avewright/proofwiki-math huggingface dataset:
Theorem: Multiple Rule for Sequence in Normed Vector Space

Let $\Bbb F$ be a subfield of $\C$. 
Let $\struct {X, \norm \cdot}$ be a normed vector space over $\Bbb F$. 
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $X$ converging to $x$. 
Let $\lambda \in \Bbb F$. 
Then: 
:$\lambda x_n \to \lambda x$

Proof:
The case $\lambda = 0$ follows from Constant Sequence in Normed Vector Space Converges.
Now take $\lambda \ne 0$. 
Since $\sequence {x_n}_{n \mathop \in \N}$ converges to $x$, for each $\epsilon > 0$ there exists $N \in \N$ such that: 
:$\ds \norm {x_n - x} < \frac \epsilon {\cmod \lambda}$ for all $n \ge N$.
Noting that: 
:$\ds \cmod \lambda \norm {x_n - x} = \norm {\lambda x_n - \lambda x}$
from the norm axioms, we have:
:$\ds \norm {\lambda x_n - \lambda x} < \epsilon$ for all $n \ge N$. 
Since $\epsilon$ was arbitrary, it follows that: 
:$\lambda x_n \to \lambda x$
{{qed}}
Category:Convergent Sequences (Normed Vector Spaces)
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Theorem: Constant Sequence in Normed Vector Space Converges

Let $\Bbb F$ be a subfield of $\C$. 
Let $\struct {X, \norm \cdot}$ be a normed vector space over $\Bbb F$. 
Let $x \in X$.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence with $x_n = x$ for all $n \in \N$. 
Then: 
:$x_n \to x$

Proof:
We have: 
:$\norm {x_n - x} = 0$ 
for all $n \in \N$. 
So, for all $\epsilon > 0$, we have: 
:$\norm {x_n - x} < \epsilon$ for all $n \in \N$.
So:
:$x_n \to x$
{{qed}}
Category:Convergent Sequences (Normed Vector Spaces)
fzpigffj0s6m8ayucmia3ww46youo3z
Theorem: Banach Isomorphism Theorem

Let $\struct {X, \norm \cdot_X}$ and $\struct {Y, \norm \cdot_Y}$ be Banach spaces.
Let $T : X \to Y$ be a bijective bounded linear transformation.
Then the inverse of $T$ is a bounded linear transformation.
That is, $T$ is a normed vector space isomorphism.

Proof:
Let $B_X^-$ be the closed unit ball of $X$.
Let $B_Y^-$ be the closed unit ball of $Y$.
Let $T^{-1} : Y \to X$ be the inverse of $T$.
From Inverse of Linear Transformation is Linear Transformation, $T^{-1} : Y \to X$ is a linear transformation.
It remains to show that $T^{-1}$ is bounded.
Since $T$ is bijective, it is surjective.
So, by the Banach-Schauder Theorem:
:$T$ is an open mapping.
From Characterization of Open Linear Transformation between Normed Vector Spaces, there exists $\delta > 0$ such that: 
:$\delta B_Y^- \subseteq T \sqbrk {B_X^-}$
Then we have from Image of Subset under Mapping is Subset of Image and Image of Preimage under Mapping: 
:$T^{-1} \sqbrk {\delta B_Y^-} \subseteq B_X^-$
From Image of Dilation of Set under Linear Transformation is Dilation of Image, we have: 
:$T^{-1} \sqbrk {B_Y^-} \subseteq \delta^{-1} B_X^-$
From Norm of Bounded Linear Transformation in terms of Closed Unit Ball, $T^{-1}$ is therefore bounded.
{{qed}}
Theorem: Countable Set is Well-Orderable

Let $S$ be a countable set.
Then $S$ is well-orderable.

Proof:
By the Well-Ordering Principle, the set of natural numbers $\N$ under the usual ordering $\le$ forms a well-ordered set.
By definition of countable set, there exists an injection:
:$f: S \to \N$
Let $V$ be a basic universe.
By definition of basic universe:
:$S \in V$
and:
:$\N \in V$
By the {{axiom-link|Transitivity}}, both $S$ and $\N$ are classes.
From Class which has Injection to Subclass of Well-Orderable Class is Well-Orderable, it follows that $S$ is well-orderable.
Hence the result.
{{qed}}
Theorem: Rational Numbers are Well-Orderable

The set $\Q$ of rational numbers is well-orderable.

Proof:
From Rational Numbers are Countably Infinite, $\Q$ is a countable set.
The result follows from Countable Set is Well-Orderable.
{{qed}}
Theorem: Rational Numbers are Well-Orderable

The set $\Q$ of rational numbers is well-orderable.

Proof:
The rational numbers are arranged thus:
:$\dfrac 0 1, \dfrac 1 1, \dfrac {-1} 1, \dfrac 1 2, \dfrac {-1} 2, \dfrac 2 1, \dfrac {-2} 1, \dfrac 1 3, \dfrac 2 3,  \dfrac {-1} 3, \dfrac {-2} 3, \dfrac 3 1, \dfrac 3 2, \dfrac {-3} 1, \dfrac {-3} 2, \dfrac 1 4, \dfrac 3 4, \dfrac {-1} 4, \dfrac {-3} 4, \dfrac 4 1, \dfrac 4 3, \dfrac {-4} 1, \dfrac {-4} 3 \ldots$
It is clear that every rational number will appear somewhere in this list.
Thus it is possible to set up a bijection $f: \Q \to \N$ between each rational number and its position in the list, which is an element of $\N$.
We set up an ordering $\preccurlyeq$ defined as:
:$\forall x, y \in \Q: x \preccurlyeq y \iff \map f x \le \map f y$
where $\le$ is the usual ordering on $\N$.
It remains to demonstrate that $\preccurlyeq$ is a well-ordering.
{{finish|Tedious exercise of limited utility and maximum futility}}
Theorem: Well-Ordering is not necessarily Usual Ordering

Let $S$ be a set of numbers.
According to Zermelo's Well-Ordering Theorem, $S$ can be well-ordered.
However, the usual ordering on $S$ may not necessarily be a well-ordering.

Proof:
From Rational Numbers are Well-Orderable, it is possible to apply a well-ordering to the set of rational numbers $\Q$.
However, the usual ordering on $\Q$ is not a well-ordering.
Indeed:
:$\set {x \in \Q: x \le 0}$
has no smallest element.
{{qed}}
Theorem: Poincaré Recurrence Theorem

Let $\struct {X, \BB, \mu, T}$ be a measure-preserving dynamical system.
Then for each $A \in \BB$:
:$\ds \map \mu {A \setminus \bigcap_{N \mathop = 1}^\infty \bigcup_{n \mathop = N}^\infty T^{-n} \sqbrk A} = 0$
That is, for $\mu$-almost all $x\in A$ there are integers $0 < n_1 < n_2 < \cdots$ such that $\map {T^{n_i} }x \in A$.

Proof:
Let:
:$\ds A_\infty := \bigcap_{N \mathop = 1}^\infty \bigcup_{n \mathop = N}^\infty T^{-n} \sqbrk A$
For $N \in \N$, let:
:$\ds A_N := \bigcup_{n \mathop = N}^\infty T^{-n} \sqbrk A$
so that:
:$\ds A_\infty = \bigcap _{N \mathop = 1} ^\infty A_N$
Now, we need to show:
:$\ds \map \mu {A \setminus A_\infty} = 0$
First, for all $N \in \N$:
{{begin-eqn}}
{{eqn | l = A_N
      | r = \bigcup_{n \mathop = N}^\infty T^{-n} \sqbrk A
}}
{{eqn | o = \supseteq
      | r = \bigcup_{n \mathop = N+1}^\infty T^{-n} \sqbrk A
      | c = 
}}
{{eqn | o = \supseteq
      | r = A_{N + 1}
}}
{{end-eqn}}
On the other hand, for all $N \in \N$:
{{begin-eqn}}
{{eqn | l = T^{-1} \sqbrk {A_N}
      | r = T^{-1} \sqbrk {\bigcup_{n \mathop = N}^\infty T^{-n} \sqbrk A}
}}
{{eqn | r = \bigcup_{n \mathop = N}^\infty T^{-1} \sqbrk {T^{-n} \sqbrk A}
      | c = Preimage of Union under Mapping/Family of Sets
}}
{{eqn | r = \bigcup_{n \mathop = N+1}^\infty T^{-n} \sqbrk A
}}
{{eqn | r = A_{N + 1}
}}
{{end-eqn}}
Hence, because $T$ is a $\mu$-preserving transformation:
:$\map \mu {A_N} = \map \mu {T^{-1} \sqbrk {A_N} } = \map \mu {A_{N + 1} }$
Therefore:
{{begin-eqn}}
{{eqn | l = \map \mu {A \setminus A_\infty}
      | r = \map \mu {A_0 \setminus A_\infty}
      | o = \le
      | c = Measure is Monotone, $A \subseteq A_0$
}}
{{eqn | r = \map \mu {A_0 \setminus \bigcap_{N \mathop = 1}^\infty A_N}
}}
{{eqn | r = \map \mu {\bigcup_{N \mathop = 1} ^\infty {A_0 \setminus A_N} }
      | c = De Morgan's Law
}}
{{eqn | o = \le
      | r = \sum_{N \mathop = 1}^\infty \map \mu {A_0 \setminus A_N}
      | c = Measure is Countably Subadditive
}}
{{eqn | r = \sum_{N \mathop = 1}^\infty \paren {\map \mu {A_0} - \map \mu {A_N} }
      | c = Measure of Set Difference with Subset
}}
{{eqn | r = 0
}}
{{end-eqn}}
{{qed}}
{{Namedfor|Jules Henri Poincaré|cat = Poincaré}}
Theorem: Characterization of Integer has Square Root in P-adic Integers/Necessary Condition

Let $\Z_p$ be the $p$-adic integers for some prime $p \ne 2$.
Let $a \in Z$ be an integer such that $p \nmid a$.
Let the exist $x \in \Z_p$ such that $x^2 = a$.
Then:
:$a$ is a quadratic residue of $p$.
That is:
:an integer $a$ not divisible by $p$ has a square root in $\Z_p$ ($p \ne 2$)
{{iff}}:
:$a$ is a quadratic residue of $p$.

Proof:
Let there exist $x$ such that $x^2 = a$.
By definition of root of polynomial:
:$\map F X$ has a root in $\Z_p$.
From Characterization of Integer Polynomial has Root in $p$-adic Integers:
:there exists an integer sequence $\sequence {a_n}$ such that:
::$(1): \quad \ds \lim_{n \mathop \to \infty} {a_n} = a$
::$(2): \quad \map F {a_n} \equiv 0 \pmod {p^{n + 1} }$
We have:
:$a_0^2 - a \equiv 0 \pmod p$
That is:
:$a_0^2 \equiv a \pmod p$
Hence by definition:
:$a$ is a quadratic residue of $p$.
Theorem: Characterization of Integer has Square Root in P-adic Integers/Sufficient Condition

Let $\Z_p$ be the $p$-adic integers for some prime $p \ne 2$.
Let $a \in Z$ be an integer such that $p \nmid a$.
Let $a$ be a quadratic residue of $p$.
Then:
:$\exists x \in \Z_p : x^2 = a$

Proof:
Let $a$ be a quadratic residue of $p$.
By definition of quadratic residue of $p$:
:$\exists b \in \Z : a \equiv b^2 \pmod p$
Then:
:$\map F b = b^2 - a \equiv 0 \pmod p$
and
:$\map {F'} b = 2b$
By hypothesis:
:$p \nmid 2$
and
:$p \nmid b^2$
From the contrapositive statement of Divisor Divides Multiple:
:$p \nmid b$
From the contrapositive statement of Euclid's Lemma for Prime Divisors:
:$p \nmid 2b$
Hence:
:$\map {F'} b = 2b \not\equiv 0 \pmod p$
From Congruence Modulo Equivalence for Integers in $p$-adic Integers:
:$\map F b \equiv 0 \pmod {p \Z}$
and
:$\map {F'} b \not\equiv 0 \pmod {p \Z}$
From Hensel's Lemma for $p$-adic Integers:
:$\exists x \in \Z_p : \map F x = 0$
That is:
:$\exists x \in \Z_p : x^2 = a$
Theorem: Categories of Elements under Well-Ordering

Let $A$ be a class.
Let $\preccurlyeq$ be a well-ordering on $A$.
Let $x \in A$ be an element of $A$.
Then $x$ falls into one of the following $3$ categories:
:$(1): \quad x$ is the smallest element of $A$ {{WRT}} $\preccurlyeq$
:$(2): \quad x$ is the immediate successor of another element $y \in A$ {{WRT}} $\preccurlyeq$
:$(3): \quad x$ is a limit element of $A$ under $\preccurlyeq$.

Proof:
Let $x \in A$.
;$(1): \quad x$ is the smallest element of $A$ {{WRT}} $\preccurlyeq$:
We note that $\preccurlyeq$ is a well-ordering on $A$.
Hence as $A \subseteq A$ we have that $A$ has a smallest element.
Hence there exists one element of $A$ which is that smallest element of $A$.
Let $x$ be that smallest element of $A$.
By definition of smallest element:
:$\forall y \in A: x \preccurlyeq y$
Hence $x$ cannot be the immediate successor of another element $y \in A$ {{WRT}} $\preccurlyeq$.
;$(2): \quad x$ is the immediate successor of another element $y \in A$ {{WRT}} $\preccurlyeq$:
Let $y \in A$ such that $y$ is not the greatest element of $A$ {{WRT}} $\preccurlyeq$.
Then:
:$\forall y \in A: \exists x \in A: y \preccurlyeq x$
such that $x$ is an immediate successor of $y$.
By definition of immediate successor, it follows that $x$ is not the smallest element of $A$ {{WRT}} $\preccurlyeq$.
;$(3): \quad x$ is a limit element of $A$ under $\preccurlyeq$:
Suppose there exists $x$ such that:
:$x$ is not the smallest element of $A$ {{WRT}} $\preccurlyeq$
and:
:$x$ is not the immediate successor of another element $y \in A$ {{WRT}} $\preccurlyeq$.
Then by definition $x$ is a limit element of $A$ under $\preccurlyeq$.
The result follows.
{{qed}}
Theorem: Characterization of P-adic Unit has Square Root in P-adic Units/Condition 1 implies Condition 2

Let $\Z_p$ be the $p$-adic integers for some prime $p \ne 2$.
Let $Z_p^\times$ be the set of $p$-adic units. 
Let $u \in Z_p^\times$ be a $p$-adic unit.
Let $u = c_0 + c_1p + c_2p^2 + \ldots$ be the $p$-adic expansion of $u$.
Let there exist $x \in \Z_p^\times$ such that $x^2 = u$.
Then:
:$c_0$ is a quadratic residue of $p$

Proof:
Let there exist $x \in \Z_p^\times$ such that $x^2 = u$
Then:
$x^2 \equiv c_0 \pmod {p\Z_p}$
Let $x = x_0 + x_1p + x_2p^2 + x_3p^3 + \ldots$ be the $p$-adic expansion of $x$.
From Partial Sum Congruent to P-adic Integer Modulo Power of p:
:$x \equiv x_0 \pmod {p\Z_p}$
Then:
:$x_0^2 \equiv x^2 \equiv c_0 \pmod {p\Z_p}$
By definition of quadratic residue:
:$c_0$ is a quadratic residue of $p$
Theorem: Characterization of P-adic Unit has Square Root in P-adic Units/Condition 2 implies Condition 3

Let $\Z_p$ be the $p$-adic integers for some prime $p \ne 2$.
Let $Z_p^\times$ be the set of $p$-adic units. 
Let $u \in Z_p^\times$ be a $p$-adic unit.
Let $u = c_0 + c_1p + c_2p^2 + \ldots$ be the $p$-adic expansion of $u$.
Let $c_0$ be a quadratic residue of $p$.
Then:
:$\exists y \in \Z_p : y^2 \equiv u \pmod{p\Z_p}$

Proof:
Let $c_0$ be a quadratic residue of $p$.
By definition of quadratic residue:
:$\exists x_0 \in \Z : x_0^2 \equiv c_0 \pmod {p\Z_p}$
Then:
:$x_0^2 \equiv u \pmod {p\Z_p}$
Theorem: Characterization of P-adic Unit has Square Root in P-adic Units/Condition 3 implies Condition 1

Let $\Z_p$ be the $p$-adic integers for some prime $p \ne 2$.
Let $Z_p^\times$ be the set of $p$-adic units. 
Let $u \in Z_p^\times$ be a $p$-adic unit.
Let there exist $y \in \Z_p$ such that $y^2 \equiv u \pmod {p \Z_p}$.
Then:
:$\exists x \in \Z_p^\times : x^2 = u$

Proof:
Let there exist $y \in \Z_p$ such that $y^2 \equiv u \pmod {p\Z_p}$
Let $y = y_0 + y_1 p + y_2 p^2 + y_3 p^3 + \ldots$ be the $p$-adic expansion of $y$.
By definition of $p$-adic expansion:
:$y_0 \in {0, 1, \ldots, p - 1}$
From $p$-adic Expansion of $p$-adic Unit:
:$y_0 \ne 0$
Hence:
:$y_0 \in {1, 2, \ldots, p - 1}$
It follows that:
:$p \nmid y_0$
From Partial Sum Congruent to $p$-adic Integer Modulo Power of $p$:
:$y \equiv y_0 \pmod {p \Z_p}$
Then:
:$y_0^2 \equiv y^2 \equiv u \pmod {p \Z_p}$
Let $F \in \Z \sqbrk X$ be the polynomial:
:$\map F X = X^2 - u$
By definition of formal derivative of $F$ is:
:$\map {F'} X = 2X$
We have:
:$\map F {y_0} \equiv 0 \pmod p$
and
:$\map {F'} {y_0} = 2 y_0$
By hypothesis:
:$p \nmid 2$
From the contrapositive statement of Euclid's Lemma for Prime Divisors:
:$p \nmid 2 y_0$
Hence:
:$\map {F'} {y_0} = 2 y_0 \not \equiv 0 \pmod p$ 
From Congruence Modulo Equivalence for Integers in P-adic Integers:
:$\map F {y_0} \equiv 0 \pmod {p \Z}$
and
:$\map {F'} {y_0} \not \equiv 0 \pmod {p \Z}$
From Hensel's Lemma for $p$-adic Integers:
:$\exists x \in \Z_p : \map F x = 0$
That is:
:$\exists x \in \Z_p : x^2 = u$
It remains to show that $x \in \Z_p^\times$.
We have:
{{begin-eqn}}
{{eqn | l = \norm x_p^2
      | r = \norm {x^2}_p
      | c = {{NormAxiomNonArch|2}}
}}
{{eqn | r = \norm u_p
      | c = 
}}
{{eqn | r = 1
      | c = $p$-adic Unit has Norm Equal to One
}}
{{eqn | ll= \leadsto
      | l = \norm x_p
      | r = 1
      | c =
}}
{{eqn | ll= \leadsto
      | l = x
      | o = \in
      | r = \Z_p^\times
      | c = $p$-adic Unit has Norm Equal to One
}}
{{end-eqn}}
Theorem: Lower Closure is Strict Lower Closure of Immediate Successor

Let $\struct {S, \preccurlyeq}$ be a totally ordered set.
Let $b$ be the immediate successor element of $a$:
Then:
:$a^\preccurlyeq = b^\prec$
where:
:$a^\preccurlyeq$ is the lower closure of $a$
:$b^\prec$ is the strict lower closure of $b$.

Proof:
Let:
:$x \in b^\prec$
By the definition of strict upper closure:
:$x \prec b$
By the definition of total ordering:
:$a \prec x$ or $x \preccurlyeq a$
If $a \prec x$ then $a \prec x \prec b$, contradicting the premise.
Thus:
:$x \preccurlyeq a$
and so:
:$x \in a^\preccurlyeq$
By definition of subset:
:$b^\prec \subseteq a^\preccurlyeq$
Let:
:$x \in a^\preccurlyeq$
By the definition of upper closure:
:$x \preccurlyeq a$
Since $a \prec b$, Extended Transitivity shows that $x \prec b$.
Thus:
:$x \in b^\prec$
By definition of subset:
:$a^\preccurlyeq \subseteq b^\prec$
Therefore by definition of set equality:
:$a^\succ = b^\succcurlyeq$
{{qed}}
Category:Total Orderings
Category:Lower Closures
625bsipdr2ebfsezfj95xtodty9p8e7
Theorem: Characterisation of Limit Element under Well-Ordering

Let $A$ be a class.
Let $\preccurlyeq$ be a well-ordering on $A$.
Let $x \in A$ be an element of $A$ such that $x$ is not the smallest element of $A$ under $\preccurlyeq$.
Then:
:$x$ is a limit element of $A$ under $\preccurlyeq$
{{iff}}:
:$x^\prec$ has no greatest element {{WRT}} $\preccurlyeq$
where $x^\prec$ denotes the strict lower closure of $x$ in $A$ under $\preccurlyeq$.

Proof:
Suppose $x^\prec$ has a greatest element $y$.
Then $x$ is the immediate successor of $y$, that is:
:$\nexists z \in A : y < z < x$
Hence $x$ is not a limit element.
Therefore, if $x$ is a limit element, then $x^\prec$ cannot have a greatest element.
{{qed|lemma}}
Suppose $x$ is an immediate successor.
Then the immediate predecessor of $x$ is the greatest element of $x^\prec$.
Therefore if $x^\prec$ has no greatest element then $x$ cannot be an immediate successor.
Hence if $x$ is not the smallest element of $A$, it must be a limit element.
{{qed}}
Theorem: Lp Norm is Norm

Let $\struct {X, \Sigma, \mu}$ be a measure space and let $p \in \hointr 1 \infty$.
Let $\map {L^p} {X, \Sigma, \mu}$ be the $L^p$ vector space on $\struct {X, \Sigma, \mu}$.
Let $\norm \cdot_p$ be the $L^p$ norm.
Then $\norm \cdot_p$ is a norm on $\map {L^p} {X, \Sigma, \mu}$.

Proof:
Let $\map {\mathcal L^p} {X, \Sigma, \mu}$ be the Lebesgue $p$-space. 
Let $\sim$ be the $\mu$-almost everywhere equality relation on $\map {\mathcal L^p} {X, \Sigma, \mu}$.
Let $\eqclass f \sim \in \map {L^p} {X, \Sigma, \mu}$. 
Then, we have by the definition of the $L^p$ norm we have: 
:$\norm {\eqclass f \sim}_p = \norm f_p$ 
From P-Seminorm is Seminorm, we have: 
:$\norm f_p \ge 0$
so:
:$\norm {\eqclass f \sim}_p \ge 0$
So $\norm \cdot_p$ is a map from $\map {L^p} {X, \Sigma, \mu}$ to the non-negative real numbers.
It remains to verify the norm axioms.
Theorem: Kernel of Linear Transformation between Finite-Dimensional Normed Vector Spaces is Closed

Let $m, n \in \N_{> 0}$ be natural numbers.
Let $A \in \R^{m \times n}$ be a matrix.
Let $\ker A = \set {\mathbf x \in \R^n : A \mathbf x = 0}$ be the kernel of $A$.
Then $\ker A$ is a closed subspace of $\R^n$.

Proof:
Let $T_A : \R^n \to \R^m$ be the linear transformation such that:
:$\forall \mathbf x \in \R^n : T_A \mathbf x := A \mathbf x$
By Linear Transformations between Finite-Dimensional Normed Vector Spaces are Continuous, $T_A : \R^n \to \R^m$ is continuous. 
We have that Singleton in Normed Vector Space is Closed. 
Hence, $\set {\mathbf 0}$ is closed in $\R^n$.
By the corollary of Mapping is Continuous iff Inverse Images of Open Sets are Open, the inverse image of $\set {\mathbf 0}$ under $T_A$ is closed in $\R^n$.
However:
:$\map {T^{-1}_A} {\set {\mathbf 0}} = \set {\mathbf x \in \R^n : A \mathbf x = 0} = \ker A$
Hence, $\ker A$ is closed in $\R^n$.
{{qed}}
Theorem: Equivalence of Definitions of Commutative Local Ring

Let $A$ be a commutative ring with unity.
{{TFAE|def = Commutative Local Ring}}

Proof:
=== Definition 1 implies Definition 2 ===
Let $\mathfrak m \subsetneq A$ be the unique maximal ideal.
First, $A$ is nontrivial, since $1 \notin \mathfrak m$.
Secondly, let $x, y \in A$ be non-units.
Let $\ideal x$ and $\ideal y$ be the principal ideals generated by $x$ and $y$, respectively.
In view of the unique maximality, $\ideal x \subseteq \mathfrak m$ and $\ideal y \subseteq \mathfrak m$.
In particular:
:$x, y \in \mathfrak m$
which implies:
:$x + y \in \mathfrak m$
Therefore $x + y$ is a non-unit.
{{qed|lemma}}
Theorem: Strict Lower Closure of Limit Element is Infinite

Let $A$ be a class.
Let $\preccurlyeq$ be a well-ordering on $A$.
Let $x \in A$ be a limit element of $A$ under $\preccurlyeq$.
Let $x^\prec$ denotes the strict lower closure of $x$ in $A$ under $\preccurlyeq$.
Then $x^\prec$ is an infinite set.

Proof:
Let $x \in A$ be a limit element of $A$ under $\preccurlyeq$.
From Characterisation of Limit Element under Well-Ordering it follows that $x^\prec$ has no greatest element {{WRT}} $\preccurlyeq$.
The result follows.
{{qed}}
Theorem: First Principle of Transfinite Induction

Let $A$ be a class.
Let $\preccurlyeq$ be a well-ordering on $A$.
Let $P$ be a property that satisfies the following condition:
:For all $x \in A$, if $P$ holds for every $y \prec x$, then $P$ holds for $x$.
Then $P$ holds for all $x \in A$.

Proof:
Let $s$ be the smallest element of $A$.
Then vacuously $P$ holds for every element $y \in A$ such that $y \prec s$.
Thus $P$ holds for $s$.
{{AimForCont}} $P$ fails to hold for some $z \in A$.
The class of elements of $A$ for which $P$ does not hold is therefore non-empty.
We have that $A$ is a well-ordered class.
Hence there must be some smallest element $x \in A$ for which $P$ fails to hold.
This cannot be $s$ as we have seen above.
But for every $y \prec x$, we have that $P$ holds for $y$.
But then {{hypothesis}} $P$ holds for $x$.
This contradicts our definition of $x$.
Hence there can be no such $z \in A$ for which $P$ fails to hold.
{{qed}}
Theorem: Second Principle of Transfinite Induction

Let $A$ be a class.
Let $\preccurlyeq$ be a well-ordering on $A$.
Let $P$ be a property that satisfies the following $3$ conditions:
:$(1): \quad P$ holds for the smallest element of $A$.
:$(2): \quad$ For all $x \in A$ which have an immediate successor $\map S x$, if $P$ holds for $x$, then $P$ holds for $\map S x$.
:$(3): \quad$ For all limit element $y$ of $A$, if $P$ holds for all $x \prec y$, then $P$ holds for $y$.
Then $P$ holds for all $x \in A$.

Proof:
{{AimForCont}} $P$ fails to hold for some $z \in A$.
The class of elements of $A$ for which $P$ does not hold is therefore non-empty.
We have that $A$ is a well-ordered class.
Hence there must be some smallest element $x \in A$ for which $P$ fails to hold.
By $(1)$, this cannot be the smallest element of $A$.
By $(2)$, this cannot be an immediate successor element.
By $(3)$, this cannot be a limit element.
From Categories of Elements under Well-Ordering, there are no other options for $x$.
Hence for all $x \in A$, {{hypothesis}}, $P$ holds for $x$.
This contradicts our definition of $x$.
Hence there can be no such $z \in A$ for which $P$ fails to hold.
{{qed}}
Theorem: Pointwise 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 almost-everywhere equality equivalence 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 $E_1, E_2 \in \map {\mathcal M} {X, \Sigma, \R}/\sim$.
We need to show that $E_1 \cdot E_2$ 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:
:$g \sim G$ 
So, from Pointwise Multiplication preserves A.E. Equality, we have: 
:$f \cdot g \sim F \cdot G$
That is, from Equivalence Class Equivalent Statements:
:$\eqclass {f \cdot g} \sim = \eqclass {F \cdot G} \sim$ 
which is what we aimed to show.
{{qed}}
Category:Space of Real-Valued Measurable Functions Identified by A.E. Equality
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