nyuuzyou/wikis huggingface dataset:
Unique Point of Minimal Distance to Closed Convex Subset of Hilbert Space

== Theorem ==
Let $H$ be a Hilbert space, and let $h \in H$.
Let $K \subseteq H$ be a closed, convex, non-empty subset of $H$.
Then there is a unique point $k_0 \in K$ such that:
:$\norm {h - k_0} = \map d {h, K}$
where $d$ denotes distance to a set.
== Proof ==
Let $\mathbf 0_H$ be the zero of $H$.
Since for every $k \in K$, we have:
:$\map d {h, k} = \norm {h - k} = \map d {\mathbf 0_H, k - h}$
it follows that:
:$\map d {h, K} = \map d {\mathbf 0_H, K - h}$
, we may therefore assume that $h = \mathbf 0_H$.
The problem has therefore reduced to finding $k_0 \in K$ such that:
:$\norm {k_0} = \map d {\mathbf 0_H, K} = \inf \set {\norm k : k \in K}$
Let $d = \map d {\mathbf 0_H, K}$.
By definition of infimum, there exists a sequence $\sequence {k_n}_{n \mathop \in \N}$ such that:
:$\ds \lim_{n \mathop \to \infty} \norm {k_n} = d$
By the Parallelogram Law, we have that for all $m, n \in \N$:
:$(1): \quad \norm {\dfrac {k_n - k_m} 2 } = \dfrac 1 2 \paren {\norm {k_n}^2 + \norm {k_m}^2} - \norm {\dfrac {k_n + k_m} 2 }^2$
Since $K$ is convex, $\dfrac {k_n + k_m} 2 \in K$.
Hence:
:$\norm {\dfrac {k_n + k_m} 2 }^2 \ge d^2$
Now given $\epsilon > 0$, choose $N$ such that for all $n \ge N$:
:$\norm {k_n}^2 < d^2 + \epsilon$
From $(1)$, it follows that:
:$\norm {\dfrac {k_n - k_m} 2} < d^2 + \epsilon - d^2 = \epsilon$
and hence that $\sequence {k_n}_{n \mathop \in \N}$ is a Cauchy sequence.
Since $H$ is a Hilbert space and $K$ is closed, it follows that there is a $k_0 \in K$ such that:
:$\ds \lim_{n \mathop \to \infty} k_n = k_0$
From Norm is Continuous, we infer that $\norm {k_0} = d$.
This demonstrates existence of $k_0$.
For uniqueness, suppose that $h_0 \in K$ has $\norm {h_0} = d$.
Since $K$ is convex, it follows that $\dfrac {h_0 + k_0} 2 \in K$.
This implies that $\norm {\dfrac {h_0 + k_0} 2} \ge d$.
Now from the Triangle Inequality:
:$\norm {\dfrac {h_0 + k_0} 2} \le \dfrac {\norm {h_0} + \norm {k_0} } 2 = d$
meaning that $\norm {\dfrac {h_0 + k_0} 2} = d$.
Thus, the Parallelogram Law implies that:
:$d^2 = \norm {\dfrac {h_0 + k_0} 2}^2 = d^2 - \norm {\dfrac {h_0 - k_0} 2}^2$
from which we conclude that $h_0 = k_0$.
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 2.$ Orthogonality: Theorem $2.5$
Category:Hilbert Spaces
Category:Convex Sets (Vector Spaces)
Unique Point of Minimal Distance to Closed Linear Subspace of Hilbert Space iff Orthogonal

== Theorem ==
Let $H$ be a Hilbert space, and let $h \in H$.
Let $K \subseteq H$ be a closed linear subspace of $H$.
Then the unique point $k_0 \in K$ from Unique Point of Minimal Distance to Closed Convex Subset of Hilbert Space such that:
:$\norm {h - k_0} = \map d {h, K}$
where $d$ denotes distance to a set, is characterised by:
:$\norm {h - k_0} = \map d {h, K} \iff \paren {h - k_0} \perp K$
where $\perp$ signifies orthogonality.
== Proof ==
=== Necessary Condition ===
Suppose that:
:$\ds \norm {h - k_0} = \map d {h, K} = \inf_{k \in K} \norm {h - k}$
Let $k \in K$. 
Since $K$ is a linear subspace, we have $k_0 + k \in K$. 
Hence, we have:
:$\norm {h - k_0}^2 \le \norm {h - \paren {k_0 + k} }^2 = \norm {\paren {h - k_0} - k}^2$
From Square of Inner Product Norm of Sum, we have:
:$\norm {\paren {h - k_0} - k}^2 = \norm {h - k_0}^2 - 2 \map \Re {\innerprod {h - k_0} k} + \norm k^2$ 
Hence:
:$\norm {h - k_0}^2 \le \norm {h - k_0}^2 - 2 \map \Re {\innerprod {h - k_0} k} + \norm k^2$
so that:
:$2 \map \Re {\innerprod {h - k_0} k} \le \norm k^2$ 
for each $k \in K$. 
Let:
:$r = \cmod {\innerprod {h - k_0} k}$
and:
:$\theta = \map \arg {\innerprod {h - k_0} k}$
so that:
:$\innerprod {h - k_0} k = r e^{i \theta}$ 
Let $t > 0$. 
Now, we have from Inner Product is Sesquilinear:
:$\innerprod {h - k_0} {t e^{i \theta} k} = t e^{-i \theta} \paren {r e^{i \theta} } = t r$ 
Since $\norm {t e^{i \theta} k }^2 = t^2 \norm k^2$ from , we therefore have:
:$t r \le t^2 \norm k^2$ 
Dividing by $t$ we have:
:$r \le t \norm k^2$ 
Taking $t \to 0^+$, we have $r \le 0$. 
Since we also have $r \ge 0$, we have $r = 0$. 
Hence we have $\innerprod {h - k_0} k = 0$ so that $h - k_0 \perp k$. 
Since $k \in K$ was arbitrary, we have $h - k_0 \perp K$. 
=== Sufficient Condition ===
Suppose that $h - k_0 \perp K$ for $k_0 \in K$. 
Let $k \in K$. 
Since $K$ is a linear subspace, we have $k_0 - k \in K$ and hence $h - k_0 \perp k_0 - k$. 
Then we have, by Pythagoras's Theorem (Inner Product Space):
:$\norm {h - k}^2 = \norm {\paren {h - k_0} + \paren {k_0 - k} }^2 = \norm {h - k_0}^2 + \norm {k_0 - k}^2$ 
Hence, since $\norm {k_0 - k}^2 \ge 0$, we have:
:$\norm {h - k} \ge \norm {h - k_0}$
for each $k \in K$. 
Since $k_0 \in K$, we have:
:$\ds \norm {h - k_0} = \inf_{k \in K} \norm {h - k}$
That is:
:$\norm {h - k_0} = \map d {h, K}$
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 2.$ Orthogonality: Theorem $2.6$
Category:Hilbert Spaces
Category:Convex Sets (Vector Spaces)
Unique Quotient in Natural Numbers

== Definition ==
Let $\N$ be the natural numbers.
Let $n \in \N$ and $m \in \N_{>0}$ such that:
:$m \divides n$
where $m \divides n$ denotes that $m$ is a divisor of $n$.
Then there exists exactly one element $p \in \N$ such that $m \times p = n$.
== Proof ==
Let $n = m \times p$.
Such a $p$ exists because $m$ is a divisor of $n$.
Suppose that $n = 0$.
Then from Natural Numbers have No Proper Zero Divisors it follows that $p = 0$.
Thus in this case the unique value of $p$ is zero.
Now suppose $n \ne 0$.
Let $n = m \times p = m \times q$ for $p, q \in \N$.
Then from Natural Number Multiplication is Cancellable it follows that $p = q$.
Hence the result.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers
Category:Natural Numbers
Unique Readability for Prefix Notation

== Theorem ==
Let $\AA$ be an alphabet.
Then prefix notation for $\AA$ has the unique readability property.
== Proof ==
Let $\phi$ be a WFF of prefix notation for $\AA$.
Apply the Principle of Mathematical Induction on the length of $\phi$ to prove:
:$(1): \quad$ No prefix of $\phi$ is a WFF, except $\phi$ itself;
:$(2): \quad$ If the first symbol of $\phi$ has arity $n$, then there exist unique WFFs $\phi_1, \ldots, \phi_n$ such that $\phi = s \phi_1 \cdots \phi_n$.
Let $\phi'$ be a WFF that is a prefix of $\phi$.
Because all WFFs have at least length $1$, it follows that:
:$\phi' = s \phi'_1 \cdots \phi'_n$
Now it must be that either $\phi_1$ is a prefix of $\phi'_1$ or vice versa.
By the induction hypothesis on $(1)$, it must be that $\phi_1 = \phi'_1$ because $\phi_1$ and $\phi'_1$ are both strictly shorter than $\phi$.
Now if, given $1 < j \le n$:
:$\phi_i = \phi'_i$ for all $i < j$
it follows that $\phi_j$ and $\phi'_j$ start at the same position in $\phi$.
Hence again, inductively, it follows that $\phi_j = \phi'_j$.
Thus $\phi_i = \phi'_i$ for all $1 \le i \le n$; that is:
:$\phi = \phi'$
The result follows by the Principle of Mathematical Induction.
== Remark ==
It is not possible to use the Principle of Structural Induction, because we still need to establish that the constituent WFFs are uniquely determined.
== Sources ==
* : $\text {II}.4$ Polish Notation: Lemma $\text {II}.4.3$
Category:Prefix Notation
Category:Collations
Unique Representation by Ordered Basis

== Theorem ==
Let $G$ be a unitary $R$-module.
Then $\sequence {a_n}$ is an ordered basis of $G$ :
:For every $x \in G$ there exists one and only one sequence $\sequence {\lambda_n}$ of scalars such that $\ds x = \sum_{k \mathop = 1}^n \lambda_k a_k$.
== Proof ==
=== Necessary Condition ===
Let $\sequence {a_n}$ be an ordered basis of $G$.
Then every element of $G$ is a linear combination of $\set {a_1, \ldots, a_n}$, which is a generator of $G$, by Generated Submodule is Linear Combinations.
Thus there exists at least one such sequence of scalar.
Now suppose there were two such sequences of scalars: $\sequence {\lambda_n}$ and $\sequence {\mu_n}$.
That is, suppose $\ds \sum_{k \mathop = 1}^n \lambda_k a_k = \sum_{k \mathop = 1}^n \mu_k a_k$.
Then:
So $\lambda_k = \mu_k$ for all $k \in \closedint 1 n$ as $\sequence {a_n}$ is a linearly independent sequence.
=== Sufficient Condition ===
Now suppose there is one and only one sequence $\sequence {\lambda_n}$ such that the condition holds.
It is clear that $\set {a_1, \ldots, a_n}$ generates $G$.
Suppose $\ds \sum_{k \mathop = 1}^n \lambda_k a_k = 0$.
Then, since also $\ds \sum_{k \mathop = 1}^n 0 a_k = 0$, we have, :
:$\forall k \in \closedint 1 n: \lambda_k = 0$
Therefore $\sequence {a_n}$ is a linearly independent sequence.
== Also see ==
* Expression of Vector as Linear Combination from Basis is Unique
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Theorem $27.4$
Category:Unitary Modules
Category:Linear Algebra
Category:Bases of Modules
Unique Representation in Polynomial Forms

== Theorem ==
Let $\struct {R, +, \circ}$ be a commutative ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $\struct {D, +, \circ}$ be an integral subdomain of $R$.
Let $X \in R$ be transcendental over $D$.
Let $D \sqbrk X$ be the ring of polynomials in $X$ over $D$.
Then each non-zero member of $D \left[{X}\right]$ can be expressed in just one way in the form:
:$\ds f \in D \sqbrk X: f = \sum_{k \mathop = 0}^n {a_k \circ X^k}$
=== General Result ===
== Proof ==
Suppose $f \in D \sqbrk X \setminus \set {0_R}$ has more than one way of being expressed in the above form.
Then you would be able to subtract one from the other and get a polynomial in $D \sqbrk X$ equal to zero.
As $f$ is transcendental, the result follows.
== Sources ==
* : $\S 64.2$ Polynomial rings over an integral domain
Category:Polynomial Theory
Unique Representation in Polynomial Forms/General Result

== Theorem ==
Let $f$ be a polynomial form in the indeterminates $\set {X_j: j \in J}$ such that $f: \mathbf X^k \mapsto a_k$.
For $r \in \R$, $\mathbf X^k \in M$, let $r \mathbf X^k$ denote the polynomial form that takes the value $r$ on $\mathbf X^k$ and zero on all other monomials.
Let $Z$ denote the set of all multiindices indexed by $J$.
Then the sum representation:
:$\ds \hat f = \sum_{k \mathop \in Z} a_k \mathbf X^k$
has only finitely many non-zero terms.
Moreover it is everywhere equal to $f$, and is the unique such sum.
=== Corollary ===
== Proof ==
Suppose that the sum has infinitely many non-zero terms.
Then infinitely many $a_k$ are non-zero, which contradicts the definition of a polynomial.
Therefore the sum consists of finitely many non-zero terms.
Let $\mathbf X^m \in M$ be arbitrary.
Then:
So $\hat f = f$.
Finally suppose that:
:$\ds \tilde f = \sum_{k \mathop \in Z} b_k \mathbf X^k$
is another such representation with $b_m \ne a_m$ for some $m \in Z$.
Then:
:$\map {\tilde f} {\mathbf X^m} = b_m \ne a_m = \map f {\mathbf X^m}$
Therefore $\hat f$ as defined above is the only such representation.
Category:Polynomial Theory
Unique Representation in Polynomial Forms/General Result/Corollary

== Theorem ==
Dropping the zero terms from the sum we can write the polynomial $f$ as
:$f = a_{k_1} \mathbf X^{k_1} + \cdots + a_{k_r} \mathbf X^{k_r}$
for some $a_{k_i}\in R$, $i = 1, \ldots, r$.
== Proof ==
Follows directly from Unique Representation in Polynomial Forms: General Result
Category:Polynomial Theory
Unique Representation of Complex Number in Spherical Form

== Theorem ==
Let $\PP$ be the complex plane.
Let $\mathbb S$ be the unit sphere which is tangent to $\PP$ at $\tuple {0, 0}$ (that is, where $z = 0$).
Let the diameter of $\mathbb S$ perpendicular to $\PP$ through $\tuple {0, 0}$ be $NS$ where $S$ is the point $\tuple {0, 0}$.
Let the point $N$ be referred to as the north pole of $\mathbb S$ and $S$ be referred to as the south pole of $\mathbb S$.
:
Let $A$ be a point on $P$.
Let the line $NA$ be constructed.
Then $NA$ passes through exactly one point $A'$ on the surface of $\mathbb S$ apart from $N$.
Similarly, let $A'$ be a point on the surface of $\mathbb S$ apart from $N$.
Let the line $NA'$ be constructed.
Then $NA'$ passes through exactly one point $A$ on $P$.
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Spherical Representation of Complex Numbers. Stereographic Projection
Category:Complex Analysis
Unique Sequence of Consecutive Odd Numbers which are Prime

== Theorem ==
Let $n \in \Z$ be an integer such that $n > 3$.
Then $n$, $n + 2$, $n + 4$ cannot all be prime.
That is, the only set of $3$ consecutive odd numbers all of which are prime is $\set {3, 5, 7}$.
== Proof ==
Let $n \in \Z_{>0}$.
For $n \le 3$ the cases can be examined in turn.
:$\set {1, 3, 5}$ are not all prime, as $1$ is not classified as prime.
:$\set {2, 4, 6}$ are not all prime, as $4$ and $6$ both have $2$ as a divisor.
:$\set {3, 5, 7}$ is the set of three consecutive odd primes which is asserted as being the unique set of odd numbers all of which are prime.
Suppose $n > 3$.
If $n$ is even then it has $2$ as a divisor, and so the set $\set {n, n + 2, n + 4}$ contains only composite numbers.
So, suppose now that $n$ is an odd integer such that $n > 3$.
Any integer $n$ can be represented as either:
If $n = 3 k$, then $n$ is not prime because $3 \divides 3 k$.
If $n = 3 k + 1$, then $n + 2$ is not prime because $3 \divides 3 k + 3$.
If $n = 3 k + 2$, then $n + 4$ is not prime because $3 \divides 3 k + 6$.
Therefore no such $n$ exists for which $n$, $n + 2$, and $n + 4$ are all prime.
== Also see ==
* No Arithmetic Sequence of 4 Primes with Common Difference 2
== Sources ==
* : $\text I$: The Series of Primes: $1.4$ The sequence of primes
Category:Prime Numbers
Unique Sequence of Consecutive Odd Numbers which are Prime/Mistake

== Source Work ==
:
:$\text I$: The Series of Primes:
::$1.4$ The sequence of primes
== Mistake ==
:The numbers $p$, $p + 2$, $p + 4$ cannot all be prime, since one of them must be divisible by $3$; ...
== Correction ==
Note should be made of the triple $3$, $5$, $7$, all of which are prime.
== Sources ==
* : $\text I$: The Series of Primes: $1.4$ The sequence of primes
Category:Mistakes/Mathematical Analysis: A Straightforward Approach
Unique Subgroup of a Given Order is Normal

== Theorem ==
Let a group $G$ have only one subgroup of a given order.
Then that subgroup is normal.
== Proof ==
Let $H \le G$, where $\le$ denotes that $H$ is a subgroup of $G$.
Let $H$ be the only subgroup of $G$ whose order is $\order H$.
Let $g \in G$.
From Conjugate of Subgroup is Subgroup:
:$g H g^{-1} \le G$
From Order of Conjugate of Subgroup:
: $\order {g H g^{-1} } = \order H$
But $H$ is the only subgroup of $G$ of order $\order H$.
Hence any subgroup whose order is $\order H$ must in fact be $H$.
That is, $g H g^{-1} = H$.
The result follows from Subgroup equals Conjugate iff Normal.
== Sources ==
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $8 \ \text{(i)}$
Category:Normal Subgroups
Unique Tree of Order 2

== Theorem ==
There exists exactly one tree of order $2$:
:
== Proof ==
By definition, a graph of order $2$ has exactly $2$ nodes.
By Finite Connected Simple Graph is Tree iff Size is One Less than Order, such a tree has one edge.
There is trivially only one way to connect $2$ nodes with one edge.
Hence the result.
Category:Tree Theory
Uniqueness Condition for Relation Value

== Theorem ==
Let $\RR$ be a relation.
Let $\tuple {x, y} \in \RR$.
Let:
:$\exists ! y: \tuple {x, y} \in \RR$
Then:
:$\map \RR x = y$
where $\map \RR x$ denotes the image of $\RR$ at $x$.
If $y$ is not unique, then:
:$\map \RR x = \O$
== Proof ==
Conversely:
Generalizing:
:$\forall z: \paren {z \in y \iff z \in \map \RR x}$
Therefore:
:$y = \map \RR x$
by the definition of class equality.
Suppose that $\neg \exists ! y: \tuple {x, y} \in \RR$.
Then:
Thus:
:$\forall z: z \notin \map \RR x$
Therefore:
:$\map \RR x = \O$
== Sources ==
* : $\S 6.12$
Category:Relation Theory
Uniqueness of Analytic Continuation

== Theorem ==
Let $U \subset V \subset \C$ be open subsets of the complex plane.
Let $V$ be connected.
Suppose:
:$(1): \quad F_1, F_2$ are functions defined on $V$
:$(2): \quad f$ is a function defined on $U$.
Let $F_1$ and $F_2$ be analytic continuations of $f$ to $V$.
Then $F_1 = F_2$.
== Proof ==
Let $\map g z = \map {F_1} z - \map {F_2} z$.
Then:
:$\forall z \in U: \map g z = 0$
Because Zeroes of Analytic Function are Isolated, and the zeroes of $g$ are not isolated, $g$ must be constant everywhere in its domain.
Since $\map g z = 0$ for some $z$, it follows that $\map g z = 0$ for all $z$.
Hence:
:$\forall z \in V: \map {F_1} z - \map {F_2} z = 0$ and so $F_1 = F_2$.
Category:Analytic Continuations
Uniqueness of Continuously Differentiable Solution to Initial Value Problem

== Theorem ==
Let $D \subseteq \R^2$ be a region containing $\tuple {a, b}$.
Let $f: D \to \R$ be real-valued mapping such that $f$ and $\dfrac {\partial f} {\partial x}$ are continuous on $D$.
Consider the initial value problem:
:$\dfrac {\d x} {\d t} = \map f {t, x}$
:$\map x a = b$
Suppose the initial value problem above has a solution $\phi$ for all $x$ in some interval $J$ containing $a$.
Then the solution $\phi$ is unique on $J$.
== Proof ==
== Sources ==
* : $\S 2.4.3$
Category:First Order ODEs
Uniqueness of Jordan Decomposition

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a signed measure on $\struct {X, \Sigma}$.
Let $\tuple {P_1, N_1}$ and $\tuple {P_2, N_2}$ be Hahn decompositions of $\mu$. 
Let $\tuple {\mu^+_1, \mu^-_1}$ be the Jordan decomposition of $\mu$ corresponding to $\tuple {P_1, N_1}$. 
Let $\tuple {\mu^+_2, \mu^-_2}$ be the Jordan decomposition of $\mu$ corresponding to $\tuple {P_2, N_2}$. 
Then: 
:$\mu^+_1 = \mu^+_2$
and:
:$\mu^-_1 = \mu^-_2$
So:
:the Jordan decomposition of a signed measure is unique.
== Proof ==
We first prove two useful identities.
=== Lemma ===
From the lemma, we have:
:$\map {\mu^+_1} A = \sup \set {\map \mu B : B \in \Sigma \text { and } B \subseteq A}$
and:
:$\map {\mu^+_2} A = \sup \set {\map \mu B : B \in \Sigma \text { and } B \subseteq A}$
for each $A \in \Sigma$.
This gives:
:$\mu^+_1 = \mu^+_2$
We also obtain: 
:$\map {\mu^-_1} A = \sup \set {-\map \mu B : B \in \Sigma \text { and } B \subseteq A}$
and:
:$\map {\mu^-_2} A = \sup \set {-\map \mu B : B \in \Sigma \text { and } B \subseteq A}$
for each $A \in \Sigma$, so:
:$\mu^-_1 = \mu^-_2$
Hence the demand.
== Sources ==
* : $4.1$
Category:Signed Measures
Category:Uniqueness of Jordan Decomposition
Uniqueness of Jordan Decomposition/Lemma

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a signed measure on $\struct {X, \Sigma}$.
Let $\tuple {P, N}$ be a Hahn decomposition of $\mu$.
Let $\tuple {\mu^+, \mu^-}$ be the Jordan decomposition of $\mu$ corresponding to $\tuple {P, N}$. 
Then, for each $A \in \Sigma$, we have: 
:$\map {\mu^+} A = \sup \set {\map \mu B : B \in \Sigma \text { and } B \subseteq A}$
and:
:$\map {\mu^-} A = \sup \set {-\map \mu B : B \in \Sigma \text { and } B \subseteq A}$
 
== Proof ==
Since $\tuple {\mu^+, \mu^-}$ is a Jordan decomposition of $\mu$, we have: 
:$\mu = \mu^+ - \mu^-$
with $\mu^+$ and $\mu^-$ measures.
Let $A \in \Sigma$.
We first show: 
:$\map {\mu^+} A = \sup \set {\map \mu B : B \in \Sigma \text { and } B \subseteq A}$
Let $B \in \Sigma$ have $B \subseteq A$.
We have:
So:
:$\map {\mu^+} A$ is an upper bound for $\set {\map \mu B : B \in \Sigma \text { and } B \subseteq A}$.
Note that from Intersection is Subset, we have: 
:$A \cap P \subseteq A$
From Sigma-Algebra Closed under Countable Intersection, we also have $A \cap P \in \Sigma$ so: 
:$\map \mu {A \cap P} \in \set {\map \mu B : B \in \Sigma \text { and } B \subseteq A}$
By definition of the Jordan decomposition: 
:$\map {\mu^+} A = \map \mu {A \cap P}$
so:
:$\map {\mu^+} A \in \set {\map \mu B : B \in \Sigma \text { and } B \subseteq A}$
So:
:$\map {\mu^+} A$ is the greatest element of $\set {\map \mu B : B \in \Sigma \text { and } B \subseteq A}$.
From Greatest Element is Supremum, we therefore have: 
:$\map {\mu^+} A = \sup \set {\map \mu B : B \in \Sigma \text { and } B \subseteq A}$
We now show that: 
:$\map {\mu^-} A = \sup \set {-\map \mu B : B \in \Sigma \text { and } B \subseteq A}$
Let $B \in \Sigma$ have $B \subseteq A$.
We have:
So:
:$\map {\mu^-} A$ is an upper bound for $\set {-\map \mu B : B \in \Sigma \text { and } B \subseteq A}$.
Note from Intersection is Subset, we have: 
:$A \cap N \subseteq A$
From Sigma-Algebra Closed under Countable Intersection, we also have $A \cap N \in \Sigma$ so: 
:$-\map \mu {A \cap N} \in \set {-\map \mu B : B \in \Sigma \text { and } B \subseteq A}$
By definition of the Jordan decomposition: 
:$\map {\mu^-} A = -\map \mu {A \cap N}$
so:
:$\map {\mu^-} A \in \set {-\map \mu B : B \in \Sigma \text { and } B \subseteq A}$
So:
:$\map {\mu^-} A$ is the greatest element of $\set {-\map \mu B : B \in \Sigma \text { and } B \subseteq A}$.
From Greatest Element is Supremum, we have: 
:$\map {\mu^-} A = \sup \set {-\map \mu B : B \in \Sigma \text { and } B \subseteq A}$
hence the result.
Category:Uniqueness of Jordan Decomposition
Uniqueness of Measures

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\GG \subseteq \powerset X$ be a generator for $\Sigma$; that is, $\Sigma = \map \sigma \GG$.
Suppose that $\GG$ satisfies the following conditions:
:$(1):\quad \forall G, H \in \GG: G \cap H \in \GG$
:$(2):\quad$ There exists an exhausting sequence $\sequence {G_n}_{n \mathop \in \N} \uparrow X$ in $\GG$
Let $\mu, \nu$ be measures on $\struct {X, \Sigma}$, and suppose that:
:$(3):\quad \forall G \in \GG: \map \mu G = \map \nu G$
:$(4):\quad \forall n \in \N: \map \mu {G_n}$ is finite
Then:
:$\mu = \nu$
Alternatively, by Countable Cover induces Exhausting Sequence, the exhausting sequence in $(2)$ may be replaced by a countable $\GG$-cover $\sequence {G_n}_{n \mathop \in \N}$, still subject to $(4)$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $5.7$
Category:Measures
Category:Uniqueness of Measures
Uniqueness of Measures/Proof 1

== Theorem ==
== Proof ==
Define, for all $n \in \N$, $\DD_n$ by:
:$\DD_n := \set {E \in \Sigma: \map \mu {G_n \cap E} = \map \nu {G_n \cap E} }$
Let us show that $\DD_n$ is a Dynkin system.
By Intersection with Subset is Subset, $G_n \cap X = G_n$, whence $(3)$ implies that $X \in \DD_n$.
Now, let $D \in \DD_n$. Then:
Therefore, $X \setminus D \in \DD_n$.
Finally, let $\sequence {D_m}_{m \mathop \in \N}$ be a sequence of pairwise disjoint sets in $\DD_n$.
Then:
Therefore:
:$\ds \bigcup_{m \mathop \in \N} D_m \in \DD_n$.
Thus, we have shown that $\DD_n$ is a Dynkin system.
Combining $(1)$ and $(3)$, it follows that:
:$\forall n \in \N: \GG \subseteq \DD_n$
From $(1)$ and Dynkin System with Generator Closed under Intersection is Sigma-Algebra:
:$\map \delta \GG = \map \sigma \GG = \Sigma$
where $\delta$ denotes generated Dynkin system.
By definition of $\map \delta \GG$, this means:
:$\forall n \in \N: \map \delta \GG \subseteq \DD_n$
That is:
:$\forall n \in \N: \Sigma \subseteq \DD_n \subseteq \Sigma$
By definition of set equality:
:$\Sigma = \DD_n$ for all $n \in \N$
Thus, for all $n \in \N$ and $E \in \Sigma$:
:$\map \mu {G_n \cap E} = \map \nu {G_n \cap E}$
Now, from Set Intersection Preserves Subsets, $E_n := G_n \cap E$ defines an increasing sequence of sets with limit:
:$\ds \bigcup_{n \mathop \in \N} \paren {G_n \cap E} = \paren {\bigcup_{n \mathop \in \N} G_n} \cap E = X \cap E = E$
from Intersection Distributes over Union and Intersection with Subset is Subset.
Thus, for all $E \in \Sigma$:
That is to say, $\mu = \nu$.
== Sources ==
* $5.7$
Category:Uniqueness of Measures
Uniqueness of Normal Coordinates

== Theorem ==
Let $\struct {M, g}$ be an $n$-dimensional Riemannian or pseudo-Riemannian manifold.
Let $U_p$ be the normal neighborhood for $p \in M$.
Then for every normal coordinate chart on $U_p$ centered at $p \in M$ the coordinate basis is orthonormal at $p$.
Furthermore, for every orthonormal basis $\tuple {b_i}$ for $T_p M$, there is a unique normal coordinate chart $\tuple {x^i}$ on $U_p$ such that:
:$\ds \forall i \in \N_{> 0} : i \le n : \valueat{\partial_i} p = b_i$
== Proof ==
== Sources ==
* : $\S 5$: The Levi-Civita Connection. Normal Neighborhoods and Normal Coordinates
Category:Normal Neighborhoods