avewright/proofwiki-math huggingface dataset:
Theorem: Set of Idempotent Elements of Entropic Structure is Closed

Let $\struct {S, \odot}$ be an algebraic structure such that $\odot$ is entropic.
Let $T \subseteq S$ be the set of idempotent elements of $S$.
Then $\struct {T, \odot}$ is closed.

Proof:
{{begin-eqn}}
{{eqn | q = \forall a, b \in T
      | l = \paren {a \odot b} \odot \paren {a \odot b}
      | r = \paren {a \odot a} \odot \paren {b \odot b}
      | c = {{Defof|Entropic Operation}}
}}
{{eqn  | r = a \odot b
       | c = {{Defof|Idempotent Operation}}
}}
{{eqn | ll= \leadsto
      | l = a \odot b
      | o = \in
      | r = T
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Endomorphisms on Entropic Structure whose Pointwise Product is Identity Automorphism

Let $\struct {S, \odot}$ be an entropic structure.
Let $\alpha$ and $\beta$ be endomorphisms on $S$ such that:
:$\alpha \odot \beta$ is the identity automorphism on $S$
where $\alpha \odot \beta$ denotes the pointwise product of $\alpha$ with $\beta$:
:$\forall x \in S: \map {\paren {\alpha \odot \beta} } x = \map \alpha x \odot \map \beta x$
Let $\otimes$ be the operation on $S$ defined as:
:$\forall x, y \in S: x \otimes y := \map \alpha x \odot \map \beta y$
Then $\struct {S, \otimes}$ is an entropic idempotent structure, and hence self-distributive.

Proof:
Let $a, b, c, d \in S$ be arbitrary.
{{begin-eqn}}
{{eqn | q = 
      | l = a \otimes a
      | r = \map \alpha a \odot \map \beta a
      | c = Definition of $\otimes$
}}
{{eqn | r = \map {\paren {\alpha \odot \beta} } a
      | c = {{Defof|Pointwise Operation}}
}}
{{eqn | r = a
      | c = {{Defof|Identity Mapping}}
}}
{{end-eqn}}
Hence $\struct {S, \otimes}$ is an idempotent structure.
{{begin-eqn}}
{{eqn | q = 
      | l = \paren {a \otimes b} \otimes \paren {c \otimes d}
      | r = \map \alpha {a \otimes b} \odot \map \beta {c \otimes d}
      | c = Definition of $\otimes$
}}
{{eqn | r = \map \alpha {\map \alpha a \odot \map \beta b} \odot \map \beta {\map \alpha c \odot \map \beta d}
      | c = Definition of $\otimes$
}}
{{eqn | r = \paren {\map \alpha {\map \alpha a} \odot \map \alpha {\map \beta b} } \odot \paren {\map \beta {\map \alpha c} \odot \map \beta {\map \beta d} }
      | c = {{Defof|Endomorphism}}
}}
{{eqn | r = \paren {\map \alpha {\map \alpha a} \odot \map \beta {\map \alpha c} } \odot \paren {\map \alpha {\map \beta b} \odot \map \beta {\map \beta d} }
      | c = {{Defof|Entropic Structure}}
}}
{{eqn | r = \paren {\map \alpha a \otimes \map \alpha c} \odot \paren {\map \beta b \otimes \map \beta d}
      | c = Definition of $\otimes$
}}
{{end-eqn}}
{{finish|Can't see where to go from here. If we can prove $\map \alpha a \otimes \map \alpha c {{=}} \map \alpha {a \otimes c}$ we're home and dry, but it's not clear that $\alpha$ and $\beta$ are homomorphisms for $\struct {S, \otimes}$ like they are for $\struct {S, \odot}$.}}
Theorem: Naturally Ordered Semigroup is Unique/Isomorphism is Unique

Let $\struct {S, \circ, \preceq}$ and $\struct {S', \circ', \preceq'}$ be naturally ordered semigroups.
Let:
:$0'$ be the smallest element of $S'$
:$1'$ be the smallest element of $S' \setminus \set {0'} = S'^*$.
Then the isomorphism $g: \struct {S, \circ, \preceq} \to \struct {S', \circ', \preceq'}$ defined as:
:$\forall a \in S: \map g a = \circ'^a 1'$
is unique.

Proof:
Let $f: S \to S'$ be another isomorphism different from $g$.
{{AimForCont}} $\map f 1 \ne 1'$.
We show by induction that $1' \notin \Cdm f$.
{{finish}}
... Thus $1' \notin \Cdm f$ which is a contradiction.
Thus $\map f 1 = 1$ and it follows
{{finish}}
that $f = g$.
Thus the isomorphism $g$ is unique.
Theorem: Naturally Ordered Semigroup is Unique/Existence of Isomorphism

Let $\struct {S, \circ, \preceq}$ and $\struct {S', \circ', \preceq'}$ be naturally ordered semigroups.
Let:
:$0'$ be the smallest element of $S'$
:$1'$ be the smallest element of $S' \setminus \set {0'} = S'^*$.
Then the mapping $g: S \to S'$ defined as:
:$\forall a \in S: \map g a = \circ'^a 1'$
is an isomorphism from $\struct {S, \circ, \preceq}$ to $\struct {S', \circ', \preceq'}$.

Proof:
Let $T' = \Cdm g$, that is, the codomain of $g$.
By Zero is Identity in Naturally Ordered Semigroup, $0'$ is the identity for $\circ'$.
Thus:
:$\map g 0 = \circ'^0 1' = 0'$
and so $0' \in T'$
Suppose $x' \in T'$.
Then:
:$\map g n = x'$
and so:
{{begin-eqn}}
{{eqn | l = x' \circ' 1'
      | r = \circ'^n 1' \circ' 1'
      | c = 
}}
{{eqn | r = \paren {\circ'^n 1'} \circ' \paren {\circ'^1 1'}
      | c = 
}}
{{eqn | r = \circ'^{\paren {n \circ 1} } 1'
      | c = 
}}
{{eqn | r = \map g {n \circ 1}
      | c = 
}}
{{end-eqn}}
So:
:$x' \in T' \implies x' \circ' 1' \in T'$
Thus, by the Principle of Mathematical Induction applied to $S'$:
:$T' = S'$
So:
:$\forall a' \in S': \exists a \in S: \map g a = a'$
and so $g$ is surjective by definition.
From Index Laws for Semigroup: Sum of Indices:
:$\map g {a \circ b} = \map g a \circ' \map g b$
and therefore $g$ is a homomorphism from $\struct {S, \circ}$ to $\struct {S', \circ'}$.
Now:
{{begin-eqn}}
{{eqn | q = \forall p \in S
      | l = \circ'^p 1'
      | o = \prec'
      | r = \circ'^p 1' \circ' 1'
      | c = Sum with One is Immediate Successor in Naturally Ordered Semigroup
}}
{{eqn | r = \paren {\circ'^p 1'} \circ' \paren {\circ'^1 1'}
      | c = 
}}
{{eqn | r = \circ'^{\paren {p \circ 1} } 1'
      | c = Index Laws for Semigroup: Sum of Indices
}}
{{eqn | ll= \leadsto
      | l = \circ'^p 1'
      | o = \prec'
      | r = \circ'^{\paren {p \circ 1} } 1'
      | c = 
}}
{{end-eqn}}
For $p \in S$, let $S_p$ be the initial segment of $S$:
:$S_p = \set {x \in S: x \prec p}$
Let:
:$T = \set {p \in S: \forall a \in S_p: \circ'^a 1' \prec' \circ'^p 1'}$
Now $S_0 = \O \implies 0 \in T$.
Suppose $p \in T$.
Then:
:$a \prec p \circ 1 \implies a \preceq p$
By Strict Lower Closure of Sum with One, either of these is the case:
:$(1): \quad a \prec p: p \in T \implies \circ'^a 1' \prec' \circ'^p 1' \prec' \circ'^{\paren {p \circ 1} } 1'$
:$(2): \quad a = p: \circ'^a 1' = \circ'^p 1' \prec' \circ'^{\paren {p \circ 1} } 1'$
In either case, we have:
:$p \in T \implies p \circ 1 \in T$, and by the Principle of Mathematical Induction:
:$T = S$
So $n \prec p \implies \circ'^n 1' \prec' \circ'^p 1'$.
Thus $g$ is a surjective monomorphism and therefore is an isomorphism from $\struct {S, \circ, \preceq}$ to $\struct {S', \circ', \preceq'}$.
Theorem: Operations with Identities which Distribute over each other are Idempotent

Let $S$ be a set.
Let $\odot$ and $\otimes$ be operations on $S$ which have identity elements $e$ and $u$ respectively.
Let $\odot$ and $\otimes$ be distributive over each other.
Then $\odot$ and $\otimes$ are both idempotent.

Proof:
{{begin-eqn}}
{{eqn | l = e \otimes \paren {u \odot e}
      | r = \paren {e \otimes u} \odot \paren {e \otimes e}
      | c = as $\otimes$ is distributive over $\odot$
}}
{{eqn | ll= \leadsto
      | l = e \otimes u
      | r = \paren {e \otimes u} \odot \paren {e \otimes e}
      | c = as $e$ is the identity of $\odot$
}}
{{eqn | ll= \leadsto
      | l = e
      | r = e \odot \paren {e \otimes e}
      | c = as $u$ is the identity of $\otimes$
}}
{{eqn | ll= \leadsto
      | l = e
      | r = e \otimes e
      | c = as $e$ is the identity of $\odot$
}}
{{end-eqn}}
In the same way:
{{begin-eqn}}
{{eqn | l = u \odot \paren {e \otimes u}
      | r = \paren {u \odot e} \otimes \paren {u \odot u}
      | c = as $\odot$ is distributive over $\otimes$
}}
{{eqn | ll= \leadsto
      | l = u \odot e
      | r = \paren {u \odot e} \otimes \paren {u \odot u}
      | c = as $u$ is the identity of $\otimes$
}}
{{eqn | ll= \leadsto
      | l = u
      | r = u \otimes \paren {u \odot u}
      | c = as $e$ is the identity of $\odot$
}}
{{eqn | ll= \leadsto
      | l = u
      | r = u \odot u
      | c = as $u$ is the identity of $\otimes$
}}
{{end-eqn}}
Then we have:
{{begin-eqn}}
{{eqn | q = \forall a \in S
      | l = a \otimes a
      | r = \paren {a \odot e} \otimes \paren {a \odot e}
      | c = as $e$ is the identity of $\odot$
}}
{{eqn | r = a \odot \paren {e \otimes e}
      | c = as $\odot$ is distributive over $\otimes$
}}
{{eqn | r = a \odot e
      | c = {{apriori}}
}}
{{eqn | r = a
      | c = as $e$ is the identity of $\odot$
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | q = \forall a \in S
      | l = a \odot a
      | r = \paren {a \otimes u} \odot \paren {a \otimes u}
      | c = as $u$ is the identity of $\otimes$
}}
{{eqn | r = a \otimes \paren {u \odot u}
      | c = as $\otimes$ is distributive over $\odot$
}}
{{eqn | r = a \otimes u
      | c = {{apriori}}
}}
{{eqn | r = a
      | c = as $u$ is the identity of $\otimes$
}}
{{end-eqn}}
Hence the result by definition of idempotent operation.
{{qed}}
Theorem: Condition for Algebraic Structure to be Self-Distributive Quasigroup

Let $\struct {S, \odot}$ be an algebraic structure.
Then:
:$\struct {S, \odot}$ is a self-distributive quasigroup
{{iff}}:
:for every $a \in S$, the left and right regular representations $\lambda_a$ and $\rho_a$ on $S$ are automorphisms of $\struct {S, \odot}$.

Proof:
=== Sufficient Condition ===
Let $\struct {S, \odot}$ be a self-distributive quasigroup.
By definition of quasigroup, we have that $\lambda_a$ and $\rho_a$ are both permutations on $S$.
It remains for the morphism property to be demonstrated.
Indeed:
{{begin-eqn}}
{{eqn | q = \forall x, y \in S
      | l = \map {\lambda_a} x \odot \map {\lambda_a} y
      | r = \paren {a \odot x} \odot \paren {a \odot y}
      | c = {{Defof|Left Regular Representation}}
}}
{{eqn | r = a \odot \paren {x \odot y}
      | c = {{Defof|Self-Distributive Structure}}
}}
{{eqn | r = \map {\lambda_a} {x \odot y}
      | c = {{Defof|Left Regular Representation}}
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | q = \forall x, y \in S
      | l = \map {\rho_a} x \odot \map {\rho_a} y
      | r = \paren {x \odot a} \odot \paren {y \odot a}
      | c = {{Defof|Right Regular Representation}}
}}
{{eqn | r = \paren {x \odot y} \odot a
      | c = {{Defof|Self-Distributive Structure}}
}}
{{eqn | r = \map {\rho_a} {x \odot y}
      | c = {{Defof|Right Regular Representation}}
}}
{{end-eqn}}
Thus $\lambda_a$ and $\rho_a$ have been shown to be automorphisms of $\struct {S, \odot}$.
{{qed|lemma}}
Theorem: Self-Distributive Quasigroup is Idempotent

Let $\struct {S, \odot}$ be a self-distributive quasigroup.
Then $\odot$ is an idempotent operation.

Proof:
Because $S$ is a quasigroup:
:$\forall a, b \in S: \exists ! x \in S: x \odot a = b$
:$\forall a, b \in S: \exists ! y \in S: a \odot y = b$
In particular:
:$\exists ! x \in S: x \circ \paren {a \odot b} = y$
for arbitrary $a, b \in S$.
Let $y = a \odot \paren {a \odot b}$.
Because $\struct {S, \odot}$ is self-distributive:
:$y = \paren {a \odot a} \odot \paren {a \odot b}$
Because $S$ is a quasigroup, we have {{apriori}} that:
:$a = a \odot a$
As $a$ is arbitrary, the result follows.
{{qed}}
Theorem: Self-Distributive Quasigroup with at least Two Elements is not Associative

Let $\struct {S, \odot}$ be a self-distributive quasigroup.
Let $S$ have at least $2$ elements.
Then $\odot$ is not an associative operation.

Proof:
{{AimForCont}} $\odot$ is associative operation.
Let $a, b \in S$ such that $a \ne b$.
Then:
{{begin-eqn}}
{{eqn | q = \forall a, b \in S
      | l = \paren {a \odot a} \odot \paren {a \odot b}
      | r = a \odot \paren {a \odot b}
      | c = {{Defof|Self-Distributive Structure}}
}}
{{eqn | r = \paren {a \odot a} \odot b
      | c = {{Defof|Associative Operation}}
}}
{{eqn | r = \paren {a \odot b} \odot \paren {a \odot b}
      | c = {{Defof|Self-Distributive Structure}}
}}
{{eqn | ll= \leadsto
      | l = a \odot a
      | r = a \odot b
      | c = {{Defof|Quasigroup}}
}}
{{eqn | ll= \leadsto
      | l = a
      | r = b
      | c = {{Defof|Quasigroup}}
}}
{{end-eqn}}
which contradicts our supposition that $a \ne b$.
{{qed}}
Theorem: Self-Distributive Quasigroup with at least Two Elements has no Identity

Let $\struct {S, \odot}$ be a self-distributive quasigroup.
Let $S$ have at least $2$ elements.
Then $\struct {S, \odot}$ has no identity element.

Proof:
{{AimForCont}} $S$ has an identity element $e$ and another element $a$ such that $a \ne e$.
Recall the definition of quasigroup:
:$\forall a, b \in S: \exists ! x \in S: x \circ a = b$
That is:
We have:
{{begin-eqn}}
{{eqn | l = a \circ a
      | r = a
      | c = Self-Distributive Quasigroup is Idempotent
}}
{{eqn | l = e \circ a
      | r = a
      | c = {{Defof|Identity Element}}
}}
{{end-eqn}}
That is, there are two $x \in S$ such that $x \circ a = a$.
This contradicts our assertion that $\struct {S, \odot}$ is a quasigroup.
Hence there can be no such element.
{{qed}}
Theorem: P-adic Number is Power of p Times P-adic Unit

Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $\Z_p^\times$ be the $p$-adic units.
Let $\nu_p: \Q_p \to \Z \cup \set {+\infty}$ denote the $p$-adic valuaton on the $p$-adic numbers.
Let $a \in \Q_p$.
Then there exists $u \in \Z_p^\times$ such that:
:$a = p^{\map {\nu_p} a} \cdot u$

Proof:
From P-adic Number times P-adic Norm is P-adic Unit, there exists $n \in \Z$ such that:
:$p^n a \in \Z_p^\times$
where
:$p^n = \norm a_p$
We have:
{{begin-eqn}}
{{eqn | l = \map {\nu_p} a
      | r = -\log_p \norm a_p
      | c = {{Defof|P-adic Valuation on P-adic Numbers}}
}}
{{eqn | r = -\log_p p^n
}}
{{eqn | r = -n
}}
{{end-eqn}}
Let $u = p^n a$.
Then:
{{begin-eqn}}
{{eqn | l = a
      | r = p^{-n} u
      | c = 
}}
{{eqn | r = p^{\map {\nu_p} a} \cdot u
}}
{{end-eqn}}
{{qed}}
Category:P-adic Number Theory
Category:P-adic Units
3qfnh0t0ykc7agf4aw46pwnsbbhrem6
Theorem: Polynomials of Congruent Ring Elements are Congruent

Let $R$ be a commutative ring with unity.
Let $I$ be an ideal of $R$.
Let $x, y \in R$.
Let:
:$x \equiv y \pmod I$
where the notation indicates congruence modulo $I$.
Let $\map F X \in R \sqbrk X$ be a polynomial in one variable over $R$.
Then:
:$\ds \map F x \equiv \map F y \pmod I$

Proof:
Let $\map F X = \ds \sum_{k \mathop = 0}^r a_k X^k$ where $X$ is the indeterminate and $a_0, a_1, \ldots, a_r \in R$.
It has to be shown:
:$\ds \sum_{k \mathop = 0}^r a_k x^k \equiv \sum_{k \mathop = 0}^r a_k y^k \pmod I$
From Left Cosets are Equal iff Product with Inverse in Subgroup:
:$\forall a, b \in R: a \equiv b \pmod I  \iff a + I = b + I$
We have:
{{begin-eqn}}
{{eqn | l = \paren {\sum_{k \mathop = 0}^r a_k x^k} + I
      | r = \sum_{k \mathop = 0}^r \paren {a_k x^k + I}
      | c = Quotient Ring Addition is Well-Defined
}}
{{eqn | r = \sum_{k \mathop = 0}^r \paren {\paren{a + I} \paren{x+ I}^k}
      | c = Quotient Ring Product is Well-Defined
}}
{{eqn | r = \sum_{k \mathop = 0}^r \paren {\paren{a + I} \paren{y+ I}^k}
      | c = Left Cosets are Equal iff Product with Inverse in Subgroup
}}
{{eqn | r = \sum_{k \mathop = 0}^r \paren {a_k y^k + I}
      | c = Quotient Ring Product is Well-Defined
}}
{{eqn | r = \paren {\sum_{k \mathop = 0}^r a_k r^k} + I
      | c = Quotient Ring Addition is Well-Defined
}}
{{eqn | ll = \leadsto
      | l = \sum_{k \mathop = 0}^r a_k x^k
      | o = \equiv
      | r = \sum_{k \mathop = 0}^r a_k y^k
      | rr= \pmod I
      | c = Left Cosets are Equal iff Product with Inverse in Subgroup
}}
{{end-eqn}}
{{qed}}
Category:Polynomial Theory
Category:Ideal Theory
jnofq56sfdfmxs7absbeu9i4l49ztj1
Theorem: Binary Relation is Subclass of Product of Domain with Range

Let $V$ be a basic universe.
Let $\RR \subseteq V \times V$ be a relation.
Let $\Dom \RR$ denote the domain of $\RR$.
Let $\Img \RR$ denote the image of $\RR$.
Then:
:$\RR$ is a subclass of $\Dom \RR \times \Img \RR$

Proof:
Let $\tuple {x, y} \in \RR$.
Then by definition of domain of $\RR$:
:$x \in \Dom \RR$
and by definition of image of $\RR$:
:$y \in \Img \RR$
Hence by definition of Cartesian product:
:$\tuple {x, y} \in \Dom \RR \times \Img \RR$
Hence the result by definition of subclass:
:$\RR \subseteq \paren {\Dom \RR \times \Img \RR}$
{{qed}}
Theorem: Equivalent Statements for Transitive Set

Let $a$ be a set.
{{TFAE}}
:$(1): \quad \bigcup a \subseteq a$
:$(2): \quad a \subseteq \powerset a$
:$(3): \quad a$ is a transitive set

Proof:
=== $(1) \iff (3)$ ===
This is shown in Class is Transitive iff Union is Subclass.
{{qed|lemma}}
Theorem: Union of Sigma-Algebras may not be Sigma-Algebra

Let $X$ be a set.
Let $\struct {X, \Sigma_1}$ and $\struct {X, \Sigma_2}$ be $\sigma$-algebras.
Then $\Sigma_1 \cup \Sigma_2$ may not be a $\sigma$-algebra on $X$.

Proof:
Let $X = \set {1, 2, 3}$. 
Let: 
:$\Sigma_1 = \set {\O, \set 1, \set {2, 3}, \set {1, 2, 3} }$
and:
:$\Sigma_2 = \set {\O, \set 2, \set {1, 3}, \set {1, 2, 3} }$
Then $\Sigma_1$ and $\Sigma_2$ are $\sigma$-algebras on $X$.
We have: 
:$\Sigma_1 \cup \Sigma_2 = \set {\O, \set 1, \set 2, \set {1, 3}, \set {2, 3}, \set {1, 2, 3} }$
However, this does not contain: 
:$\set 1 \cup \set 2 = \set {1, 2}$
Since $\sigma$-algebras are closed under countable union, $\Sigma_1 \cup \Sigma_2$ therefore cannot be a $\sigma$-algebra. 
{{qed}}
Category:Sigma-Algebras
29t8maxe9hb4i3kf1ilpdtmgmox0g90
Theorem: Characterization of Measures that Share Null Sets

Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ and $\nu$ be $\sigma$-finite measures on $\struct {X, \Sigma}$.
{{TFAE}}
:$(1) \quad$ $\nu \ll \mu$ and $\mu \ll \nu$, where $\ll$ denotes absolute continuity
:$(2) \quad$ $A \in \Sigma$ is a $\mu$-null set {{iff}} it is a $\nu$-null set
:$(3) \quad$ there exists a positive $\Sigma$-measurable function $g : X \to \R$ such that: 
::::$\ds \map \nu A = \int_A g \rd \mu$
:::for each $A \in \Sigma$.

Proof:
=== $(1)$ implies $(2)$ ===
Suppose that $\nu \ll \mu$ and $\mu \ll \nu$. 
Let $A \in \Sigma$ have $\map \mu A = 0$.
Then, since $\nu \ll \mu$, we have $\map \nu A = 0$ from the definition of absolute continuity.
So if a $\Sigma$-measurable set is $\mu$-null, it is $\nu$-null.
Similarly, let $B \in \Sigma$ have $\map \nu B = 0$.
Then, since $\mu \ll \nu$, we have $\map \mu B = 0$ from the definition of absolute continuity.
So if a $\Sigma$-measurable set is $\nu$-null, it is $\mu$-null.
So a $\Sigma$-measurable set is $\nu$-null {{iff}} it is $\mu$-null.
Hence $(2)$.
{{qed|lemma}}
Theorem: Principle of Recursive Definition/Strong Version

Let $\omega$ denote the natural numbers as defined by the von Neumann construction.
Let $A$ be a class.
Let $c \in A$.
Let $g: \omega \times A \to A$ be a mapping.
Then there exists exactly one mapping $f: \omega \to A$ such that:
:<nowiki>$\forall x \in \omega: \map f x = \begin{cases}
c & : x = \O \\
\map g {n, \map f n} & : x = n^+
\end{cases}$</nowiki>

Proof:
From Von Neumann Construction of Natural Numbers is Minimally Inductive, $\omega$ is a minimally inductive class under the successor mapping.
First an '''admissible mapping''' is defined.
Let $n \in \omega$.
The mapping $h: n \to A$ is defined as an '''admissible mapping''' for $n$ {{iff}}:
:<nowiki>$\forall r \subseteq n: \map h r = \begin{cases}
c & : r = 0 \\
\map g {y, \map h y} & : r = y^+
\end{cases}$</nowiki>
Define:
:$S = \set {x \in \omega: x \text { has exactly one admissible mapping} }$
We now use the Principle of Finite Induction for Peano Structure to prove that $S = \omega$.
Theorem: Relation between Volume Forms of Conformally Related Metrics on Oriented Riemannian Manifold

Let $M$ be an $n$-dimensional oriented Riemannian manifold.
Let $g_1$ and $g_2$ be conformally related Riemannian metrics.
That is, suppose:
:$g_2 = f g_1$
where $f \in \map {C^\infty} M$ is a positive smooth real function.
Let $\rd V_{g_1}$ and $\rd V_{g_2}$ be Riemannian volume forms associated with $g_1$ and $g_2$.
Then:
:$\ds \rd V_{g_2} = f^{\frac n 2} \rd V_{g_1}$

Proof:
Let $\tuple {x_1, \ldots, x_n}$ be a set of local oriented coordinates.
Let $g_{i j}$ be a local form of the metric $g_1$.
Then $f g_{i j}$ is the local form of the metric $g_2$.
{{begin-eqn}}
{{eqn | l = \rd V_{g_2}
      | r = \sqrt {\map \det {f g_{i j} } } \rd x^1 \wedge \ldots \wedge \rd x^n
      | c = {{Defof|Riemannian Volume Form|index = 3}}
}}
{{eqn | r = \sqrt {f^n \map \det {g_{i j} } } \rd x^1 \wedge \ldots \wedge \rd x^n
}}
{{eqn | r = f^{\frac n 2} \sqrt {\map \det {g_{i j} } } \rd x^1 \wedge \ldots \wedge \rd x^n
}}
{{eqn | r = f^{\frac n 2} \rd V_{g_1}
      | c = {{Defof|Riemannian Volume Form|index = 3}}
}}
{{end-eqn}}
{{qed}}
Theorem: Hensel's Lemma/P-adic Integers/Lemma 6

:$x \equiv 0 \pmod {p^k \Z_p} \implies \exists y \in \Z_p : x = y p^k$

Proof:
Let:
:$x \equiv 0 \pmod {p^k \Z_p}$
By definition of congruence modulo ideal:
:$x \in p^k \Z_p$
By definition of principal ideal:
:$\exists y \in \Z_p : x = y p^k$
{{qed}}
Category:Hensel's Lemma
gri91j3fizduk4zsihea7lqyawt7pkn
Theorem: Hensel's Lemma/P-adic Integers/Lemma 7

:$x \in \Z_p \implies \exists y \in T : y p^k \equiv x p^k \pmod {p^{k+1}\Z_p}$

Proof:
Let $x \in \Z_p$.
From $p$-adic Integer is Limit of Unique $p$-adic Expansion, let:
:$x = \ds \sum_{n \mathop = 0}^\infty d_n p^n$
We have:
{{begin-eqn}}
{{eqn | l = x - d_0
      | r = \paren{d_0 + d_1p + d_2p^2 + d_3p^3 + \ldots} - d_0
}}
{{eqn | r = \paren{d_1p + d_2p^2 + d_3p^3 + \ldots}
      | c = $d_0$ terms cancel
}}
{{eqn | r = p \paren{d_1 + d_2p + d_3p^2 + \ldots}
      | c = Factor $p$ from each term
}}
{{eqn | o = \in
      | r = p\Z_p
      | c = $d_1 + d_2p + d_3p^2 + \ldots$ is the $p$-adic expansion of a $p$-adic integer
}}
{{eqn | ll= \leadsto
      | l = xp^k - d_0p^k
      | r = \paren{x - d_0}p^k
}}
{{eqn | o = \in
      | r = p^{k+1}\Z_p
}}
{{eqn | ll= \leadsto
      | l = xp^k
      | o = \equiv
      | r = d_0p^k \pmod {p^{k+1}\Z_p}
      | c = {{Defof|Congruence Modulo Ideal}}
}}
{{end-eqn}}
Let $y = d_0$.
The result follows.
{{qed}}
Category:Hensel's Lemma
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Theorem: Hensel's Lemma/P-adic Integers/Lemma 8

:$x, y \in \Z_p \implies \map F {x + y p ^k} \equiv \map F x + y p^k \map {F'} x \pmod {p^{k+1}\Z_p}$

Proof:
Let $\ds \map F X = \sum_{j \mathop = 0}^r c_j X^j$ where $X$ is the indeterminate and $c_0, c_1, \ldots, c_r \in \Z_p$.
Then:
:$\ds \map {F'} X = \sum_{j \mathop = 1}^r j c_j X^j$
We have:
{{begin-eqn}}
{{eqn | l = \map F {x + yp^k}
      | r = \sum_{j \mathop = 0}^r c_j \paren{x + yp^k}^j
      | c = Definition of $\map F X$
}}
{{eqn | r = c_0 + \sum_{j \mathop = 1}^r c_j \paren{x^j + jx^{j-1}yp^k + \text{terms divisible by } p^{k+1} }
      | c = Binomial Theorem
}}
{{eqn | r = c_0 + \sum_{j \mathop = 1}^r c_j x^j + \sum_{j \mathop = 1}^r j c_j x^{j-1}yp^k + \text{terms divisible by } p^{k+1}
      | c = Rearrange terms
}}
{{eqn | r = \sum_{j \mathop = 0}^r c_j x^j + \sum_{j \mathop = 1}^r j c_j x^{j-1}yp^k + \text{terms divisible by } p^{k+1}
      | c = Combine first term with first sum
}}
{{eqn | r = \sum_{j \mathop = 0}^r c_j x^j + yp^k \sum_{j \mathop = 1}^r j c_j x^{j-1} + \text{terms divisible by } p^{k+1}
      | c = Extract common factor $yp^k$ from second sum
}}
{{eqn | r = \map F x + y p^k \map {F'} x + \text{terms divisible by } p^{k+1}
      | c = Definitions of $\map F X$ and $\map {F'} X$
}}
{{eqn | o = \equiv
      | r = \map F x + y p^k \map {F'} a
      | rr= \pmod{ p^{k+1}\Z_p} 
      | c = {{Defof|Congruence Modulo Ideal}}
}}
{{end-eqn}}
{{qed}}
Category:Hensel's Lemma
6vy6wzidvm6dachfxw94tdl1kyecl41
Theorem: Hensel's Lemma/P-adic Integers/Lemma 9

:$\paren{c, d \in T : \tuple{b_0, b_1, \ldots, b_{k-1}, c}, \tuple{b_0, b_1, \ldots, b_{k-1}, d} \in S_{k+1}} \implies c = d$

Proof:
=== Lemma 7 ===
{{:Hensel's Lemma/P-adic Integers/Lemma 7}}{{qed|lemma}}
Theorem: Poincaré Conjecture/Dimension 1

Let $\Sigma^1$ be a smooth $1$-manifold.
Let $\Sigma^1$ satisfy:
:$H_0 \struct {\Sigma; \Z} = 0$
and:
:$H_m \struct {\Sigma; \Z} = \Z$
Then $\Sigma^1$ is homeomorphic to the $1$-sphere $\Bbb S^1$.
{{explain|Definition of the notation $H_0 \struct {\Sigma; \Z}$, nature of $H_0$ and $H_m$}}

Proof:
Follows from the Classification of Compact One-Manifolds.
Theorem: Poincaré Conjecture/Dimension 2

Let $\Sigma^2$ be a smooth $2$-manifold.
Let $\Sigma^2$ satisfy:
:$H_0 \struct {\Sigma; \Z} = 0$
and:
:$H_2 \struct {\Sigma; \Z} = \Z$
Then $\Sigma^2$ is homeomorphic to the $2$-sphere $\Bbb S^2$.
{{explain|Definition of the notation $H_0 \struct {\Sigma; \Z}$, nature of $H_0$ and $H_m$}}

Proof:
Follows from the Classification of Compact Two-Manifolds.