avewright/proofwiki-math huggingface dataset:
Theorem: Lp Space is Vector Space

Let $\struct {X, \Sigma, \mu}$ be a measure space, and let $p \in \closedint 1 \infty$.
Let $\map \MM {X, \Sigma, \R}/\sim_\mu$ be the space of real-valued measurable functions identified by $\mu$-A.E. equality.
Let $\map {L^p} {X, \Sigma, \mu}$ be the $L^p$ space of $\struct {X, \Sigma, \mu}$.
Let $+$ denote pointwise addition on $\map \MM {X, \Sigma, \R}/\sim_\mu$.
Let $\cdot$ be pointwise scalar multiplication on $\map \MM {X, \Sigma, \R}/\sim_\mu$.
Then: 
:$\struct {\map {L^p} {X, \Sigma, \mu}, +, \cdot}_\R$
forms a vector space.

Proof:
It is sufficient to show that $\struct {\map {L^p} {X, \Sigma, \mu}, +, \cdot}_\R$ is a vector subspace of $\struct {\map \MM {X, \Sigma, \R}/\sim_\mu, +, \cdot}_\R$.
From $L^p$ Space is Subset of Space of Measurable Functions Identified by A.E. Equality:
:$\map {L^p} {X, \Sigma, \mu} \subseteq \map \MM {X, \Sigma, \R}/\sim_\mu$
Let:
:$\eqclass f {\sim_\mu}, \eqclass g {\sim_\mu} \in \map {L^p} {X, \Sigma, \mu}$
and:
:$\lambda \in \R$
Then:
:$f, g \in \map {\LL^p} {X, \Sigma, \mu}$
where $\map {\LL^p} {X, \Sigma, \mu}$ is the Lebesgue $p$-space $\struct {X, \Sigma, \mu}$. 
From the definition of pointwise addition on $\map \MM {X, \Sigma, \R}/\sim_\mu$: 
:$\eqclass f {\sim_\mu} + \lambda \eqclass g {\sim_\mu} = \eqclass {f + \lambda g} {\sim_\mu}$
From Lebesgue Space is Vector Space:
:$f + \lambda g \in \map {\LL^p} {X, \Sigma, \mu}$
So, from $L^p$ Space is Subset of Space of Measurable Functions Identified by A.E. Equality:
:$\eqclass {f + \lambda g} {\sim_\mu} \in \map {L^p} {X, \Sigma, \mu}$.
So, from the One-Step Vector Subspace Test:
:$\struct {\map {L^p} {X, \Sigma, \mu}, +, \cdot}_\R$ is a vector subspace of $\struct {\map \MM {X, \Sigma, \R}/\sim_\mu, +, \cdot}_\R$.
So $\struct {\map {L^p} {X, \Sigma, \mu}, +, \cdot}_\R$ forms a vector space as required.
{{qed}}
Category:Lp Spaces
Category:Vector Spaces
4srd75nr4rtfn421yx6uf0e414ehjar
Theorem: Ordering where every Subclass has Smallest Element is Well-Ordering

Let $A$ be a class.
Let $\RR$ be an ordering on $A$.
Let $\RR$ be such that every subclass of $A$ has a smallest element under $\RR$.
Then $\RR$ is a well-ordering.

Proof:
We are given that $\RR$ is an ordering.
It will be shown that $\RR$ is a total ordering, from which the fact that it is a well-ordering follows directly {{hypothesis}}.
Let $x$ and $y$ be elements of $A$.
Then $\set {x, y}$ is a subclass of $A$.
Hence {{hypothesis}} $\set {x, y}$ has a smallest element.
That is, either $x \mathrel \RR y$ or $y \mathrel \RR x$.
As $x$ and $y$ are arbitrary, it follows that:
:$\forall x, y \in A: x \mathrel \RR y \lor y \mathrel \RR x$
Hence, by definition, $\RR$ is a total ordering.
So, we have that $\RR$ is a total ordering on $A$ such that every subclass of $A$ has a smallest element under $\RR$.
That is, $\RR$ is a well-ordering.
{{qed}}
Theorem: Survival Function preserves Inequality

Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f, g : X \to \overline \R$ be $\Sigma$-measurable functions such that: 
:$\size f \le \size g$ $\mu$-almost everywhere.
Let $F_f$ and $F_g$ be the survival functions of $f$ and $g$ respectively.
Then: 
:$F_f \le F_g$

Proof:
We aim to show that: 
:$\map {F_f} \alpha \le \map {F_g} \alpha$ for all $\alpha \in \hointr 0 \infty$.
That is: 
:$\map \mu {\set {x \in X : \size {\map f x} \ge \alpha} } \le \map \mu {\set {x \in X : \size {\map g x} \ge \alpha} }$ for all $\alpha \in \hointr 0 \infty$.
Since $f \le g$ $\mu$-almost everywhere, there exists a $\mu$-null set such that: 
:if $x \in X$ has $\size {\map f x} > \size {\map g x}$ then $x \in N$.
From Measurable Functions Determine Measurable Sets, in particular we have: 
:$\set {x \in X : \size {\map f x} > \size {\map g x} }$ is $\mu$-null
from Null Sets Closed under Subset.
Let $\alpha \in \hointr 0 \infty$.
We have: 
{{begin-eqn}}
{{eqn	| l = \map \mu {\set {x \in X : \size {\map f x} \ge \alpha} }
		| r = \map \mu {\set {x \in X \setminus N : \size {\map f x} \ge \alpha} \cup \set {x \in N : \size {\map f x} \ge \alpha} }
}}
{{eqn	| r = \map \mu {\set {x \in X \setminus N : \size {\map f x} \ge \alpha} } + \map \mu {\set {x \in N : \size {\map f x} \ge \alpha} }
		| c = from the countable additivity of $\mu$
}}
{{eqn	| o = \le
		| r = \map \mu {\set {x \in X \setminus N : \size {\map f x} \ge \alpha} } + \map \mu N
		| c = Measure is Monotone
}}
{{eqn	| r = \map \mu {\set {x \in X \setminus N : \size {\map f x} \ge \alpha} }
}}
{{end-eqn}}
Swapping $f$ for $g$ in this computation also gives: 
:$\ds \map \mu {\set {x \in X : \size {\map g x} \ge \alpha} } = \map \mu {\set {x \in X \setminus N : \size {\map g x} \ge \alpha} }$
So we aim to show that: 
:$\ds \map \mu {\set {x \in X \setminus N : \size {\map f x} \ge \alpha} } \le \map \mu {\set {x \in X \setminus N : \size {\map g x} \ge \alpha} }$
From Measure is Monotone, it suffices to show:
:$\set {x \in X \setminus N : \size {\map f x} \ge \alpha} \subseteq \set {x \in X \setminus N : \size {\map g x} \ge \alpha}$
Let $x \in X \setminus N$ have: 
:$\size {\map f x} \ge \alpha$
Then, we have: 
:$\alpha \le \size {\map f x} \le \size {\map g x}$
So we have: 
:$\size {\map g x} \ge \alpha$
So, from the definition of set inclusion, we have: 
:$\set {x \in X \setminus N : \size {\map f x} \ge \alpha} \subseteq \set {x \in X \setminus N : \size {\map g x} \ge \alpha}$ for all $\alpha \in \hointr 0 \infty$
and hence the demand.
{{qed}}
Theorem: Bound on Survival Function of Pointwise Sum

Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f, g : X \to \overline \R$ be $\Sigma$-measurable functions such that the pointwise sum $f + g$ is well-defined.
Let $F_{f + g}$ be the survival function of the pointwise scalar multiple $f + g$. 
Let $F_f$ and $F_g$ be the survival functions of $f$ and $g$ respectively. 
Then: 
:$\ds \map {F_{f + g} } {\alpha + \beta} \le \map {F_f} \alpha + \map {F_g} \beta$ for all $\alpha, \beta \in \hointr 0 \infty$.

Proof:
Let $\alpha, \beta \in \hointr 0 \infty$.
We show that: 
:$\set {x \in X : \size {\map f x + \map g x} \ge \alpha + \beta} \subseteq \set {x \in X : \size {\map f x} \ge \alpha} \cup \set {x \in X : \size {\map g x} \ge \beta}$ 
Let $x \in X$ be such that $\size {\map f x + \map g x} \ge \alpha + \beta$.
If $\size {\map f x + \map g x} = \infty$, we have $\map f x + \map g x = \infty$ or $\map f x + \map g x = -\infty$. 
In the former case, we have $\map f x = \infty$ or $\map g x = \infty$. 
In the latter case, we have $\map f x = -\infty$ or $\map g x = -\infty$.
So in either case we have $\size {\map f x} = \infty$ or $\size {\map g x} = \infty$. 
So we either have $\size {\map f x} \ge \alpha$ or $\size {\map g x} \ge \beta$. 
So:
:$x \in \set {x \in X : \size {\map f x} \ge \alpha} \cup \set {x \in X : \size {\map g x} \ge \beta}$
in this case.
Now suppose that $\size {\map f x + \map g x} < \infty$. 
Suppose that $\size {\map f x} < \alpha$ and $\size {\map g x} < \beta$. 
Then from the Triangle Inequality, we have: 
:$\size {\map f x + \map g x} \le \size {\map f x} + \size {\map g x} < \alpha + \beta$ 
contradicting that:
:$\size {\map f x + \map g x} \ge \alpha + \beta$ 
So we must have $\size {\map f x} \ge \alpha$ or $\size {\map g x} \ge \beta$. 
So, we have: 
:$x \in \set {x \in X : \size {\map f x} \ge \alpha} \cup \set {x \in X : \size {\map g x} \ge \beta}$
again in this case.
So:
:$\set {x \in X : \size {\map f x + \map g x} \ge \alpha + \beta} \subseteq \set {x \in X : \size {\map f x} \ge \alpha} \cup \set {x \in X : \size {\map g x} \ge \beta}$ for all $\alpha, \beta \in \hointr 0 \infty$. 
by the definition of set inclusion.
Then we have: 
{{begin-eqn}}
{{eqn	| l = \map {F_{f + g} } {\alpha + \beta}
		| r = \map \mu {\set {x \in X : \size {\map f x + \map g x} \ge \alpha + \beta} }
		| c = {{Defof|Survival Function}}
}}
{{eqn	| o = \le
		| r = \map \mu {\set {x \in X : \size {\map f x} \ge \alpha} \cup \set {x \in X : \size {\map g x} \ge \beta} }
		| c = Measure is Monotone
}}
{{eqn	| o = \le
		| r = \map \mu {\set {x \in X : \size {\map f x} \ge \alpha} } + \map \mu {\set {x \in X : \size {\map g x} \ge \beta} }
		| c = Measure is Subadditive
}}
{{eqn	| r = \map {F_f} \alpha + \map {F_g} \beta
		| c = {{Defof|Survival Function}}
}}
{{end-eqn}}
{{qed}}
Theorem: Survival Function is Decreasing

Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f : X \to \overline \R$ be a $\Sigma$-measurable function.
Let $F_f$ be the survival function of $f$.
Then $F_f$ is a decreasing function.

Proof:
Let $\alpha, \beta \in \hointr 0 \infty$ with $\alpha \le \beta$.
We show: 
:$\map {F_f} \beta \le \map {F_f} \alpha$
We first show: 
:$\set {x \in X : \size {\map f x} \ge \beta} \subseteq \set {x \in X : \size {\map f x} \ge \alpha}$
Let $x \in X$ have $\size {\map f x} \ge \beta$.
Then, since $\beta \ge \alpha$ we have $\size {\map f x} \ge \alpha$.
So by the definition of set inclusion, we have: 
:$\set {x \in X : \size {\map f x} \ge \beta} \subseteq \set {x \in X : \size {\map f x} \ge \alpha}$
So, by Measure is Monotone we have: 
:$\map \mu {\set {x \in X : \size {\map f x} \ge \beta} } \le \map \mu {\set {x \in X : \size {\map f x} \ge \alpha} }$
So by the definition of the survival function we have: 
:$\map {F_f} \beta \le \map {F_f} \alpha$
So $F_f$ is a decreasing function.
{{qed}}
Theorem: Survival Function is Well-Defined

Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f: X \to \overline \R$ be a $\Sigma$-measurable function.
Then the survival function $F_f$ is well-defined.

Proof:
From Absolute Value of Measurable Function is Measurable:
:$\size f$ is $\Sigma$-measurable.
Then from Characterization of Measurable Functions, we have: 
:$\set {x \in X : \size {\map f x} \ge \alpha} \in \Sigma$ for all $\alpha \in \hointr 0 \infty$.
So:
:$\map \mu {\set {x \in X : \size {\map f x} \ge \alpha} }$
is well-understood, and the survival function $F_f$ is well-defined.
{{qed}}
Category:Survival Functions
Category:Examples of Well-Defined Mappings
2igr2ppyc0p07o2gnjs2xkr5ojzx4o2
Theorem: Restriction of Norm on Vector Space to Subspace is Norm

Let $\Bbb F$ be a subfield of $\C$. 
Let $X$ be a vector space over $\Bbb F$.
Let $\norm \cdot_X : X \to \hointr 0 \infty$ be a norm on $X$.
Let $Y$ be a vector subspace of $X$.
Let $\norm \cdot_Y$ denote the restriction of $\norm \cdot_X$ to $Y$.
Then $\norm \cdot_Y$ is a norm on $Y$.

Proof:
Since $\norm x_X \ge 0$ for any $x \in X$, we have $\norm y_Y \ge 0$ for any $y \in Y$.
We verify each of the axioms for a norm on a vector space.
Theorem: Closed Subspace of Banach Space forms Banach Space

Let $\struct {X, \norm \cdot_X}$ be a Banach space.
Let $Y$ be a closed linear subspace of $X$.
Let $\norm \cdot_Y$ be the restriction of $\norm \cdot_X$ to $Y$. 
Then $\struct {Y, \norm \cdot_Y}$ is a Banach space.

Proof:
From Restriction of Norm on Vector Space to Subspace is Norm, $\struct {Y, \norm \cdot_Y}$ is a normed vector space.
We now show that $\struct {Y, \norm \cdot_Y}$ is complete.
Let $\sequence {y_n}_{n \mathop \in \N}$ be a Cauchy sequence in $\struct {Y, \norm \cdot_Y}$. 
Then, for each $\epsilon > 0$ there exists $N \in \N$ such that: 
:$\norm {y_n - y_m}_Y < \epsilon$ 
for all $n, m \ge N$.
Since $\norm \cdot_Y$ is the restriction of $\norm \cdot_X$ to $Y$, we have: 
:$\norm {y_n - y_m}_Y = \norm {y_n - y_m}_X$
so:
:$\norm {y_n - y_m}_X < \epsilon$
So $\sequence {y_n}_{n \mathop \in \N}$ is a Cauchy sequence in $\struct {X, \norm \cdot_X}$.
Since $\struct {X, \norm \cdot_X}$ is Banach, we have that $\sequence {y_n}_{n \mathop \in \N}$ converges to $y \in X$.
Since $Y$ is closed in $\struct {X, \norm \cdot_X}$, we have that $Y$ contains all its limit points.
So $y \in Y$. 
From Sequence in Normed Vector Space Convergent to Limit iff Norm of Sequence minus Limit is Null Sequence, we therefore have: 
:$\norm {y_n - y}_X \to 0$
with $y_n - y \in Y$ for each $n \in \N$. 
Since $\norm \cdot_Y$ is the restriction of $\norm \cdot_X$ to $Y$, we have: 
:$\norm {y_n - y}_Y \to 0$
So $\sequence {y_n}_{n \mathop \in \N}$ converges to $y$ in $\struct {Y, \norm \cdot_Y}$. 
Since $\sequence {y_n}_{n \mathop \in \N}$ was arbitrary, we have that all Cauchy sequence in $\struct {Y, \norm \cdot_Y}$ converge.
{{qed}}
Theorem: Condition for Equivalence of Norms that induce Banach Spaces

Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a vector space over $\Bbb F$.
Let $\norm \cdot_1$ and $\norm \cdot_2$ be norms on $X$ such that $\struct {X, \norm \cdot_1}$ and $\struct {X, \norm \cdot_2}$ are Banach spaces.
Suppose that, for some real number $C > 0$: 
:$\norm x_2 \le C \norm x_1$ for all $x \in X$. 
Then $\norm \cdot_1$ and $\norm \cdot_2$ are equivalent.

Proof:
Consider $I : \struct {X, \norm \cdot_1} \to \struct {X, \norm \cdot_2}$ defined by: 
:$I x = x$ for each $x \in X$. 
Then for $x, y \in X$ and $\lambda \in \Bbb F$. 
Then we have: 
:$\map I {\lambda x + y} = \lambda x + y = \lambda I x + I y$
so $I$ is linear. 
Similarly, we have:
:$\norm {I x}_2 = \norm x_2 \le C \norm x_1$
so $I$ is a bounded linear transformation.
Clearly $I$ is a bijection, so by the Banach Isomorphism Theorem, the inverse $I^{-1} : \struct {X, \norm \cdot_2} \to \struct {X, \norm \cdot_1}$ is bounded.
So, there exists a real number $C' > 0$ such that: 
:$\norm {I x}_1 = \norm x_1 \le C' \norm x_2$
Then we have: 
:$\norm x_1 \le C' \norm x_2 \le C C' \norm x_1$
so that: 
:$\ds \frac 1 {C'} \norm x_1 \le \norm x_2 \le C \norm x_1$
for $C, C' > 0$. 
So $\norm \cdot_1$ and $\norm \cdot_2$ are equivalent.
{{qed}}
Category:Banach Spaces
f4a4x7yhhpop5aehyi3xbzibet1ezeh
Theorem: Condition for Total Ordering to be Well-Ordering

Let $A$ be a non-empty class under a total ordering $\preccurlyeq$.
Then $\preccurlyeq$ is a well-ordering on $A$ {{iff}} the following $3$ conditions are fulfilled:
:$(1): \quad A$ has a smallest element $x_0$
:$(2): \quad$ Every element of $A$ except the greatest (if there is one) has an immediate successor
:$(3): \quad$ For every proper lower section $L$ of $A$, if $L$ has no greatest element, then there is a smallest element of $A$ which is not in $L$.

Proof:
=== Sufficient Condition ===
Let $\preccurlyeq$ be a well-ordering on $A$.
Then by definition of well-ordering, there exists a smallest element of $A$ which we may call $x_0$.
Hence condition $(1)$ is fulfilled.
Let there exist an element $x$ of $A$ such that $x$ has no immediate successor.
Let $S$ be the subclass of $A$ defined as:
:$S = \set {y \in A: x \prec y}$
Suppose $S \ne \O$.
As $S \subset A$ it follows by definition of well-ordering that there exists a smallest element $z$ of $S$.
That is, there exists $z \in A$ such that $x \prec z$ but that there exists no $y \in A$ such that $x \prec y \prec z$.
Hence there can be no element $y \in A$ such that $x \prec y$ and so $S = \O$.
It follows that $x$ must be the greatest element of $A$.
Thus every element of $A$ except the greatest (if there is one) has an immediate successor.
That is, condition $(2)$ is fulfilled.
Let $L$ be a proper lower section of $A$ which has no greatest element.
As $L$ is a proper lower section of $A$, $L$ does not equal $A$.
Therefore the difference $A \setminus L \ne \O$.
But as $A \setminus L \subseteq A$ it follows by the fact that $\preccurlyeq$ is a well-ordering that $A \setminus L$ has a smallest element $z$.
This $z$ is the smallest element of $A$ which is not in $L$.
That is, condition $(3)$ is fulfilled.
{{qed|lemma}}
Theorem: Backward Path of Well-Ordering is Finite

Let $A$ be a class under a well-ordering $\preccurlyeq$.
Let $S$ be a backward path of $\preccurlyeq$ in $A$.
Then $S$ is finite.

Proof:
First suppose that $S$ has no terms.
Then $S$ is finite by default.
Hence let it be assumed that $S$ has at least $1$ term.
Let $T = \set {x: x \in \sequence {x_n} }$ be the range of $S$.
Then by definition $T$ is a subclass of $A$.
{{AimForCont}} $S$ is infinite.
By definition of well-ordering, $T$ has a smallest element, which can be referred to as $y$.
Hence:
:$\exists n \in \N: y = x_n$
But then by definition of backward path:
:$x_{n + 1} \prec x_n$
and so $y$ is not the smallest element of $T$ after all.
Hence $T$ can have no such smallest element.
From this contradiction we conclude that $S$ cannot be an infinite sequence.
Hence the result.
{{qed}}
Theorem: P-adic Norm forms Non-Archimedean Valued Field/Rational Numbers

The $p$-adic norm $\norm{\,\cdot\,}_p$ forms a non-Archimedean norm on the rational numbers $\Q$.
The rational numbers $\struct{\Q, \norm{\,\cdot\,}_p}$ with the $p$-adic norm is a valued field with a non-Archimedean norm.

Proof:
First we note that the $p$-adic norm is a norm.
Let $\nu_p$ denote the $p$-adic valuation on the rational numbers.
Recall the definition of the $p$-adic norm:
:<nowiki>$\forall q \in \Q: \norm q_p := \begin{cases}
  0 & : q = 0 \\
  p^{-\map {\nu_p} q} & : q \ne 0
\end{cases}$
</nowiki>
We must show the following holds for all $x, y \in \Q$:
:$\norm {x + y}_p \le \max \set {\norm x_p, \norm y_p}$
If $x = 0$ or $y = 0$, or $x + y = 0$, the result is trivial, as follows:
Let $x = 0$.
Then:
{{begin-eqn}}
{{eqn | l = x
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \norm x_p
      | r = 0
      | c = {{Defof|P-adic Norm|$p$-adic Norm}}
}}
{{eqn | ll= \leadsto
      | l = \max \set {\norm x_p, \norm y_p}
      | r = \norm y_p
      | c = as $\norm y_p \ge 0 = \norm x_p$ from {{NormAxiomNonArch|1}}
}}
{{eqn | r = \norm {x + y}_p
      | c = 
}}
{{end-eqn}}
and so $\norm {x + y}_p \le \max \left( \norm x_p, \norm y_p \right)$
The same argument holds for $y = 0$.
Let $x + y = 0$.
{{begin-eqn}}
{{eqn | l = x + y
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \norm {x + y}_p
      | r = 0
      | c = {{Defof|P-adic Norm|$p$-adic Norm}}
}}
{{eqn | o = \le
      | r = \max \set {\norm x_p, \norm y_p}
      | c = as $\norm x_p \ge 0$ and $\norm y_p \ge 0$ from {{NormAxiomNonArch|1}}
}}
{{end-eqn}}
Let $x, y, x + y \in \Q_{\ne 0}$.
From $p$-adic Valuation is Valuation:
:$\map {\nu_p} {x + y} \ge \min \set {\map {\nu_p} x, \map {\nu_p} y}$
Then:
{{begin-eqn}}
{{eqn | l = \norm {x + y}_p
      | r = p^{-\map {\nu_p} {x + y} }
      | c = {{Defof|P-adic Norm|$p$-adic Norm}}
}}
{{eqn | o = \le
      | r = \max \set {p^{- \map {\nu_p} x}, p^{-\map {\nu_p} y} }
}}
{{eqn | r = \max \set {\norm x_p, \norm y_p}
      | c = {{Defof|P-adic Norm|$p$-adic Norm}}
}}
{{end-eqn}}
{{qed}}
Theorem: P-adic Norm forms Non-Archimedean Valued Field/P-adic Numbers

Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Then:
:$\struct{\Q_p, \norm {\,\cdot\,}_p}$ is a valued field
:$\norm {\,\cdot\,}_p$ is a non-Archimedean norm
That is, the $p$-adic numbers $\struct {\Q_p, \norm {\,\cdot\,}_p}$ form a valued field with a non-Archimedean norm.

Proof:
Let $\norm {\,\cdot\,}^\Q_p$ be the $p$-adic norm on the rationals $\Q$.
From $p$-adic Norm on Rational Numbers is Non-Archimedean Norm:
:$\struct{Q, \norm {\,\cdot\,}^\Q_p}$ is a valued field with non-Archimedean norm $\norm {\,\cdot\,}_p$ 
By definition of the $p$-adic numbers:
:$\Q_p$ is the quotient ring $\CC \, \big / \NN$
where:
:$\CC$ is the commutative ring of Cauchy sequences over $\struct {\Q, \norm {\,\cdot\,}^\Q_p}$.
and
:$\NN$ is the set of null sequences in $\struct {\Q, \norm {\,\cdot\,}^\Q_p}$.
From {{Corollary|Quotient Ring of Cauchy Sequences is Normed Division Ring|1}}:
:$\struct {\Q_p, \norm {\, \cdot \,}_p}$ is a valued field.
From Completion of Normed Division Ring:
:$\struct {\Q_p, \norm {\, \cdot \,}_p}$ is a normed division ring completion of $\struct {\Q, \norm {\,\cdot\,}^\Q_p}$
From Non-Archimedean Division Ring iff Non-Archimedean Completion:
:$\norm {\, \cdot \,}_p$ on $\Q_p$ is a non-Archimedean norm.
{{qed}}
Category:P-adic Norm forms Non-Archimedean Valued Field
nxep6gda8gj632wacpgdlsvx9rnv3jp
Theorem: Composition of Linear Transformations is Linear Transformation

Let $K$ be a field.
Let $X, Y, Z$ be vector spaces over $K$.
Let $T_1 : X \to Y$ and $T_2 : Y \to Z$ be linear transformations.
Then the composition $T_2 \circ T_1 : X \to Z$ is a linear transformation.

Proof:
Let $\lambda \in K$ and $u, v \in X$. 
Then, we have: 
{{begin-eqn}}
{{eqn	| l = \map {\paren {T_2 \circ T_1} } {\lambda u + v}
		| r = \map {T_2} {\map {T_1} {\lambda u + v} }
}}
{{eqn	| r = \map {T_2} {\lambda T_1 u + T_1 v}
		| c = {{Defof|Linear Transformation}}
}}
{{eqn	| r = \lambda \paren {T_2 T_1} u + \paren {T_2 T_1} v
}}
{{eqn	| r = \lambda \paren {T_2 \circ T_1} u + \paren {T_2 \circ T_1} v
}}
{{end-eqn}}
so $T_2 \circ T_1 : X \to Z$ is a linear transformation.
{{qed}}
Category:Linear Transformations
coif8omw2w4177q0lpw4ar1pvgeyx60
Theorem: Image of Vector Subspace under Linear Transformation is Vector Subspace

Let $K$ be a field.
Let $X$ and $Y$ be vector spaces over $K$. 
Let $U$ be a vector subspace of $X$.
Let $T : X \to Y$ be a linear transformation.
Then $T \sqbrk U$ is a vector subspace of $Y$.

Proof:
Since $U$ is a non-empty set, we can apply the One-Step Vector Subspace Test to $T \sqbrk U$.
Let $\lambda \in K$ and $u, v \in T \sqbrk U$.
Then there exists $x, y \in U$ such that: 
:$u = T x$
and:
:$v = T y$ 
Then: 
{{begin-eqn}}
{{eqn	| l = \lambda u + v
		| r = \lambda T x + T y
}}
{{eqn	| r = \map T {\lambda x + y}
		| c = {{Defof|Linear Transformation}}
}}
{{end-eqn}}
Since $U$ is a vector subspace, we have $\lambda x + y \in U$.
So $\map T {\lambda x + y} \in T \sqbrk U$.
That is, $\lambda u + v \in T \sqbrk U$.
So $T \sqbrk U$ is a vector subspace of $Y$.
{{qed}}
Category:Vector Subspaces
Category:Linear Transformations
dx3qjl342byhotlfphi9zm3joouwrit
Theorem: Equivalence of Formulations of Axiom of Choice/Formulation 1 iff Formulation 4

The following formulation of the Axiom of Choice:

Proof:
We note from Set equals Union of Power Set that:
:$x = \ds \map \bigcup {\powerset x}$
Setting $\powerset A =: s$, we see that from Formulation $1$:
:$\ds \paren {\O \notin \powerset A \implies \exists \paren {f: \powerset A \to \bigcup \powerset A}: \forall x \in \powerset A: \map f x \in x}$ 
That is:
:for every non-empty proper subset of $A$
there exists a mapping $f: \powerset A \to A$ such that:
:for every non-empty proper subset $x$ of $A$: $\map f x \in x$.
That is Formulation $4$ of the Axiom of Choice
Theorem: Equivalence of Definitions of Limit Point of Set/Definition from Open Neighborhood iff Definition from Relative Complement

Let $T = \struct {S, \tau}$ be a topological space.
Let $A \subseteq S$.
{{TFAE|def = Limit Point of Set|view = limit point}}
=== Definition from Open Neighborhood===
{{:Definition:Limit Point of Set/Definition from Open Neighborhood}}
=== Definition from Relative Complement===
{{:Definition:Limit Point of Set/Definition from Relative Complement}}

Proof:
The following equivalence holds:
{{begin-eqn}}
{{eqn | o =
      | c = There exists an open neighborhood $U$ of $x$ such that $A \cap \paren {U \setminus \set x} = \O$
}}
{{eqn | o = \leadstoandfrom
      | c = There exists an open neighborhood $U$ of $x$ such that $U \subseteq \paren{S \setminus A} \cup \set x$
      | cc = Modus Ponendo Tollens
}}
{{eqn | o = \leadstoandfrom
      | c = $\paren {S \setminus A} \cup \set x$ is a neighborhood of $x$
      | cc = {{Defof|Neighborhood of Point}}
}}
{{end-eqn}}
The result follows from the Rule of Transposition.
{{qed}}