nyuuzyou/wikis huggingface dataset:
Existence and Uniqueness of Dynkin System Generated by Collection of Subsets

== Theorem ==
Let $X$ be a set.
Let $\GG \subseteq \powerset X$ be a collection of subsets of $X$.
Then $\map \delta \GG$, the Dynkin system generated by $\GG$, exists and is unique.
== Proof ==
=== Existence ===
By Power Set is Dynkin System, there exists at least one Dynkin system containing $\GG$.
Next, let $\Bbb D$ be the collection of Dynkin systems containing $\GG$:
:$\Bbb D := \set {\DD': \GG \subseteq \DD', \text{$\DD'$ is a Dynkin system} }$
By Intersection of Dynkin Systems is Dynkin System, $\DD := \bigcap \Bbb D$ is a Dynkin system.
Also, by Set Intersection Preserves Subsets:
:$\GG \subseteq \DD$
Now let $\DD'$ be a Dynkin system containing $\GG$.
By construction of $\DD$, and Intersection is Subset: General Result:
:$\DD \subseteq \DD'$
=== Uniqueness ===
Suppose both $\DD_1$ and $\DD_2$ are Dynkin systems generated by $\GG$.
Then property $(2)$ for these Dynkin systems implies both $\DD_1 \subseteq \DD_2$ and $\DD_2 \subseteq \DD_1$.
By definition of set equality:
:$\DD_1 = \DD_2$
== Sources ==
* : $5.3$
Category:Dynkin Systems
Existence and Uniqueness of Generated Topology

== Theorem ==
Let $X$ be a set.
Let $\SS \subseteq \powerset X$ be a subset of the power set of $X$.
Then there exists a unique topology $\map \tau \SS$ on $X$ such that:
:$(1): \quad \SS \subseteq \map \tau \SS$.
:$(2): \quad$ For any topology $\TT$ on $X$, the implication $\SS \subseteq \TT \implies \map \tau \SS \subseteq \TT$ holds.
== Proof ==
=== Existence ===
Define:
:$\mathbb E = \leftset {\TT \subseteq \powerset X: \SS \subseteq \TT}$ and $\TT$ is a topology on $\rightset X$
Since $\powerset X$ is a topology on $X$, it follows that $\mathbb E$ is non-empty.
Hence, we can define:
:$\ds \map \tau \SS = \bigcap \mathbb E$
It follows that $\map \tau \SS$ is a topology on $X$.
By Intersection is Largest Subset: General Result, it follows that $\SS \subseteq \map \tau \SS$.
By Intersection is Subset: General Result, it follows that if $\SS \subseteq \TT$ and $\TT$ is a topology on $X$, then $\map \tau \SS \subseteq \TT$.
=== Uniqueness ===
Suppose that $\TT_1$ and $\TT_2$ are both topologies on $X$ satisfying conditions $(1)$ and $(2)$.
By condition $(1)$, we have $\SS \subseteq \TT_2$; hence, we can apply condition $(2)$ to conclude that:
:$\TT_1 \subseteq \TT_2$
Similarly:
:$\TT_2 \subseteq \TT_1$
By definition of set equality:
:$\TT_1 = \TT_2$
== Also see ==
* Definition:Topology Generated by Synthetic Sub-Basis
Category:Topology
Existence and Uniqueness of Geodesics

== Theorem ==
Let $M$ be a smooth manifold.
Let $TM$ be the tangent bundle of $M$.
Let $T_p M$ be the tangent space at $p \in M$.
Let $\nabla$ be a connection in $TM$.
Let $I \subseteq \R$ be an open real interval.
Let $\gamma : I \to M$ be a geodesic and $\gamma'$ its velocity.
Then:
:$\forall p \in M : \forall w \in T_p M : \forall t_0 \in \R : \exists I \subseteq \R : t_0 \in I : \exists \gamma : \paren {\map \gamma {t_0} = p} \land \paren {\map {\gamma'} {t_0} = w}$
Furthermore, any two geodesics agree on their common domain.
== Proof ==
== Sources ==
* : $\S 4$: Connections. Geodesics
Category:Differential Equations
Category:Geodesics
Existence and Uniqueness of Identification Topology

== Theorem ==
Let $T_1 = \struct {S_1, \tau_1}$ be a topological space.
Let $S_2$ be a set.
Let $f: S_1 \to S_2$ be a mapping.
Let $\tau_2$ be the identification topology on $S_2$ with respect to $f$ and $\struct {S_1, \tau_1}$.
Then the identification topology on $S_2$ with respect to $f$ and $\struct {S_1, \tau_1}$ always exists and is unique.
== Proof ==
Let $\tau_2$ be the identification topology on $S_2$ with respect to $f$ and $\struct {S_1, \tau_1}$.
By definition:
:$\tau_2 := \set {V \in \powerset {S_2}: f^{-1} \sqbrk V \in \tau_1} \subseteq \powerset {S_2}$
where $\powerset {S_2}$ is the power set of $S_2$.
Let $V \subseteq S_2$.
Then either:
:$f^{-1} \sqbrk V \in \tau_1 \implies V \in \tau_2$
or:
:$f^{-1} \sqbrk V \notin \tau_1 \implies V \notin \tau_2$
In particular:
:$f^{-1} \sqbrk {S_2} = S_1 \in \tau_1 \implies S_2 \in \tau_2$
Thus $\tau_2$ exists.
By the same coin, given any $V \subseteq S_2$, either $V \in \tau_2$ or $V \notin \tau_2$, and so the set:
:$\tau_2 := \set {V \in \powerset {S_2}: f^{-1} \sqbrk V \in \tau_1}$
consists of a uniquely defined set of elements.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.8$: Quotient spaces
Category:Identification Topology
Existence and Uniqueness of Image of Relation

== Theorem ==
Let $V$ be a basic universe.
Let $\RR \subseteq V \times V$ be a relation.
Then the image $\Img \RR$ of $\RR$ exists and is unique.
== Proof ==
By the Axiom of Specification we can create the subclass of $V$:
:$\Img \RR = \set {y \in V: \exists x \in V: \tuple {x, y} \in \RR}$
Hence $\Img \RR$ exists.
Suppose $\QQ \subseteq V$ such that $\QQ$ and $\Img \RR$ are both the image of $\RR$.
Then:
:$\QQ = \set {y \in V: \exists x \in V: \tuple {x, y} \in \RR}$
Thus:
:$x \in \QQ \implies x \in \Img \RR$
and:
:$x \in \Img \RR \implies x \in \QQ$
Hence by the Axiom of Extension:
:$\QQ = \Img \RR$
and uniqueness has been demonstrated.
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 8$ Relations
Category:Relation Theory
Existence and Uniqueness of Lebesgue Measure

== Theorem ==
Let $\lambda^n$ be the Lebesgue pre-measure on the half-open $n$-rectangles $\JJ_{ho}^n$.
Then Lebesgue measure, the extension of $\lambda^n$ to the Borel $\sigma$-algebra $\BB \left({\R^n}\right)$, exists and is unique.
== Proof ==
From Lebesgue Pre-Measure is Pre-Measure, $\lambda^n$ is a pre-measure on $\JJ_{ho}^n$.
By Half-Open Rectangles form Semiring of Sets, $\JJ_{ho}^n$ is a semiring of sets.
Also, from Characterization of Euclidean Borel Sigma-Algebra, $\map \BB {\R^n} = \map \sigma {\JJ_{ho}^n}$.
Now observe that the half-open $n$-rectangles $\horectr {-n} n$ form an increasing sequence of sets with limit $\R^n$.
Also, by definition of $\lambda^n$, have:
:$\map {\lambda^n} {\horectr {-n} n} = \ds \prod_{i \mathop = 1}^n n - \paren {-n} = \paren {2 n}^n < +\infty$
Hence, Carathéodory's Theorem and its corollary apply.
These yield existence and uniqueness of Lebesgue measure, the extension of $\lambda^n$ to $\map \BB {\R^n}$.
== Sources ==
* : $4.9$
* : $6.6$
Category:Lebesgue Measure
Existence and Uniqueness of Magma of Sets Generated by Collection of Subsets

== Theorem ==
Let $X$ be a set, and let $\Phi := \set {\phi_i: i \in I}$ be a collection of partial mappings with codomain $\powerset X$, the power set of $X$.
Let $\GG \subseteq \powerset X$ be a collection of subsets of $X$.
Then the magma of sets generated by $\GG$ exists and is unique.
== Proof ==
=== Uniqueness ===
Suppose that both $\SS$ and $\TT$ are magmas of sets for $\Phi$ generated by $\GG$.
Applying condition $(2)$ for these twice, we obtain:
:$\SS \subseteq \TT$
:$\TT \subseteq \SS$
By definition of set equality:
:$\SS = \TT$
=== Existence ===
Define $\SS$ by:
:$\SS := \bigcap \set {\TT: \GG \subseteq \TT}$
where $\TT$ ranges over the magmas of sets for $\Phi$ on $X$.
From Power Set is Magma of Sets, there intersection is at least one such $\TT$.
From Intersection of Magmas of Sets is Magma of Sets, $\SS$ is a magma of sets for $\Phi$.
By Intersection Preserves Subsets: General Result: Corollary, $\GG \subseteq \SS$.
By Intersection is Subset: General Result, $\SS \subseteq \TT$ for every other magma of sets $\TT$ containing $\GG$.
Thus $\SS$ is a magma of sets for $\Phi$ generated by $\GG$.
Category:Magmas of Sets
Existence and Uniqueness of Maximal Geodesic

== Theorem ==
Let $M$ be a smooth manifold.
Let $TM$ be the tangent bundle of $M$.
Let $T_p M$ be the tangent space at $p \in M$.
Let $\nabla$ be a connection in $TM$.
Let $I \subseteq \R$ be an open real interval.
Let $\gamma : I \to M$ be a geodesic and $\gamma'$ its velocity.
Then $\forall p \in M$ and $\forall v \in T_p M$ there is a unique maximal geodesic $\gamma$ with:
:$\map \gamma 0 = p$
:$\map {\gamma'} 0 = v$
defined on some $I$ with $0 \in I$.
== Proof ==
== Sources ==
* : $\S 4$: Connections. Geodesics
Category:Differential Equations
Category:Geodesics
Existence and Uniqueness of Monotone Class Generated by Collection of Subsets

== Theorem ==
Let $X$ be a set.
Let $\GG \subseteq \powerset X$ be a collection of subsets of $X$.
Then $\map {\mathfrak m} \GG$, the monotone class generated by $\GG$, exists and is unique.
== Proof ==
=== Existence ===
By Power Set is Monotone Class, there is at least one monotone class containing $\GG$.
Now let $\Bbb M$ be the collection of monotone classes containing $\GG$:
:$\Bbb M := \set {\mathfrak m': \GG \subseteq \mathfrak m', \mathfrak m' \text{ is a monotone class} }$
By Intersection of Monotone Classes is Monotone Class, $\mathfrak m := \bigcap \Bbb M$ is a monotone class.
Also, by Set Intersection Preserves Subsets:
:$\GG \subseteq \mathfrak m$
Now let $\mathfrak m'$ be a monotone class containing $\GG$.
Then by construction of $\mathfrak m$, and Intersection is Subset: General Result:
:$\mathfrak m \subseteq \mathfrak m'$
=== Uniqueness ===
Suppose $\mathfrak m_1$ and $\mathfrak m_2$ are both monotone classes generated by $\GG$.
Then property $(2)$ for these monotone classes implies both $\mathfrak m_1 \subseteq \mathfrak m_2$ and $\mathfrak m_2 \subseteq \mathfrak m_1$.
By definition of set equality:
:$\mathfrak m_1 = \mathfrak m_2$
== Sources ==
* : $\S 3$: Problem $11 \ \text{(i)}$
Category:Monotone Classes
Existence and Uniqueness of Outward-Pointing Normal

== Theorem ==
Let $\struct {M, g}$ be a smooth Riemannian manifold with boundary $\partial M$.
Then the normal bundle to $\partial M$ is a rank-$1$ smooth vector bundle over $\partial M$.
Furthermore, there is a unique smooth outward-pointing unit normal vector field along all of $\partial M$.
== Proof ==
== Sources ==
* : $\S 2$: Riemannian Metrics. Methods for Constructing Riemannian Metrics
Category:Riemannian Geometry
Category:Unit Normals
Existence and Uniqueness of Positive Root of Positive Real Number

== Theorem ==
Let $x \in \R$ be a real number such that $x \ge 0$.
Let $n \in \Z$ be an integer such that $n \ne 0$.
Then there always exists a unique $y \in \R: \paren {y \ge 0} \land \paren {y^n = x}$.
Hence the justification for the terminology the positive $n$th root of $x$ and the notation $x^{1/n}$.
=== Positive Exponent ===
=== Negative Exponent ===
== Proof ==
The result follows from Existence of Positive Root of Positive Real Number and Uniqueness of Positive Root of Positive Real Number.
== Sources ==
* : $\S 12.11$
Category:Existence and Uniqueness of Positive Root of Positive Real Number
Category:Roots of Numbers
Category:Real Numbers
Existence and Uniqueness of Positive Root of Positive Real Number/Negative Exponent

== Theorem ==
Let $x \in \R$ be a real number such that $x \ge 0$.
Let $n \in \Z$ be an integer such that $n < 0$.
Then there always exists a unique $y \in \R: \paren {y \ge 0} \land \paren {y^n = x}$.
== Proof ==
=== Proof of Existence ===
=== Proof of Uniqueness ===
Category:Existence and Uniqueness of Positive Root of Positive Real Number
Existence and Uniqueness of Positive Root of Positive Real Number/Positive Exponent

== Theorem ==
Let $x \in \R$ be a real number such that $x \ge 0$.
Let $n \in \Z$ be an integer such that $n > 0$.
Then there always exists a unique $y \in \R: \paren {y \ge 0} \land \paren {y^n = x}$.
== Proof ==
=== Proof of Existence ===
=== Proof of Uniqueness ===
== Sources ==
* : $\S 1.2.2$: Numbers, Powers, and Logarithms
* : $\S 1.2.2$: Numbers, Powers, and Logarithms: Exercise $8$
Category:Existence and Uniqueness of Positive Root of Positive Real Number
Existence and Uniqueness of Reduced Form of Group Word

== Theorem ==
Let $X$ be a set.
Let $w$ be a group word on $X$.
Then $w$ has a unique reduced form.
== Outline of proof ==
For existence, we proceed by induction on the length.
For uniqueness, we again proceed by induction on the length, by observing that two distinct elementary reductions have a common elementary reduction.
== Proof ==
=== Existence ===
By induction on the length of $w$.
==== Basis for the induction ====
Let $w$ have length $0$.
Then $w$ is the empty word.
By Empty Group Word is Reduced, $w$ is reduced.
==== Induction step ====
Let $w$ have length $n \geq 1$.
If $w$ is reduced, we are done.
Otherwise, let $k \in \{1, \ldots, n-1\}$ be such that $w_{k+1} = w_k^{-1}$.
Then $w$ has an elementary reduction of length $n-2$.
By the induction hypothesis, $w$ has a reduction.
The result follows by induction.
=== Uniqueness ===
By induction on the length of $w$.
==== Basis for the induction ====
Let $w$ have length $0$.
By Length of Reduced Form of Group Word is at most Length of Word, a reduced form of $w$ must have length $0$.
Thus a reduced form of $w$ can only be $w$ itself.
==== Induction step ====
Let $w$ have length $n \geq 1$.
Let $u$ and $v$ be reduced forms of $w$.
Let $u_1$ and $v_1$ be elementary reductions of $w$ with reduced forms $u$ and $v$.
By Distinct Elementary Reductions of Group Word have Equal Elementary Reduction, there exists a word $t$ which is an elementary reduction of $u_1$ and $v_1$.
Let $s$ be a reduced form of $t$.
Then $s$ is a reduced form of $u_1$ and $v_1$.
By the induction hypothesis, $s=u=v$.
The result follows by induction.
Category:Group Words
Category:Proofs by Induction
Existence and Uniqueness of Sigma-Algebra Generated by Collection of Mappings

== Theorem ==
Let $I$ be an indexing set.
Let $\family {\struct {X_i, \Sigma_i} }_{i \mathop \in I}$ be a family of measurable spaces.
Let $X$ be a set.
Let $\family {f_i: X \to X_i}_{i \mathop \in I}$ be a family of mappings.
Then $\map \sigma {f_i: i \in I}$, the $\sigma$-algebra generated by $\family {f_i}_{i \mathop \in I}$, exists and is unique.
== Proof ==
By Characterization of Sigma-Algebra Generated by Collection of Mappings:
:$\ds \map \sigma {f_i: i \in I} = \map \sigma {\bigcup_{i \mathop \in I} \map {f_i^{-1} } {\Sigma_i} }$
where the second is a $\sigma$-algebra generated by a collection of subsets.
The result follows from applying Existence and Uniqueness of Sigma-Algebra Generated by Collection of Subsets.
== Sources ==
* : $7.5$
Category:Sigma-Algebras Generated by Collection of Mappings
Existence and Uniqueness of Sigma-Algebra Generated by Collection of Subsets

== Theorem ==
Let $X$ be a set.
Let $\GG \subseteq \powerset X$ be a collection of subsets of $X$.
Then $\map \sigma \GG$, the $\sigma$-algebra generated by $\GG$, exists and is unique.
== Proof ==
=== Existence ===
By Power Set is Sigma-Algebra, there is at least one $\sigma$-algebra containing $\GG$.
Next, let $\Bbb E$ be the collection of $\sigma$-algebras containing $\GG$:
:$\Bbb E := \set {\Sigma': \GG \subseteq \Sigma', \text{$\Sigma'$ is a $\sigma$-algebra} }$
By Intersection of Sigma-Algebras, $\Sigma := \bigcap \Bbb E$ is a $\sigma$-algebra.
Also, by Set Intersection Preserves Subsets:
:$\GG \subseteq \Sigma$
Now let $\Sigma'$ be a $\sigma$-algebra containing $\GG$.
By construction of $\Sigma$, and Intersection is Subset: General Result:
:$\Sigma \subseteq \Sigma'$
=== Uniqueness ===
Suppose both $\Sigma_1$ and $\Sigma_2$ are $\sigma$-algebras generated by $\GG$.
Then property $(2)$ for these $\sigma$-algebras implies both $\Sigma_1 \subseteq \Sigma_2$ and $\Sigma_2 \subseteq \Sigma_1$. 
By definition of set equality:
:$\Sigma_1 = \Sigma_2$
== Sources ==
* : $3.4 \ \text{(ii)}$
* : $1.1$: Algebras and Sigma-Algebras
Category:Sigma-Algebras Generated by Collection of Subsets
Existence and Uniqueness of Solution for Linear Second Order ODE with two Initial Conditions

== Theorem ==
Let $\map P x$, $\map Q x$ and $\map R x$ be continuous real functions on a closed real interval $\closedint a b$.
Let $x_0$ be any point in $\closedint a b$.
Let $y_0$ and ${y_0}'$ be real numbers.
Then the linear second order ordinary differential equation:
:$(1): \quad \dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = \map R x$
has a unique particular solution $\map y x$ on $\closedint a b$ such that:
:$\map y {x_0} = y_0$
and:
:$\map {y'} {x_0} = {y_0}'$
== Proof ==
== Sources ==
* : $\S 3.14$: Second Order Linear Equations: Introduction: Theorem $\text {A}$
Category:Linear Second Order ODEs
Existence and Uniqueness of Solution for Linear Second Order ODE with two Initial Conditions/Proof

== Theorem ==
== Proof ==
Let $z = \dfrac {\d y} {\d x}$.
Then a solution to $(1)$ will yield a particular solution to:
:$(2): \quad \begin{cases}
\dfrac {\d y} {\d x} = z & , \map y {x_0} = y_0 \\
& \\
\dfrac {\d z} {\d x} = -\map P x \dfrac {\d y} {\d x} - \map Q x y + \map R x & , \map z {x_0} = {y_0}'
\end{cases}$
From Lipschitz Condition on Linear ODE of Continuous Functions, $(2)$ satisfies the Lipschitz condition.
Hence Picard's Existence Theorem applies.
Hence the result.
== Sources ==
* : $\S 11.57$: Systems. The Second Order Linear Equation: Theorem $\text{A}$
Category:Linear Second Order ODEs
Existence of Abscissa of Absolute Convergence

== Theorem ==
Let $\ds \map f s = \sum_{n \mathop = 1}^\infty a_n n^{-s}$ be a Dirichlet series.
Let the series $\ds \sum_{n \mathop = 1}^\infty \size { a_n n^{-s} }$ not converge for all $s \in \C$, or diverge for all $s \in \C$.
Then there exists a real number $\sigma_a$ such that $\map f s$ converges absolutely for all $s = \sigma + it$ with $\sigma > \sigma_a$, and does not converge absolutely for all $s$ with $\sigma < \sigma_a$.
We call $\sigma_a$ the abscissa of absolute convergence of the Dirichlet series.
== Proof ==
Let $S$ be the set of all complex numbers $s$ such that $\map f s$ converges absolutely.
By hypothesis, there is some $s_0 = \sigma_0 + it_0 \in \C$ such that $\map f {s_0}$ converges absolutely, so $S$ is not empty.
Moreover, $S$ is bounded below, for otherwise it follows from Dirichlet Series Absolute Convergence Lemma that $\map f s$ converges absolutely for all $s \in \C$, a contradiction of our assumptions.
Therefore the infimum:
:$\sigma_a = \inf \set {\sigma: s = \sigma + i t \in S} \in \R$
is well defined.
Now if $s = \sigma + it$ with $\sigma > \sigma_a$, then there is $s' = \sigma' + i t' \in S$ with $\sigma' < \sigma$, and $\map f {s'}$ is absolutely convergent.
Then it follows from Dirichlet Series Absolute Convergence Lemma that $\map f s$ is absolutely convergent.
If $s = \sigma + it$ with $\sigma < \sigma_a$, and $\map f s$ is absolutely convergent then $s$ contradicts the definition of $\sigma_a$.
Therefore, $\sigma_a$ has the claimed properties.
== Note ==
It is conventional to set $\sigma_a = -\infty$ if the series $\map f s$ is absolutely convergent for all $s \in \C$, and $\sigma_a = \infty$ if the series converges absolutely for no $s \in \C$.
Therefore, allowing $\sigma_a$ to be an extended real number, $\sigma_a$ is defined for all Dirichlet series.
== Also see ==
* Abscissa of Convergence
* Definition:Dirichlet Series
* Dirichlet Series Absolute Convergence Lemma
== Sources ==
* : $\S 11.2$: Theorem $11.1$
Category:Dirichlet Series
Existence of Abscissa of Convergence

== Theorem ==
Let $\ds \map f s = \sum_{n \mathop = 1}^\infty a_n n^{-s}$ be a Dirichlet series.
Let the series $\ds \sum_{n \mathop = 1}^\infty \size {a_n n^{-s} }$ not converge for all $s \in \C$, or diverge for all $s \in \C$.
Then there exists a real number $\sigma_c$ such that $\map f s$ converges for all $s = \sigma + i t$ with $\sigma > \sigma_c$, and does not converge for all $s$ with $\sigma < \sigma_c$.
We call $\sigma_c$ the abscissa of convergence of the Dirichlet series.
=== General ===
== Proof ==
Let $S$ be the set of all complex numbers $s$ such that $\map f s$ converges.
By hypothesis, there is some $s_0 = \sigma_0 + i t_0 \in \C$ such that $\map f {s_0}$ converges, so $S$ is not empty.
Moreover, $S$ is bounded below, for otherwise it follows from Dirichlet Series Convergence Lemma that $\map f s$ converges for all $s \in \C$, a contradiction of our assumptions.
Therefore the infimum:
:$\sigma_c = \inf \set {\sigma: s = \sigma + i t \in S} \in \R$
is well defined.
Now if $s = \sigma + it$ with $\sigma > \sigma_c$, then there is $s' = \sigma' + i t' \in S$ with $\sigma' < \sigma$, and $\map f {s'}$ is convergent.
Then it follows from Dirichlet Series Convergence Lemma that $\map f s$ is convergent.
If $s = \sigma + it$ with $\sigma < \sigma_c$, and $\map f s$ is convergent then $s$ contradicts the definition of $\sigma_c$.
Therefore, $\sigma_c$ has the claimed properties.
== Note ==
It is conventional to set $\sigma_c = -\infty$ if the series $\map f s$ is convergent for all $s \in \C$, and $\sigma_c = \infty$ if the series converges for no $s \in \C$.
Therefore, allowing $\sigma_c$ to be an extended real number, $\sigma_c$ is defined for all Dirichlet series.
== Also see ==
* Definition:Dirichlet Series
* Dirichlet Series Convergence Lemma
== Sources ==
* : $\S 11.6$: Theorem $11.9$
Category:Dirichlet Series
Existence of Abscissa of Convergence/General

== Theorem ==
Let $s = \sigma + i t$
Let $\ds \map f s = \sum_{n \mathop = 1}^\infty a_n e^{-\lambda_n s}$ be a general Dirichlet series.
Then there exists a extended real number, $\sigma_0$, such that
:$(1): \quad$ For $\sigma < \sigma_0$, $\map f s$ diverges
:$(2): \quad$ For $\sigma > \sigma_0$, $\map f s$ converges.
== Proof ==
If there does not exist an $s_0$ such that $\map f {s_0}$ converges, then $\sigma_0 = \infty$ and the theorem is vacuously true
If there does exist such an $s_0$, let $\sigma_0$ be the infimum of the real part of all such $s_0$, where $\sigma_0= -\infty$ if the set is not bounded from below.
*It is clear that $\map f s$ diverges if $\sigma < \sigma_0$ by construction of $\sigma_0$ as the infimum of all convergent $s$
*As $\sigma_0$ is the infimum of the the real part of the set all s such that $\map f s$ converges, for any $\epsilon > 0$, there exists an $s$ such that $\map f s$ converges and $\sigma_0 \le \sigma < \sigma_0+ \epsilon$ as otherwise $\sigma_0 + \epsilon$ would be the infimum of the set. 
:Therefore, for any $s$ such that $\sigma > \sigma_0$, we can pick $\epsilon = \sigma - \sigma_0$ and find a $q$ such that $\map f q$ converges and $\map \Re q < \sigma$.
:By Dirichlet Series Convergence Lemma, We therefore have that $\map f s$ converges.
Category:General Dirichlet Series
Existence of Adapted Orthonormal Frames

== Theorem ==
Let $\struct {\tilde M, \tilde g}$ be a Riemannian manifold without boundary.
Let $M \subseteq \tilde M$ be an embedded smooth submanifold with or without boundary. 
Let $p \in M$ be a point.
Then there exists a neighborhood $\tilde U$ of $p$ in $\tilde M$ and a smooth orthonormal frame for $\tilde M$ on $\tilde U$ that is adapted to $M$.
== Proof ==
== Sources ==
* : $\S 2$: Riemannian Metrics. Methods for Constructing Riemannian Metrics
Category:Orthonormal Frames