avewright/proofwiki-math huggingface dataset:
Theorem: Factors of Polynomial with Integer Coefficients have Integer Coefficients

Let $\Q \sqbrk X$ be the ring of polynomial forms over the field of rational numbers in the indeterminate $X$.
Let $\map h X \in \Q \sqbrk X$ have coefficients all of which are integers.
Let it be possible to express $\map h X$ as:
:$\map h X = \map f X \, \map g X$
where $\map f X, \map g X \in \Q \sqbrk X$.
Then it is also possible to express $\map h X$ as:
:$\map h X = \map {f'} X \, \map {g'} X$
where:
:$\map {f'} X, \map {g'} X \in \Q \sqbrk X$
:the coefficients of $\map {f'} X$ and $\map {g'} X$ are all integers
:$\map {f'} X = a \map f X$ and $\map {g'} X = b \map f X$, for $a, b \in \Q$.

Proof:
Let $\cont h$ denote the content of $\map h X$.
From Polynomial has Integer Coefficients iff Content is Integer:
:$\cont h \in \Z$
Let $\map h X = \map f X \, \map g X$ as suggested.
Then from Rational Polynomial is Content Times Primitive Polynomial:
{{begin-eqn}}
{{eqn | l = \map h X
      | r = \cont f \cont g \cdot \map {f'} X \, \map {g'} X
      | c = Rational Polynomial is Content Times Primitive Polynomial
}}
{{eqn | r = \cont h \cdot \map {f'} X \, \map {g'} X
      | c = Content of Rational Polynomial is Multiplicative
}}
{{end-eqn}}
From the above, $\map {f'} X$ and $\map {g'} X$ are primitive.
Hence by definition:
:$\cont {f'} = \cont {g'} = 1$
From Polynomial has Integer Coefficients iff Content is Integer, both $\map {f'} X$ and $\map {g'} X$ have coefficients which are all integers.
We also have by definition of content that $\cont f$ and $\cont g$ are rational numbers.
The result follows.
{{qed}}
Theorem: Polynomial which is Irreducible over Integers is Irreducible over Rationals

Let $\Z \sqbrk X$ be the ring of polynomial forms over the integers in the indeterminate $X$.
Let $\Q \sqbrk X$ be the ring of polynomial forms over the field of rational numbers in the indeterminate $X$.
Let $\map f X \in \Z \sqbrk X$ be irreducible in $\Z \sqbrk X$.
Then $\map f X$ is also irreducible in $\Q \sqbrk X$.

Proof:
{{AimForCont}} $\map f X = \map g X \, \map h X$ for some $\map g X, \map h X \in \Q \sqbrk X$.
By hypothesis:
:$\map f X \in \Z \sqbrk X$
and so by definition has coefficients all of which are integers.
But from Factors of Polynomial with Integer Coefficients have Integer Coefficients it follows that $\map f X$ can be expressed as:
:$\map f X = \map {g'} X \, \map {h'} X$
where both $\map {g'} X$ and $\map {h'} X$ are elements of $\Q \sqbrk X$ which have coefficients all of which are integers.
That is:
:$\map {g'} X \in \Z \sqbrk X$
and:
:$\map {h'} X \in \Z \sqbrk X$
This contradicts the statement that $\map f X$ is irreducible in $\Z \sqbrk X$.
Hence the result by Proof by Contradiction.
{{qed}}
Theorem: Dirichlet Function is Periodic

Let $D: \R \to \R$ be a Dirichlet function:
:$\forall x \in \R: \map D x = \begin{cases} c & : x \in \Q \\ d & : x \notin \Q \end{cases}$
Then $D$ is periodic.
Namely, every non-zero rational number is a periodic element of $D$.

Proof:
Let $x \in \R$.
Let $L \in \Q$.
If $x \in \Q$, then:
{{begin-eqn}}
{{eqn | l = \map D {x + L}
      | r = c
      | c = Rational Addition is Closed
}}
{{eqn | r = \map D x
}}
{{end-eqn}}
If $x \notin \Q$, then:
{{begin-eqn}}
{{eqn | l = \map D {x + L}
      | r = d
      | c = Rational Number plus Irrational Number is Irrational
}}
{{eqn | r = \map D x
}}
{{end-eqn}}
Combining the above two shows that:
:$\forall x \in \R: \map D x = \map D {x + L}$
Hence the result.
{{qed}}
Category:Periodic Functions
Category:Dirichlet Functions
lmk5v0tg77gtjenhxexcv37egu6ri8h
Theorem: Dirichlet Function has no Period

The Dirichlet functions are periodic by Dirichlet Function is Periodic.
However, they do not admit a period.
That is, there does not exist a smallest value $L \in \R_{> 0}$ such that:
:$\forall x \in \R: \map D x = \map D {x + L}$

Proof:
Let $D: \R \to \R$ be a Dirichlet function.
In proving that the Dirichlet Function is Periodic, it was shown that every non-zero rational number is a periodic element of $D$.
Therefore, the period of $D$ must be the smallest element of $\Q_{> 0}$.
But from Rational Numbers are not Well-Ordered under Conventional Ordering there cannot exist such an element.
Hence the result.
{{qed}}
Category:Periodic Functions
Category:Dirichlet Functions
8o3dnxc9ae9nm9m865w3ri9z18xfb93
Theorem: Existence of Nonconstant Periodic Function with no Period

There exists a real, non-constant function $f$ such that:
:$(1): \quad f$ is periodic.
:$(2): \quad f$ does '''not''' have a period.

Proof:
By Dirichlet Function is Periodic and Dirichlet Function has no Period, it is seen that the Dirichlet functions are such an example.
{{qed}}
Category:Periodic Functions
ouf4sp5a3l3br6w4gpr3o3zkq48s3a0
Theorem: Vector Augend plus Addend equals Augend implies Addend is Zero

Let $V$ be a vector space over a field $F$.
Let $\mathbf a, \mathbf b \in V$.
Let $\mathbf a + \mathbf b = \mathbf a$.
Then:
:$\mathbf b = \bszero$
where $\bszero$ is the zero vector of $V$.

Proof:
{{begin-eqn}}
{{eqn | l = \mathbf a + \mathbf b
      | r = \mathbf a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {-\mathbf a} + \paren {\mathbf a + \mathbf b}
      | r = \paren {-\mathbf a} + \mathbf a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {\paren {-\mathbf a} + \mathbf a} + \mathbf b
      | r = \paren {-\mathbf a} + \mathbf a
      | c = {{Abelian-group-axiom|1||underlying}}
}}
{{eqn | ll= \leadsto
      | l = \bszero + \mathbf b
      | r = \bszero
      | c = {{Abelian-group-axiom|3||underlying}}
}}
{{eqn | ll= \leadsto
      | l = \mathbf b
      | r = \bszero
      | c = {{Abelian-group-axiom|2||underlying}}
}}
{{end-eqn}}
{{qed}}
Theorem: Vector Space of All Mappings is Vector Space

Let $\struct {K, +, \circ}$ be a division ring.
Let $\struct {G, +_G, \circ}_K$ be a $K$-vector space.
Let $S$ be a set.
Let $\struct {G^S, +_G', \circ}_R$ be the vector space of all mappings from $S$ to $G$.
Then $\struct {G^S, +_G', \circ}_K$ is a $K$-vector space.

Proof:
Follows directly from Module of All Mappings is Module and the definition of vector space.
Theorem: Unitary Module of All Mappings is Unitary Module

Let $\struct {R, +_R, \times_R}$ be a ring with unity whose unity is $1_R$.
Let $\struct {G, +_G, \circ}_R$ be a unitary $R$-module.
Let $S$ be a set.
Let $\struct {G^S, +_G', \circ}_R$ be the module of all mappings from $S$ to $G$.
Then $\struct {G^S, +_G', \circ}_R$ is a unitary module.

Proof:
From Module of All Mappings is Module, we have that $\struct {G^S, +_G', \circ}_R$ is an $R$-module.
To show that $\struct {G^S, +_G', \circ}_R$ is a unitary $R$-module, we verify the following:
:$\forall f \in G^S: 1_R \circ f = f$
Let $\struct {G, +_G, \circ}_R$ be a unitary $R$-module.
Then:
{{begin-eqn}}
{{eqn | q = \forall x \in G
      | l = 1_R \circ x
      | r = x
      | c = {{left-module-axiom|4}} for $\struct {G, +_G, \circ}_R$
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | q = \forall f \in G^S, \forall x \in G
      | l = \map {\paren {1_R \circ f} } x
      | r = 1_R \circ \paren {\map f x}
      | c = {{Defof|Module of All Mappings}}
}}
{{eqn | r = \map f x
      | c = {{Defof|Unity of Ring}}
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Finite Direct Product of Unitary Modules is Unitary Module

Let $\struct {R, +_R, \times_R}$ be a ring.
Let $\struct {G_1, +_1, \circ_1}_R, \struct {G_2, +_2, \circ_2}_R, \ldots, \struct {G_n, +_n, \circ_n}_R$ be unitary $R$-modules.
Let:
:$\ds G = \prod_{k \mathop = 1}^n G_k$
be their direct product.
Then $G$ is a unitary module.

Proof:
This is a special case of Direct Product of Unitary Modules is Unitary Module.
Theorem: Finite Direct Product of Unitary Modules is Unitary Module

Let $\struct {R, +_R, \times_R}$ be a ring.
Let $\struct {G_1, +_1, \circ_1}_R, \struct {G_2, +_2, \circ_2}_R, \ldots, \struct {G_n, +_n, \circ_n}_R$ be unitary $R$-modules.
Let:
:$\ds G = \prod_{k \mathop = 1}^n G_k$
be their direct product.
Then $G$ is a unitary module.

Proof:
From Finite Direct Product of Modules is Module we have that $G$ is a module.
It remains to be shown that:
:$\forall x \in G: 1_R \circ x = x$
Let $x = \tuple {x_1, x_2, \ldots, x_n} \in G$.
Then:
{{begin-eqn}}
{{eqn | l = 1_R \circ x
      | r = 1_R \circ \tuple {x_1, x_2, \ldots, x_n}
      | c = 
}}
{{eqn | r = \tuple {1_R \circ x_1, 1_R \circ x_2, \ldots, 1_R \circ x_n}
      | c = 
}}
{{eqn | r = \tuple {x_1, x_2, \ldots, x_n}
      | c = 
}}
{{eqn | r = x
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Nonconstant Periodic Function with no Period is Discontinuous Everywhere

Let $f$ be a real periodic function that does not have a period.
Then $f$ is either constant or discontinuous everywhere.

Proof:
Let $f$ be a real periodic function that does not have a period.
Let $x \in \R$.
If there is no $y \in \R$ such that $\map f x \ne \map f y$, then the result is verified by Constant Function has no Period.
If $f$ is non-constant, then let $y$ be such a value.
Let $\LL_{f > 0}$ be the set of all positive periodic elements of $f$.
This set is non-empty by Absolute Value of Periodic Element is Peroidic.
It is seen that $<$ forms a right-total relation on $\LL_{f > 0}$, for if not then $f$ would have a period.
By the Axiom of Dependent Choice there must exist a strictly-decreasing sequence $\sequence {L_n}$ in $\LL_{f > 0}$.
Since this sequence is bounded below by zero, it follows via Monotone Convergence Theorem that this sequence converges.
Also, by Convergent Sequence is Cauchy Sequence the sequence is Cauchy.
Since $\sequence {L_n}$ is Cauchy, the sequence $\sequence {d_n}_{n \mathop \ge 1}$ formed by taking $d_n = L_n - L_{n - 1}$ is null.
From Combination of Periodic Elements it follows that $\sequence {d_n}_{n \mathop \ge 1}$ is contained in $\LL_{f > 0}$, for every $d_n$ is a periodic element of $f$.
Consider the sequence $\sequence {\paren {x - y} \bmod d_n}_{n \mathop \ge 1}$.
It is seen by Limit of Modulo Operation that this sequence is also null.
Let $\epsilon \in \R_{> 0}$ such that $\epsilon < \size {\map f x - \map f y}$.
For any $\delta \in \R_{> 0}$, there is a $n \in \N_{> 0}$ such that:
{{begin-eqn}}
{{eqn | l = \paren {x - y} \bmod d_n
      | r = \paren {x - y} - d_n \floor {\frac {x - y} {d_n} }
      | c = {{Defof|Modulo Operation}}
}}
{{eqn | r = x - \paren {y + d_n \floor {\frac {x - y} {d_n} } }
}}
{{eqn | o = <
      | r = \delta
}}
{{end-eqn}}
But:
{{begin-eqn}}
{{eqn | l = \size {\map f x - \map f {y + d_n \floor {\frac {x - y} {d_n} } } }
      | r = \size {\map f x - \map f y}
      | c = General Periodicity Property
}}
{{eqn | o = >
      | r = \epsilon
}}
{{end-eqn}}
And so $f$ is not continuous at $x$.
But our choice of $x$ was completely arbitrary.
Hence the result.
{{qed}}
{{ADC}}
Category:Periodic Functions
87p6o0hif1alx50l4ak9vyhfftmqnob
Theorem: Module on Cartesian Product is Module

Let $\struct {R, +_R, \times_R}$ be a ring.
Let $n \in \N_{>0}$.
Let $\struct {R^n, +, \times}_R$ be the '''$R$-module $R^n$'''.
Then $\struct {R^n, +, \times}_R$ is an $R$-module.

Proof:
This is a special case of Direct Product of Modules is Module.
{{qed}}
Theorem: Module on Cartesian Product is Module

Let $\struct {R, +_R, \times_R}$ be a ring.
Let $n \in \N_{>0}$.
Let $\struct {R^n, +, \times}_R$ be the '''$R$-module $R^n$'''.
Then $\struct {R^n, +, \times}_R$ is an $R$-module.

Proof:
This is a special case of the Module of All Mappings, where $S$ is the set $\closedint 1 n \subset \N_{>0}$.
{{qed}}
Theorem: Module on Cartesian Product is Module

Let $\struct {R, +_R, \times_R}$ be a ring.
Let $n \in \N_{>0}$.
Let $\struct {R^n, +, \times}_R$ be the '''$R$-module $R^n$'''.
Then $\struct {R^n, +, \times}_R$ is an $R$-module.

Proof:
This is a special case of a Finite Direct Product of Modules is Module where each of the $G_k$ is the $R$-module $R$.
{{qed}}
Theorem: Module on Cartesian Product of Ring with Unity is Unitary Module

Let $\struct {R, +_R, \times_R}$ be a ring with unity.
Let $n \in \N_{>0}$.
Let $\struct {R^n, +, \times}_R$ be the '''$R$-module $R^n$'''.
Then $\struct {R^n, +, \times}_R$ is a unitary $R$-module.

Proof:
This is a special case of Direct Product of Unitary Modules is Unitary Module.
{{qed}}
Theorem: Module on Cartesian Product of Ring with Unity is Unitary Module

Let $\struct {R, +_R, \times_R}$ be a ring with unity.
Let $n \in \N_{>0}$.
Let $\struct {R^n, +, \times}_R$ be the '''$R$-module $R^n$'''.
Then $\struct {R^n, +, \times}_R$ is a unitary $R$-module.

Proof:
This is a special case of the Unitary Module of All Mappings is Unitary Module, where $S$ is the set $\closedint 1 n \subset \N_{>0}$.
{{qed}}
Theorem: Module on Cartesian Product of Ring with Unity is Unitary Module

Let $\struct {R, +_R, \times_R}$ be a ring with unity.
Let $n \in \N_{>0}$.
Let $\struct {R^n, +, \times}_R$ be the '''$R$-module $R^n$'''.
Then $\struct {R^n, +, \times}_R$ is a unitary $R$-module.

Proof:
This is a special case of a Finite Direct Product of Unitary Modules is Unitary Module where each of the $G_k$ is the $R$-module $R$.
{{qed}}
Theorem: Vector Space on Cartesian Product is Vector Space

Let $\struct {K, +, \circ}$ be a division ring.
Let $n \in \N_{>0}$.
Let $\struct {K^n, +, \times}_K$ be the '''$K$-vector space $K^n$'''.
Then $\struct {K^n, +, \times}_K$ is a $K$-vector space.

Proof:
This is a special case of the Vector Space of All Mappings, where $S$ is the set $\closedint 1 n \subset \N^*$.
{{qed}}
Theorem: Vector Space on Cartesian Product is Vector Space

Let $\struct {K, +, \circ}$ be a division ring.
Let $n \in \N_{>0}$.
Let $\struct {K^n, +, \times}_K$ be the '''$K$-vector space $K^n$'''.
Then $\struct {K^n, +, \times}_K$ is a $K$-vector space.

Proof:
This is a special case of a direct product of vector spaces where each of the $G_k$ is the $K$-vector space $K$.
{{qed}}
Theorem: Ring of Polynomial Forms over Field is Vector Space

Let $\struct {F, +, \times}$ be a field whose unity is $1_F$.
Let $F \sqbrk X$ be the ring of polynomials over $F$.
Then $F \sqbrk X$ is a vector space over $F$.

Proof:
Let the operation $\times': F \to F \sqbrk X$ be defined as follows.
Let $x \in F$.
Let $\mathbf y \in F \sqbrk X$ be defined as:
:$\mathbf y = \ds \sum_{k \mathop = 0}^n y_k X^k$
where $n = \map \deg {\mathbf y}$ denotes the degree of $\mathbf y$
Thus:
:$x \times' \mathbf y := \ds x \sum_{k \mathop = 0}^n y_k X^k = \sum_{k \mathop = 0}^n \paren {x \times y_k} X^k$
We have that $\times': F \to F \sqbrk X$ is an instance of polynomial multiplication where the multiplier $x$ is a polynomial of degree $0$.
Hence, let the supposed vector space over $F$ in question be denoted in full as:
:$\mathbf V = \struct {F \sqbrk X, +', \times'}_F$
where:
:$+': F \sqbrk X \to F \sqbrk X$ denotes polynomial addition
:$\times': F \to F \sqbrk X$ denotes the operation as defined above. 
We already have that $F \sqbrk X$ is an integral domain.
Thus the underlying group axioms $\text G 0$ to $\text G 3$ and $\text C$ are fulfilled.
By definition of $\times'$, it is seen that the remaining vector space axioms are fulfilled as follows:
Let $\lambda, \mu \in F$.
Let $\mathbf x, \mathbf y \in F \sqbrk X$ such that $\map \deg {\mathbf x} = m$ and $\map \deg {\mathbf y} = n$.
Theorem: No Non-Trivial Norm on Rational Numbers is Complete

No non-trivial norm on the set of the rational numbers is complete.

Proof:
By P-adic Norm not Complete on Rational Numbers, no $p$-adic norm $\norm{\,\cdot\,}_p$ on the set of the rational numbers, for any prime $p$, is complete.
By Rational Number Space is not Complete Metric Space, the absolute value $\size{\,\cdot\,}$ on the set of the rational numbers is not complete.
By Norm is Complete Iff Equivalent Norm is Complete, no norm is complete if it is equivalent to either the absolute value $\size{\,\cdot\,}$ or the $p$-adic norm $\norm{\,\cdot\,}_p$ for some prime $p$.
By Ostrowski's Theorem, every non-trivial norm is equivalent to either the absolute value $\size{\,\cdot\,}$ or the $p$-adic norm $\norm{\,\cdot\,}_p$ for some prime $p$.
The result follows.
{{qed}}
Theorem: Norm is Complete Iff Equivalent Norm is Complete

Let $R$ be a division ring.
Let $\norm {\,\cdot\,}_1$ and $\norm {\,\cdot\,}_2$ be equivalent norms on $R$.
Then:
:$\struct {R,\norm {\,\cdot\,}_1}$ is complete {{iff}} $\struct {R,\norm {\,\cdot\,}_2}$ is complete.

Proof:
By Cauchy Sequence Equivalence, for all sequences $\sequence {x_n}$ in $R$:
:$\sequence {x_n}$ is a Cauchy sequence in $\norm{\,\cdot\,}_1$ {{iff}} $\sequence {x_n}$ is a Cauchy sequence in $\norm{\,\cdot\,}_2$.
By Convergent Equivalence, for all sequences $\sequence {x_n}$ in $R$:
:$\sequence {x_n}$ converges in $\norm{\,\cdot\,}_1$ {{iff}} $\sequence {x_n}$ converges in $\norm{\,\cdot\,}_2$.
Hence:
:every Cauchy sequence in $\norm{\,\cdot\,}_1$ converges in $\norm{\,\cdot\,}_1$ {{iff}} every Cauchy sequence in $\norm{\,\cdot\,}_2$ converges in $\norm{\,\cdot\,}_2$.
The result follows.
{{qed}}
Category:Normed Division Rings
Category:Complete Metric Spaces
ekkypuxvfwisc4mndifp6qmzk0hcwwq
Theorem: Ring of Polynomial Forms over Field is Vector Space/Corollary

Let $S \subseteq F \sqbrk X$ denote the subset of $F \sqbrk X$ defined as:
:$S = \set {\mathbf x \in F \sqbrk X: \map \deg {\mathbf x} < d}$
for some $d \in \Z_{>0}$.
Then $S$ is a vector space over $F$.

Proof:
From Ring of Polynomial Forms over Field is Vector Space we note that $\struct {F, +, \times}$ is a vector space over $F$.
The remaining question is that $S$ remains closed under polynomial addition and scalar multiplication.
Let $\mathbf x, \mathbf y \in S$ such that $\map \deg {\mathbf x} = m$ and $\map \deg {\mathbf y} = n$.
We have:
{{begin-eqn}}
{{eqn | l = \mathbf x + \mathbf y
      | r = \sum_{j \mathop = 0}^m x_j X^j + \sum_{k \mathop = 0}^n y_k X^k
      | c = 
}}
{{eqn | r = \sum_{k \mathop = 0}^{\max \set {m, n} } \paren {x_k + y_k} X^k
      | c = {{Defof|Polynomial Addition}}
}}
{{end-eqn}}
As both $m < d$ and $n < d$ by dint of them belonging to $S$, it follows that $\max \set {m, n} < d$.
Hence $\mathbf x + \mathbf y \in S$ and so $S$ is closed under polynomial addition.
Let the operation $\times': F \to F \sqbrk X$ be defined as follows.
Let $\lambda \in F$.
Let $\mathbf x \in F \sqbrk X$ be defined as:
:$\mathbf x = \ds \sum_{k \mathop = 0}^n x_k X^k$
where $n = \map \deg {\mathbf x}$ denotes the degree of $\mathbf x$.
Thus:
:$\lambda \times' \mathbf x := \ds \lambda \times' \sum_{k \mathop = 0}^n x_k X^k = \sum_{k \mathop = 0}^n \paren {\lambda \times x_k} X^k$
We have that $\times': F \to F \sqbrk X$ is an instance of polynomial multiplication where the multiplier $\lambda$ is a polynomial of degree $0$.
By definition of polynomial multiplication:
{{begin-eqn}}
{{eqn | l = \map \deg {\lambda \times' \mathbf x}
      | r = \map \deg \lambda + \map \deg {\mathbf x}
      | c = 
}}
{{eqn | r = 0 + \map \deg {\mathbf x}
      | c = 
}}
{{eqn | o = <
      | r = d
      | c = as $\mathbf x \in S$
}}
{{end-eqn}}
The result follows.
{{qed}}
Theorem: P-adic Norm of p-adic Number is Power of p

Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $x \in \Q_p: x \ne 0$.
Then:
:$\exists v \in \Z: \norm x_p = p^{-v}$
== Lemma ==
{{:P-adic Norm of p-adic Number is Power of p/Lemma}}{{qed|lemma}}

Proof:
From Rational Numbers are Dense Subfield of P-adic Numbers $\Q$ is dense in $\Q_p$.
By the definition of a dense subset then $\map \cl \Q = \Q_p$.
By Closure of Subset of Metric Space by Convergent Sequence then:
:there exists a sequence $\sequence {x_n} \subseteq \Q$ that converges to $x$.
That is:
:$\ds \lim_{n \mathop \to \infty} x_n = x$
From Modulus of Limit:
:$\ds \lim_{n \mathop \to \infty} \norm{x_n}_p = \norm x_p$
By Convergent Sequence in Normed Division Ring is Cauchy Sequence, $\sequence {x_n}$ is a Cauchy sequence in $\struct {\Q_p, \norm {\,\cdot\,}_p}$.
From Sequence is Cauchy in P-adic Norm iff Cauchy in P-adic Numbers, $\sequence {x_n}$ is a Cauchy sequence in $\struct {\Q, \norm {\,\cdot\,}_p}$.
From Lemma:
:$\exists v \in \Z: \norm x_p = \lim_{n \mathop \to \infty} \norm{x_n}_p = p^{-v}$
Theorem: P-adic Norm of p-adic Number is Power of p

Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $x \in \Q_p: x \ne 0$.
Then:
:$\exists v \in \Z: \norm x_p = p^{-v}$
== Lemma ==
{{:P-adic Norm of p-adic Number is Power of p/Lemma}}{{qed|lemma}}

Proof:
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
That is, $\Q_p$ is the quotient ring $\CC \, \big / \NN$ where:
:$\CC$ denotes the commutative ring of Cauchy sequences over $\struct {\Q, \norm {\,\cdot\,}_p}$
:$\NN$ denotes the set of null sequences in $\struct {\Q, \norm {\,\cdot\,}_p}$.
Then $x$ is a left coset in $\CC \, \big / \NN$.
Let $\sequence{x_n}$ be any Cauchy sequence in $x$.
From Lemma:
:$\exists v \in \Z: \ds \lim_{n \mathop \to \infty} \norm{x_n}_p = p^{-v}$
By definition of the $p$-adic norm on the $p$-adic numbers:
:$\norm x_p = \ds \lim_{n \mathop \to \infty} \norm{x_n}_p = p^{-v}$
Theorem: Fisher-Tippett-Gnedenko Theorem

In general, for iid $\set {X_i} _{i\mathop = 1}^n$ with cdf $\map F x$ we have $\mathbb P \sqbrk {\max_{i \mathop = 1}^n \set {X_i} > x} = \map {F^n} x$.
Let $x_* := \sup \set {x: \map F x < 1}$.
Then as $n \to \infty$:
:$\ds \max_{i \mathop = 1, \mathop \ldots, n} \set {X_i} \stackrel {prob} {\to} x_*$
We normalize it by constants $a_n > 0, b_n$ such that:
:$\dfrac {\ds \max_{i \mathop = 1, \mathop \ldots, n} \set {X_i} - b_n} {a_n} \stackrel {dis} {\to} X_*$
for some non-degenerate $X_*$ with distribution $\map G x$.
We have the following three possible limits
* Frechet: If $x > 0$, $\map G x = \map \exp {-x^{-a} }$ for some $a > 0$. If $x \le 0$, $\map G x = 0$.
* Gumbel: If $x \in \R$, $\map G x = \map \exp {-e^{-\frac {x - \mu} {\sigma} } }$ for $\mu\in \R, \sigma>0$. 
* Weibull: If $x \ge 0$, $\map G x = 1 - \map \exp {-\paren {\dfrac x \lambda}^k}$ for $\lambda, k > 0$. If $x < 0$ , $\map G x = 0$.
'''Fisher-Tippet-Gnedenko theorem'''
:$(1): \quad \map G x$ will be Frechet {{iff}} $\map F x < 1$ and $\ds \lim_{t \mathop \to +\infty} \dfrac {1 - \map F {t x} } {1 - \map F t} = x^{-\theta}$ for $x > 0$ and $\theta > 0$. Here we can set $b_n := 0$ and $a_n := F^{-1} \paren {1 - \dfrac 1 n}$.
:$(2): \quad \map G x$ will be Weibull {{iff}} $x_* : = \sup \set {x: \map F x < 1} < \infty$ and $\ds \lim_{t \mathop \to +\infty} \dfrac {1 - \map F {x_* + t x} } {1 - \map F {x_* - t} } = \paren {-x}^{-\theta}$ for $x < 0$ and $\theta > 0$. Here we can set $b_n := x_*$ and $a_n := x_* - \map {F^{-1} } {1 - \dfrac 1 n}$.
:$(3): \quad G$ will be Gumbel if the density $\map f x := \dfrac \d {\d x} \map F x > 0$ and it is differentiable in $\openint {x_1} {x_*}$ for some $x_1$, and $\ds \lim_{x \mathop \to x_*} \map {\dfrac \d {\d x} } {\dfrac {1 - \map F x} {\map f x} } = 0$. Here we can set $b_n := \map {F^{-1} } {1 - \dfrac 1 n}$ and $a_n := \dfrac 1 {n \map f {b_n} }$.

Proof:
Proof under construction
{{Namedfor|Ronald Aylmer Fisher|name2 = Leonard Henry Caleb Tippett|name3 = Boris Vladimirovich Gnedenko|cat = Fisher|cat2 = Tippett|cat3 = Gnedenko}}
Category:Probability Theory
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Theorem: Borell-TIS Inequality

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:
{{ProofWanted}}
{{Namedfor|Christer Borell|name2 = Boris Tsirelson|name3 = Ildar Ibragimov|name4 = and Vladimir Sudakov|cat=Borell|cat2 = Tsirelson|cat3 = Ibragimov|cat4 = Sudakov}}
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