel \RR p$.
Since this holds for all $p \in A$, $\RR$ is antireflexive.
Category:Well-Founded Relations
Category:Antireflexive Relations
Strictly Well-Founded Relation is Antireflexive/Corollary

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Suppose that $S$ is non-empty.
Then $\preceq$ is not a strictly well-founded relation.
== Proof ==
Since $S$ is non-empty, it has an element $x$.
By the definition of ordering, $\preceq$ is a reflexive relation.
Thus $x \preceq x$.
By Strictly Well-Founded Relation is Antireflexive, $\preceq$ is not a strictly well-founded relation.
Category:Well-Founded Relations
Strictly Well-Founded Relation is Asymmetric

== Theorem ==
Let $\struct {S, \RR}$ be a relational structure, where $S$ is a set or a proper class.
Let $\RR$ be a strictly well-founded relation.
Then $\RR$ is asymmetric.
== Proof ==
Let $p, q \in S$ and suppose that $p \mathrel \RR q$.
Then $\set {p, q} \ne \O$ and $\set {p, q} \subseteq S$.
By the definition of strictly well-founded relation, $\set {p, q}$ has a strictly minimal element under $\RR$.
Since $p \mathrel \RR q$, $q$ is not an $\RR$-minimal element of $\set {p, q}$.
Thus $p$ is a strictly minimal element under $\RR$ of $\set {p, q}$.
Thus $q \not \mathrel \RR p$.
Since for all $p, q \in S$, $p \mathrel \RR q \implies q \not \mathrel \RR p$, $\RR$ is asymmetric.
Category:Well-Founded Relations
Category:Asymmetric Relations
Strictly Well-Founded Relation is Well-Founded

== Theorem ==
Let $\struct {S, \RR}$ be a relational structure.
Let $\RR$ be a strictly well-founded relation on $S$.
Then $\RR$ is a well-founded relation on $S$.
== Proof ==
We have that $\RR$ is a strictly well-founded relation on $S$.
By definition:
:$\forall T: \paren {T \subseteq S \land T \ne \O} \implies \exists y \in T: \forall z \in T: \neg \paren {z \mathrel \RR y}$
It immediately follows that:
:$\forall T: \paren {T \subseteq S \land T \ne \O} \implies \exists y \in T: \forall z \in \paren {T \setminus \set y}: \neg \paren {z \mathrel \RR y}$
which is the definition of a well-founded relation on $S$.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations
Category:Well-Founded Relations
String is Substring of Itself

== Theorem ==
Let $S$ be a string.
Then $S$ is a substring of itself.
== Proof ==
By definition, a string $T$ is a substring of $S$ in $\AA$ :
:$S = S_1 T S_2$
where:
:$S_1$ and $S_2$ are strings in $\AA$ (possibly null)
:$S_1 T S_2$ is the concatenation of $S_1$, $T$ and $S_2$.
Let $S_1$ and $S_2$ both be the null string.
Then it follows that:
:$S = T$
Hence the result.
Category:Substrings
Strong Separation Theorem

== Theorem ==
Let $C \subset \R^\ell$ be closed and convex.
Let $D = \set {\mathbf v} \subset C^c$.
Then $C$ and $D$ can be strongly separated.
== Proof ==
=== Lemma 1 ===
For any $\mathbf y \ne 0$ and any $\mathbf z \in \map {H_{\mathbf y}^<} r$, $r = \mathbf y \mathbf y$, the line segment joining $\mathbf y$ and $\mathbf z$ contains points with lengths strictly less than $\mathbf y$.
=== Proof of Lemma 1 ===
Since $\mathbf z \in \map {H_\mathbf y} r^<$:
:$\mathbf z = \mathbf x - s \mathbf y$
for some $\mathbf x \perp \mathbf y$ and $s > 0$.
Let:
:$\map f \gamma = \norm {\gamma \mathbf y + \paren {1 - \gamma} \mathbf z}^2$
for $\gamma \in \closedint 0 1$.
We show that:
:$\map f \gamma < \norm {\mathbf y}^2$
as $\gamma \to 1$.
Expand $\map f \gamma$:
Since $\map {f'} 1 = 2 \paren {1 + s} > 0$:
:$\map f \gamma < \map f 1$
as $\gamma \to 1$.
Since $\map f 1 = \mathbf y \mathbf y = \norm {\mathbf y}^2$:
:$\map f \gamma < \norm {\mathbf y}^2$
as $\gamma \to 1$.
=== Proof of Theorem ===
Translating $C$ and $D = \set {\mathbf v}$ by $-\mathbf v$:
We show that the theorem holds for $\mathbf v = 0$.
Pick $n > 0$ such that:
:$K = \map \cl {\map {B_n} {\mathbf v} } \cap C \ne \O$
Since $C$ is closed, $K$ is closed.
Since $\map {B_n} {\mathbf v}$ is bounded, $K$ is bounded.
Since $K$ is closed and bounded, $K$ is compact.
Let $\map f {\mathbf x} = \norm {\mathbf x}$.
Since $\norm {\mathbf x}^2$ and $\sqrt x: \R_+ \to \R_+$ are both continuous, $\map f {\mathbf x}$ is continuous.
Since $K$ is compact and $\map f {\mathbf x}$ is continuous, $\map f {\mathbf x}$ achieves its minimum at $\mathbf y \in C$ by the Weierstrass Theorem.
Let $\map L {\mathbf x} = \mathbf y \mathbf x$ and $r = \mathbf y \mathbf y$.
We show that:
:$C \subset \map {L^{-1} } {\hointr r \to}$
Suppose $C$ is not a subset of $\map {L^{-1} } {\hointr r \to}$.
There exists $\mathbf z \in C \cap \map {L^{-1} } {\closedint {-\infty} r}$
So:
:$\mathbf z \in \map {H_\mathbf y} r$
As $\gamma \to 1$, there exists a $0 < \gamma < 1$ such that:
:$\norm {\mathbf y \, \gamma \, \mathbf z} < \norm {\mathbf y}$
by Lemma 1.
Since $C$ is convex, $\mathbf y \in C$ and $\mathbf z \in C$, convex combination $\mathbf y \, \gamma \, \mathbf z \in C$.
Since $\mathbf y $ is a minimmizer of $\norm {\mathbf x}$:
:$\norm {\mathbf y} \le \norm {\mathbf y \, \gamma \, \mathbf z}$
which is a contradiction.
For strong separation, let $\epsilon < r / 2$.
We have:
:$C \subset \map {L^{-1} } {\hointr r \to} \subset \map {L^{-1} } {\hointr {r/2 + \epsilon} \infty}$
Since:
:$\map L {\mathbf 0} = \mathbf y \mathbf 0 = 0$
and:
:$r/2 - \epsilon > 0$
it follows that:
:$\mathbf v = \mathbf 0 \subset \map {L^{-1} } {\hointl {-\infty} {r/2 - \epsilon} }$
== Sources ==
* : $\S 5.4.a$ Theorem $5.4.1$
Category:Convex Analysis
Category:Named Theorems
Strong Twin Prime Conjecture

== Conjecture ==
It is conjectured that the number of twin primes less than or equal to $N \in \N$ is asymptotically equal to:
:$2 C_2 \ds \int_2^N \dfrac {\d x} {\paren {\ln x}^2} = \dfrac {2 C_2 N} {\paren {\ln N}^2}$
The above is true only if the Twin Prime Conjecture holds.
== Historical Note ==
== Sources ==
* : $1,32032 36316 \ldots$
* 
Category:Unproven Hypotheses
Category:Twin Primes
Category:Analytic Number Theory