avewright/proofwiki-math huggingface dataset:
Theorem: Lp Norm is Well-Defined

Let $\struct {X, \Sigma, \mu}$ be a measure space, and let $p \in \hointr 1 \infty$. 
Let $\map {\LL^p} {X, \Sigma, \mu}$ be the Lebesgue $p$-space of $\struct {X, \Sigma, \mu}$. 
Let $\sim$ be the $\mu$-almost-everywhere equality relation on $\map {\LL^p} {X, \Sigma, \mu}$.
Let $\map {L^p} {X, \Sigma, \mu}$ be the $L^p$ space on $\struct {X, \Sigma, \mu}$.
Let $\eqclass f \sim \in \map {L^p} {X, \Sigma, \mu}$. 
Then the $L^p$ norm: 
:$\ds \norm {\eqclass f \sim}_p = \paren {\norm f_p}^{1/p}$
is well-defined.

Proof:
Note that: 
:$\ds \int \size f^p \rd \mu$ 
is well-defined from the definition of the Lebesgue $p$-space. 
We show that for $E \in \map {L^p} {X, \Sigma, \mu}$, $\norm E_p$ is independent of the representative chosen for $E$.
Let:
:$E = \eqclass f \sim = \eqclass g \sim$
for $\eqclass f \sim, \eqclass g \sim \in \map {L^p} {X, \Sigma, \mu}$.
We show that: 
:$\norm f_p = \norm g_p$ 
From Equivalence Class Equivalent Statements, we have: 
:$f \sim g$ 
So, from the definition of the almost-everywhere equality relation, we have: 
:$\map f x = \map g x$ for $\mu$-almost all $x \in X$.
That is, there exists a $\mu$-null set $N$ such that: 
:if $\map f x \ne \map g x$ then $x \in N$.
If $\size {\map f x}^p \ne \size {\map g x}^p$, we have $\map f x \ne \map g x$, so:
:if $\size {\map f x}^p \ne \size {\map g x}^p$ then $x \in N$.
So:
:$\size {\map f x}^p = \size {\map g x}^p$ for $\mu$-almost all $x \in X$. 
So, from A.E. Equal Positive Measurable Functions have Equal Integrals, we have: 
:$\ds \paren {\int \size f^p \rd \mu}^{1/p} = \paren {\int \size g^p \rd \mu}^{1/p}$ 
That is: 
:$\norm f_p = \norm g_p$
as required.
{{qed}}
Category:Lp Norms
5n23fiz4984powsnw4husf30ujiol35
Theorem: Equivalence of Definitions of Left Quasi-Reflexive Relation

Let $\RR \subseteq S \times S$ be a relation in $S$.
{{TFAE|def = Left Quasi-Reflexive Relation}}

Proof:
=== $(1)$ implies $(2)$ ===
Let $\RR$ be a left quasi-reflexive relation by definition $1$.
Then by definition:
:$\forall x, y \in S: \tuple {x, y} \in \RR \implies \tuple {x, x} \in \RR$
Let $x \in \Dom \RR$ be arbitrary.
Then by definition of domain:
:$\exists y \in S: \tuple {x, y} \in \RR$
Hence {{hypothesis}}:
:$\tuple {x, x} \in \RR$
As $y$ is arbitrary:
:$\forall x \in \Dom \RR: \tuple {x, x} \in \RR$
Thus $\RR$ is a left quasi-reflexive relation by definition $2$.
{{qed|lemma}}
Theorem: Equivalence of Definitions of Right Quasi-Reflexive Relation

Let $\RR \subseteq S \times S$ be a relation in $S$.
{{TFAE|def = Right Quasi-Reflexive Relation}}

Proof:
=== $(1)$ implies $(2)$ ===
Let $\RR$ be a right quasi-reflexive relation by definition $1$.
Then by definition:
:$\forall x, y \in S: \tuple {x, y} \in \RR \implies \tuple {y, y} \in \RR$
Let $y \in \Img \RR$ be arbitrary.
Then by definition of image set:
:$\exists x \in S: \tuple {x, y} \in \RR$
Hence {{hypothesis}}:
:$\tuple {y, y} \in \RR$
As $y$ is arbitrary:
:$\forall y \in \Img \RR: \tuple {y, y} \in \RR$
Thus $\RR$ is a right quasi-reflexive relation by definition $2$.
{{qed|lemma}}
Theorem: Pointwise Addition on Space of Real-Valued Measurable Functions Identified by A.E. Equality is Well-Defined

Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\map {\mathcal M} {X, \Sigma, \R}$ be the set of real-valued $\Sigma$-measurable functions on $X$. 
Let $\sim$ be the almost-everywhere equality relation on $\map {\mathcal M} {X, \Sigma, \R}$.
Let $\map {\mathcal M} {X, \Sigma, \R}/\sim$ be the space of real-valued measurable functions identified by $\mu$-A.E. equality. 
Then pointwise addition on $\map {\mathcal M} {X, \Sigma, \R}/\sim$ is well-defined.

Proof:
Let $E_1, E_2 \in \map {\mathcal M} {X, \Sigma, \R}/\sim$.
First, we show that if $E_1 = \eqclass f \sim$ and $E_2 = \eqclass g \sim$, that $\eqclass {f + g} \sim$ is well-understood. 
This follows from Pointwise Sum of Measurable Functions is Measurable.
We now need to show that $E_1 + E_2$ is independent of the choice of representative for $E_1$ and $E_2$.
Suppose that: 
:$\eqclass f \sim = \eqclass F \sim = E_1$
and:
:$\eqclass g \sim = \eqclass G \sim = E_2$
From Equivalence Class Equivalent Statements, we have: 
:$f \sim F$
and:
:$g \sim G$
From Pointwise Addition preserves A.E. Equality, we have: 
:$f + g \sim F + G$
That is, from Equivalence Class Equivalent Statements:
:$\eqclass {f + g} \sim = \eqclass {F + G} \sim$ 
which is what we aimed to show.
{{qed}}
Category:Space of Real-Valued Measurable Functions Identified by A.E. Equality
6ukrtosi2q8tza111ir7hxr4osswmtv
Theorem: Pointwise Scalar Multiplication preserves A.E. Equality

Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f, g : X \to \overline \R$ be extended real-valued functions such that: 
:$f = g$ $\mu$-almost everywhere.
Let $\lambda \in \overline \R$.
Then: 
:$\lambda \cdot f = \lambda \cdot g$ $\mu$-almost everywhere
where $\lambda \cdot f$ denotes pointwise scalar multiplication.

Proof:
Since:
:$f = g$ $\mu$-almost everywhere
there exists a $\mu$-null set $N \subseteq X$ such that: 
:if $\map f x \ne \map g x$ then $x \in N$. 
Note that if $\map f x = \map g x$ then $\lambda \map f x = \lambda \map g x$. 
So, from the Rule of Transposition we have:
:if $\lambda \map f x \ne \lambda \map g x$ then $\map f x \ne \map g x$
so:
:if $\lambda \map f x \ne \lambda \map g x$ then $x \in N$. 
So, we have: 
:$\lambda \cdot f = \lambda \cdot g$ $\mu$-almost everywhere.
{{qed}}
Category:Almost-Everywhere Equality Relation
2c30095chs1zayatsg9lo0z40ksiksa
Theorem: Equivalence of Definitions of Quasi-Reflexive Relation

Let $\RR \subseteq S \times S$ be a relation in $S$.
{{TFAE|def = Quasi-Reflexive Relation}}

Proof:
=== $(1)$ implies $(3)$ ===
Let $\RR$ be a quasi-reflexive relation by definition $1$.
Then by definition:
:$\forall x, y \in S: \tuple {x, y} \in \RR \implies \tuple {x, x} \in \RR \land \tuple {y, y} \in \RR$
That is:
:$\forall x, y \in S: \tuple {x, y} \in \RR \implies \tuple {x, x} \in \RR$
and:
:$\forall x, y \in S: \tuple {x, y} \in \RR \implies \tuple {y, y} \in \RR$
Hence:
:$\RR$ is a left quasi-reflexive relation by definition
and:
:$\RR$ is a right quasi-reflexive relation by definition.
Thus $\RR$ is a quasi-reflexive relation by definition $3$.
{{qed|lemma}}
Theorem: Reflexive Relation is Quasi-Reflexive

Let $\RR$ be a reflexive relation on a set $S$.
Then $\RR$ is a quasi-reflexive relation on $S$.

Proof:
By definition of reflexive relation:
:$\forall x \in S: \tuple {x, x} \in \RR$
Hence by definition of domain:
:$x \in \Dom \RR$
and hence by definition of field and Set is Subset of Union:
:$x \in \Field \RR$
That is:
:$\forall x \in \Field \RR: \tuple {x, x} \in \RR$
Hence the result by definition of quasi-reflexive relation.
{{qed}}
Category:Quasi-Reflexive Relations
Category:Reflexive Relations
5kwy3jhu7mk4c9pq8qeq244g7nvumc7
Theorem: Diagonal Relation is Reflexive (Class Theory)

Let $V$ be a basic universe.
Let $\Delta_V$ denote the diagonal relation on $V$:
:$\Delta_V = \set {\tuple {x, x}: x \in V}$
$\Delta_V$ is a reflexive relation.

Proof:
{{begin-eqn}}
{{eqn | q = \forall x \in V
      | l = x
      | r = x
      | c = {{Defof|Equals}}
}}
{{eqn | ll= \leadsto
      | l = \tuple {x, x}
      | o = \in
      | r = \Delta_V
      | c = {{Defof|Diagonal Relation}}
}}
{{end-eqn}}
So $\Delta_V$ is reflexive.
{{qed}}
Theorem: Pointwise Scalar Multiplication on Space of Real-Valued Measurable Functions Identified by A.E. Equality is Well-Defined

Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\map {\mathcal M} {X, \Sigma, \R}$ be the set of real-valued $\Sigma$-measurable functions on $X$. 
Let $\sim$ be the $\mu$-almost-everywhere equality relation on $\map {\mathcal M} {X, \Sigma, \R}$.
Let $\map {\mathcal M} {X, \Sigma, \R}/\sim$ be the space of real-valued $\Sigma$-measurable functions identified by $\sim$.
Then pointwise scalar multiplication on $\map {\mathcal M} {X, \Sigma, \R}/\sim$ is well-defined.

Proof:
Let $\lambda \in \R$.
Let $E \in \map {\mathcal M} {X, \Sigma, \R}/\sim$.
First, we show that if $E = \eqclass f \sim$, then $\eqclass {\lambda f} \sim$ is well-understood.
This follows from Pointwise Scalar Multiple of Measurable Function is Measurable.
We need to show that $\lambda \cdot E$ is independent of the choice of representative for $E$.
Suppose that: 
:$\eqclass f \sim = \eqclass g \sim = E$ 
From Equivalence Class Equivalent Statements, we have:
:$f \sim g$
So, from Pointwise Scalar Multiplication preserves A.E. Equality, we have: 
:$\lambda \cdot f \sim \lambda \cdot g$
That is, from Equivalence Class Equivalent Statements:
:$\eqclass {\lambda \cdot f} \sim = \eqclass {\lambda \cdot g} \sim$ 
which is what we aimed to show.
{{qed}}
Category:Space of Real-Valued Measurable Functions Identified by A.E. Equality
sr6jwor9nkayklbai0b857fy03u4kb0
Theorem: Subset Relation is Ordering/Class Theory

Let $C$ be a class.
Then the subset relation $\subseteq$ is an ordering on $C$.

Proof:
To establish that $\subseteq$ is an ordering, we need to show that it is reflexive, antisymmetric and transitive.
So, checking in turn each of the criteria for an ordering:
Theorem: Characterization of Polynomial has Root in P-adic Integers

Let $\Z_p$ be the $p$-adic integers for some prime $p$.
Let $\map F X \in \Z_p \sqbrk X$ be a polynomial over $\Z_p$.
Let $a \in \Z_p$.
Then:
:$\map F a  = 0$
{{iff}}:
:there exists a sequence $\sequence {a_n}$ of integers:
::$(1): \quad \ds \lim_{n \mathop \to \infty} {a_n} = a$
::$(2): \quad \map F {a_n} \equiv 0 \pmod {p^{n + 1} \Z_p}$

Proof:
=== Necessary Condition ===
{{:Characterization of Polynomial has Root in P-adic Integers/Necessary Condition}}{{qed|lemma}}
Theorem: Smallest Element is Unique/Class Theory

Let $V$ be a basic universe.
Let $\RR \subseteq V \times V$ be an ordering.
Let $A$ be a subclass of the field of $\RR$.
Suppose $A$ has a smallest element $s$ {{WRT}} $\RR$.
Then $s$ is unique.

Proof:
Let $s$ and $t$ both be smallest elements of $A$.
Then by definition:
:$\forall y \in A: s \mathrel \RR y$
:$\forall y \in A: t \mathrel \RR y$
Thus it follows that:
:$s \preceq t$
:$t \preceq s$
But as $\preceq$ is an ordering, it is antisymmetric.
Hence by definition of antisymmetric, $a = b$.
{{qed}}
Theorem: Greatest Element is Unique/Class Theory

Let $V$ be a basic universe.
Let $\RR \subseteq V \times V$ be an ordering.
Let $A$ be a subclass of the field of $\RR$.
Suppose $A$ has a greatest element $g$ {{WRT}} $\RR$.
Then $g$ is unique.

Proof:
Let $g$ and $h$ both be smallest elements of $A$.
Then by definition:
:$\forall y \in A: y \mathrel \RR g$
:$\forall y \in A: y \mathrel \RR h$
Thus it follows that:
:$g \preceq h$
:$h \preceq g$
But as $\preceq$ is an ordering, it is antisymmetric.
Hence by definition of antisymmetric, $g = h$.
{{qed}}
Theorem: Characterization of Polynomial has Root in P-adic Integers/Necessary Condition

Let $\Z_p$ be the $p$-adic integers for some prime $p$.
Let $\map F X \in \Z_p \sqbrk X$ be a polynomial over $\Z_p$.
Let $a \in \Z_p$.
Let $\map F a  = 0$.
Then:
:there exists a sequence $\sequence {a_n}$ of integers: 
::$\ds (1): \quad \lim_{n \mathop \to \infty} {a_n} = a$
::$(2): \quad \map F {a_n} \equiv 0 \pmod {p^{n + 1} \Z_p}$
where $\map F {a_n} \equiv 0 \pmod {p^{n + 1} \Z_p}$ denotes congruence modulo the ideal $p^{n + 1} \Z_p$

Proof:
Let $\map F a = 0$.
Let $\ds a = \sum_{j \mathop = 0}^\infty d_j p^j$ be the $p$-adic expansion of $a$.
For all $n \in \N_{>0}$, let:
:$\ds a_n = \sum_{j \mathop = 0}^{n - 1} d_j p^j$
By definition of $p$-adic expansion:
:$\ds (1): \quad \lim_{n \mathop \to \infty} {a_n} = a$
By definition of $p$-adic expansion of a $p$-adic integer:
:$\forall n \in \N_{>0} : a_n \in \Z$
We have:
{{begin-eqn}}
{{eqn | q = \forall n \in \N_{>0}
      | l = a_n
      | o = \equiv
      | r = a
      | rr= \pmod {p^{n + 1} \Z_p}
      | c = Partial Sum Congruent to $p$-adic Integer Modulo Power of $p$
}}
{{eqn | ll= \leadsto
      | n = 2
      | q = \forall n \in \N_{>0}
      | l = \map F {a_n}
      | o = \equiv
      | r = \map F a
      | c = Polynomials of Congruent Ring Elements are Congruent
}}
{{eqn | o = \equiv
      | r = 0
      | rr= \pmod {p^{n + 1} \Z_p}
      | c = as $\map F a = 0$
}}
{{end-eqn}}
Theorem: Characterization of Polynomial has Root in P-adic Integers/Sufficient Condition

Let $\Z_p$ be the $p$-adic integers for some prime $p$.
Let $\map F X \in \Z_p \sqbrk X$ be a polynomial over $\Z_p$.
Let $a \in \Z_p$.
Let there exist a sequence $\sequence{a_n}$ of integers: 
:$\ds (1): \quad \lim_{n \mathop \to \infty} {a_n} = a$
:$(2): \quad \map F {a_n} \equiv 0 \pmod {p^{n + 1} \Z_p}$
where $\map F {a_n} \equiv 0 \pmod {p^{n + 1} \Z_p}$ denotes congruence modulo the ideal $p^{n + 1} \Z_p$.
Then:
:$\map F a  = 0$

Proof:
Let there exist a sequence $\sequence{a_n}$ of integers: 
:$\ds (1): \quad \lim_{n \mathop \to \infty} {a_n} = a$
:$(2): \quad \map F {a_n} \equiv 0 \pmod {p^{n + 1} \Z_p}$
We have:
{{begin-eqn}}
{{eqn | q = \forall n \in \N
      | l = a
      | o = \equiv
      | r = a_n
      | rr= \pmod {p^{n + 1} \Z_p}
      | c = Partial Sum Congruent to $p$-adic Integer Modulo Power of $p$
}}
{{eqn | ll= \leadsto
      | q = \forall n \in \N
      | l = \map F a
      | o = \equiv
      | r = \map F {a_n}
      | c = Polynomials of Congruent Ring Elements are Congruent
}}
{{eqn | o = \equiv
      | r = 0
      | rr= \pmod {p^{n + 1} \Z_p}
      | c = from $(2)$
}}
{{eqn | ll= \leadsto
      | q = \forall n \in \N
      | l = \map F a
      | o = \in
      | r = p^{n + 1} \Z_p
      | c = {{Defof|Congruence Modulo Ideal}}
}}
{{eqn | ll= \leadsto
      | q = \forall n \in \N
      | l = \norm {\map F a}_p
      | o = \le
      | r = p^{-n - 1}
      | c = Characterization of Closed Ball in $p$-adic Numbers
}}
{{eqn | ll= \leadsto
      | l = \norm {\map F a}_p
      | o = \le
      | r = \lim_{n \mathop \to \infty} p^{-n - 1}
      | c = Squeeze Theorem for Real Sequences
}}
{{eqn | r = 0
      | c = Sequence of Powers of Number less than One
}}
{{eqn | ll= \leadsto
      | l = \norm {\map F a}_p
      | r = 0
      | c = $p$-adic norm is a mapping from $\Q_p$ to the non-negative reals
}}
{{eqn | ll= \leadsto
      | l = \map F a
      | r = 0
      | c = {{NormAxiomNonArch|1}}
}}
{{end-eqn}}
Theorem: Subclass of Subclass is Subclass

Let $A$, $B$ and $C$ be classes.
Let $A$ be a subclass of $B$.
Let $B$ be a subclass of $C$.
Then $A$ is a subclass of $C$.

Proof:
Let $x \in A$ be arbitrary.
It follows by definition of subclass that $x \in B$.
It further follows by definition of subclass that $x \in C$.
So we have that $x \in A$ implies that $x \in C$.
As $x$ is arbitrary, the result follows.
{{qed}}
Category:Subclasses
9i86a2uac5awae3004l5cgl1wu13l7g
Theorem: Subclass of Well-Ordered Class is Well-Ordered

Let $V$ be a basic universe.
Let $\RR \subseteq V \times V$ be a relation.
Let $A$ be a subclass of $V$ which is well-ordered under $\RR$.
Let $B$ be a subclass of $A$.
Then $B$ is also well-ordered under $\RR$.

Proof:
First suppose $B$ is the empty class.
From Empty Class is Subclass of All Classes, $B$ is a subclass of $A$.
Then by Empty Class is Well-Ordered, $\O$ is well-ordered under $\RR$.
Otherwise, let $X$ be an arbitrary non-empty class subclass of $B$.
By Subclass of Subclass is Subclass, $X$ is a subclass of $A$.
Hence by definition of well-ordered class, $X$ has a smallest element under $\RR$.
As $X$ is arbitrary, it follows that every non-empty class subclass of $B$ has a smallest element under $\RR$.
Hence the result by definition of well-ordered class.
{{qed}}
Theorem: Image of Convex Set under Linear Transformation is Convex

Let $\Bbb F \in \set {\R, \C}$.
Let $X$ and $Y$ be vector spaces over $\Bbb F$.
Let $C \subseteq X$ be convex.
Let $T : X \to Y$ be a linear transformation.
Then $\map T C \subseteq Y$ is convex.

Proof:
Let $y_1, y_2 \in \map T C$ and $\lambda \in \closedint 0 1$. 
Then, there exists $x_1, x_2 \in C$ such that: 
:$y_1 = T x_1$
and:
:$y_2 = T x_2$ 
Then, we have: 
:$\lambda y_1 + \paren {1 - \lambda} y_2 = \lambda T x_1 + \paren {1 - \lambda} T x_2 = \map T {\lambda x_1 + \paren {1 - \lambda} x_2}$
Since $C$ is convex, we have: 
:$\lambda x_1 + \paren {1 - \lambda} x_2 \in C$
So, we have: 
:$\map T {\lambda x_1 + \paren {1 - \lambda} x_2} \in \map T C$
That is: 
:$\lambda y_1 + \paren {1 - \lambda} y_2 \in \map T C$
So $\map T C$ is convex.
{{qed}}
Category:Convex Sets (Vector Spaces)
Category:Linear Transformations
luyau1j5tggj5mqqo4bsk1zr8azt1g1
Theorem: Open Ball Centred at Origin in Normed Vector Space is Symmetric

Let $\struct {X, \norm \cdot}$ be a normed vector space.
Let $\map B {0, r}$ be the open ball in $X$ centered at $0$ with radius $r$.
Then $\map B {0, r}$ is symmetric.

Proof:
Let $x \in \map B {0, r}$.
Then: 
:$\norm x < r$
We then have: 
:$\norm {-x} = \cmod {-1} \norm x = \norm x < r$
So $-x \in \map B {0, r}$.
So $\map B {0, r}$ is symmetric.
{{qed}}
Category:Symmetric Subsets of Vector Spaces
Category:Open Balls
bguvmmwlmpaxhz26mj9z9nl405ex5iu
Theorem: Image of Symmetric Set under Linear Transformation is Symmetric

Let $X$ and $Y$ be vector spaces over a subfield of $\C$.
Let $C$ be a symmetric subset of $X$.
Let $T : X \to Y$ be a linear transformation.
Then $\map T C$ is a symmetric subset of $Y$.

Proof:
Let $y \in \map T C$.
Then there exists $x \in C$ such that $y = T x$.
Then from linearity we have $-y = \map T {-x}$. 
Since $C$ is symmetric, we have $-x \in C$.
So $-y \in \map T C$.
So $\map T C$ is symmetric.
{{qed}}
Category:Symmetric Subsets of Vector Spaces
bakajtjrqucx8uln7bmkjbm01ekrc4r
Theorem: Closure of Symmetric Subset of Normed Vector Space is Symmetric

Let $\struct {X, \norm \cdot}$ be a normed vector space.
Let $A \subseteq X$ be symmetric.
Then the topological closure of $A$ is symmetric.

Proof:
Let $A^-$ be the topological closure of $A$.
Let $a \in A^-$.
Then from Point in Closure of Subset of Metric Space iff Limit of Sequence, we have: 
:there exists a sequence $\sequence {a_n}_{n \mathop \in \N}$ in $A$ such that $a_n \to a$. 
Since $A$ is symmetric, we have: 
:$-a_n \in A$ for each $n \in \N$.
We also have, from Multiple Rule for Sequences in Normed Vector Space:
:$-a_n \to -a$ 
So we have: 
:$-a \in A^-$
{{qed}}
Category:Set Closures
Category:Symmetric Subsets of Vector Spaces
drv7wweya95fqcsvec9410czzdvwh9z
Theorem: Empty Set is Well-Ordered

Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a relation on $S$.
Let $\O$ denote the empty set.
Let $\RR_\O$ denote the restriction of $\RR$ to $\O$.
Then $\struct {\O, \RR_\O}$ is a well-ordered set.

Proof:
Let $V$ be a basic universe.
By definition of basic universe, $\O$ is an element of $V$.
By the Axiom of Transitivity, $\O$ is a class.
The result follows from Empty Class is Well-Ordered.
{{qed}}
Theorem: Empty Class is Well-Ordered

Let $V$ be a basic universe.
Let $\RR \subseteq V \times V$ be a relation.
Let $\O$ denote the empty class.
Then $\O$ is well-ordered under $\RR$.

Proof:
We have that $\O$ is well-ordered under $\RR$ {{iff}} every non-empty subclass of $\O$ has a smallest element under $\RR$.
But $\O$ has no non-empty subclass.
Hence this condition is satisfied vacuously.
The result follows.
{{qed}}
Category:Empty Class
Category:Well-Orderings
k0efs1sd9vqr2i3dw69066dtmt5mbfx
Theorem: Infinite Series preserves Strict Inequality

Let $\sequence {x_n}_{n \mathop \in \N}$ and $\sequence {y_n}_{n \mathop \in \N}$ be real sequences such that: 
:$x_n \le y_n$ for all $n \in \N$
and:
:$x_m < y_m$ for at least one $m \in \N$.
Suppose that: 
:$\ds \sum_{n \mathop = 1}^\infty x_n$ and $\ds \sum_{n \mathop = 1}^\infty y_n$ both converge.
Then:
:$\ds \sum_{n \mathop = 1}^\infty x_n < \sum_{n \mathop = 1}^\infty y_n$

Proof:
We have: 
:$\ds \sum_{n \mathop = 1}^N x_n \le \sum_{n \mathop = 1}^N y_n$ for each $N \in \N$.
Then: 
:$\ds \sum_{n \mathop = 1}^N \paren {y_n - x_n} \ge 0$ for each $N \in \N$. 
From the Well-Ordering Principle, there exists a least $m \in \N$ such that $x_m < y_m$. 
Then for this $m$ we have $y_m - x_m > 0$.
So for $N \ge m$ we have: 
:$\ds \sum_{n \mathop = 1}^N \paren {y_n - x_n} \ge y_m - x_m > 0$
From Linear Combination of Convergent Series:
:$\ds \sum_{n \mathop = 1}^\infty \paren {y_n - x_n}$ converges and: 
::$\ds \sum_{n \mathop = 1}^\infty \paren {y_n - x_n} = \sum_{n \mathop = 1}^\infty y_n - \sum_{n \mathop = 1}^\infty x_n$
From Limits Preserve Inequalities, we have: 
:$\ds \sum_{n \mathop = 1}^\infty \paren {y_n - x_n} \ge y_m - x_m > 0$
That is: 
:$\ds \sum_{n \mathop = 1}^\infty y_n > \sum_{n \mathop = 1}^\infty x_n$
{{qed}}
Category:Series
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Theorem: Class which has Injection to Subclass of Well-Orderable Class is Well-Orderable

Let $B$ be a well-orderable class.
Let $A$ be a class such that there exists an injection $f: A \to C$, where $C$ is a subclass of $B$.
Then $A$ is a well-orderable class.

Proof:
Let $\RR$ be a well-ordering that can be established on $B$.
This can always be done, as $B$ is a well-orderable class.
Let $F$ be an injection that maps each element $x$ of $A$ to an element $\map F x$ of $B$.
Let $\preccurlyeq$ be the class of all ordered pairs $\tuple {x, y}$ of elements of $A$ such that $\tuple {\map F x, \map F y} \in \RR$.
Thus:
:$(1): \quad x \preccurlyeq y \iff \tuple {\map F x, \map F y} \in \RR$
It will be demonstrated that $A$ is well-ordered under $\preccurlyeq$.
Let $\Img F$ denote the image of $F$.
By Subclass of Well-Ordered Class is Well-Ordered, $\Img F$ is well-ordered under $\RR$.
We have {{hypothesis}} that $\RR$ is a well-ordering
Hence {{afortiori}}:
:$\RR$ is reflexive, transitive and antisymmetric
:Every pair of elements of $B$ is comparable under $\RR$
From reflexivity:
:$\forall a \in A: \forall a \in A: \tuple {\map F a, \map F a} \in \RR$
and so from $(1)$:
:$\forall a \in A: a \preccurlyeq a$
Thus $\preccurlyeq$ is reflexive on $A$.
Let $a, b, c, \in A$ such that $a \preccurlyeq b$ and $b \preccurlyeq c$.
Then from $(1)$:
:$\tuple {\map F a, \map F b} \in \RR$
and:
:$\tuple {\map F b, \map F c} \in \RR$
As $\RR$ is transitive:
:$\tuple {\map F a, \map F c} \in \RR$
Hence from $(1)$:
:$a \preccurlyeq c$
Thus $\preccurlyeq$ is transitive on $A$.
Let $a, b \in A$ such that $a \preccurlyeq b$ and $b \preccurlyeq a$.
Then from $(1)$:
:$\tuple {\map F a, \map F b} \in \RR$
and:
:$\tuple {\map F b, \map F a} \in \RR$
As $\RR$ is antisymmetric:
:$\map F a = \map F b$
Because $F$ is an injection:
:$a = b$
Thus $\preccurlyeq$ is antisymmetric on $A$.
Let $a, b \in A$ be arbitrary.
As every pair of elements of $B$ is comparable under $\RR$:
:$\tuple {\map F a, \map F b} \in \RR$ or $\tuple {\map F b, \map F a} \in \RR$
Hence from $(1)$:
:$a \preccurlyeq b$ or $b \preccurlyeq a$
Hence every pair of elements of $A$ is comparable under $\preccurlyeq$.
Thus {{apriori}} $\preccurlyeq$ is a total ordering on $A$.
Let $C$ be an arbitrary non-empty subclass of $A$.
Let $C'$ be the class of all $\map F x$ such that $x \in C$.
We have that $C'$ is a subclass of $B$.
Then as $B$ is well-ordered by $\RR$, it follows that $C'$ has a smallest element.
This smallest element is $\map F b$ for some $b \in C$.
Let $x \in C$.
Then:
:$\map F x \in C'$
and so:
:$\tuple {\map F b, \map F x} \in \RR$
Hence by $(1)$:
:$b \preccurlyeq x$
As $x$ is arbitrary:
:$\forall x \in C: b \preccurlyeq x$
and so $b$ is the smallest element of $C$ {{WRT}} $\preccurlyeq$.
It follows by definition that $\preccurlyeq$ is a well-ordering.
Thus $A$ is well-ordered by $\preccurlyeq$.
Hence the result.
{{qed}}
Theorem: Norm of Summation

Let $\struct {X, \norm \cdot}$ be a normed vector space.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $X$.
Then: 
:$\ds \norm {\sum_{k \mathop = 1}^n x_k} \le \sum_{k \mathop = 1}^n \norm {x_k}$ for all $n \in \N$.

Proof:
The proof proceeds by induction.
For each $n \in \N$, let $\map P n$ be the proposition:
:$\ds \norm {\sum_{k \mathop = 1}^n x_k} \le \sum_{k \mathop = 1}^n \norm {x_k}$ for all $n \in \N$.