avewright/proofwiki-math huggingface dataset:
Theorem: Reduction of Explicit ODE to First Order System

Let $\map {x^{\paren n} } t = \map F {t, x, x', \ldots, x^{\paren {n - 1} } }$, $\map x {t_0} = x_0$ be an explicit ODE with $x \in \R^m$.
Let there exist $I \subseteq \R$ such that there exists a unique particular solution:
:$x: I \to \R^m$
to this ODE.
Then there exists a system of first order ODEs:
:$y' = \map {\tilde F} {t, y}$
with $y = \tuple {y_1, \ldots, y_{m n} }^T \in \R^{m n}$ such that:
:$\tuple {\map {y_1} t, \ldots, \map {y_m} t} = \map x t$
for all $t \in I$ and $\map y {t_0} = x_0$.

Proof:
Define the mappings:
:$z_1, \ldots, z_n: I \to \R^m$
by:
:$z_j = x^{\paren {j - 1} }$, $j = 1, \ldots, n$
Then:
{{begin-eqn}}
{{eqn | l = z_1'
      | r = z_2
}}
{{eqn | o = \vdots
}}
{{eqn | l = z_{n - 1}'
      | r = z_n
}}
{{eqn | l = z_n'
      | r = \map F {t, z_1, \ldots, z_n}
}}
{{end-eqn}}
This is a system of $m n$ first order ODEs.
By construction:
:$\map {z_1} t = \map x t$
for all $t \in I$ and $\map {z_1} {t_0} = x_0$.
Therefore we can take:
:$y = \begin {pmatrix} z_1 \\ \vdots \\ z_{n - 1} \\ z_n \end {pmatrix}, \quad \tilde F: \begin {pmatrix} z_1 \\ \vdots \\ z_n \end{pmatrix} \mapsto \begin {pmatrix} z_2 \\ \vdots \\ z_n \\ \map F {t, z_1, \ldots, z_n} \end {pmatrix}$
{{qed}}
Theorem: Open Continuous Image of Paracompact Space is not always Countably Metacompact

Let $T_A = \struct {X_A, \tau_A}$ be a topological space which is paracompact.
Let $T_B = \struct {X_B, \tau_B}$ be another topological space.
Let $\phi: T_A \to T_B$ be a mapping which is both continuous and open.
Then it is not necessarily even the case that $T_B$ is countably metacompact, let alone paracompact.

Proof:
Let $T_X = \struct {X, \tau}$ be a countable discrete space.
Let $T_Y = \struct {\set {0, 1}, \tau_0}$ be a Sierpiński space.
Let $T_A = T_X \times T_Y$ be the product space of $T_X$ and $T_Y$ whose product topology is $\tau_A$
Then $T_A$ is paracompact from Product of Countable Discrete Space with Sierpiński Space is Paracompact.
Let $T_B = \struct {S, \tau_p}$ be a countable particular point space.
Then $T_B$ is not countably metacompact from Infinite Particular Point Space is not Countably Metacompact
Let $\phi: T_A \to T_B$ be a mapping defined as:
:<nowiki>$\forall \tuple {x, y} \in T_A: \map \phi {x, y} = \begin {cases}
p & : y = 0 \\
x & : y = 1 \end {cases}$</nowiki>
Then $\phi$ is open and continuous as follows:
Let $U \in \tau_A$ such that $U \ne \O$.
By construction of $T_A$, $U$ is of the form $\tuple {U_A, U_B}$ where $U_A \in \tau$ and $U_B \in \tau_0$.
By definition of $\tau_0$, $U_B \in \set {\set 0, \set {0, 1} }$.
It follows that $U$ contains elements of the form $\tuple {x, 0}$.
Hence by definition of $\phi$ we have that:
:$\exists \tuple {x, y} \in U: \map \phi {x, y} = p$
and so:
:$p \in \phi \sqbrk U$ 
By definition of $T_B$ it follows that $\phi \sqbrk U$ is an open set of $T_B$.
Thus $\phi$ maps from open sets to open sets and so $\phi$ is an open mapping.
{{qed|lemma}}
Let $V \in T_B$ be an open set such that $V \ne \O$.
Then by definition:
:$p \in V$
Let $F = \phi^{-1} \sqbrk V$.
As $p \in V$ we have that $F$ is of the form:
:$F = A \times B$
where:
:$A \subseteq X$
and:
:$B \in \set {\set 0, \set {0, 1} }$
Because $T_X$ is a countable discrete space, $A$ is open in $T_X$.
By definition of $T_Y$, $B$ is open in $T_Y$.
Hence by definition of product space, $A \times B = F$ is open in $T_A$.
Thus the preimage of $V$ under $\phi$ is an open set.
Hence $\phi$ is a continuous mapping.
{{qed|lemma}}
So we have constructed a mapping $\phi$ which is both open and continuous from a paracompact space to a space which is not countably metacompact.
{{qed}}
Theorem: Fixed Point Formulation of Explicit ODE

Let $x' = \map f {t, x}$ with $\map x {t_0} = x_0$ be an explicit ODE of dimension $n$.
For $a, b \in \R$, let $\XX = \map {\CC} {\closedint a b; \R^n}$ be the space of continuous functions on the closed interval $\closedint a b$.
Let $T: \XX \to \XX$ be the map defined by:
:$\ds \map {\paren {T x} } t = x_0 + \int_{t_0}^t \map f {s, \map x s} \rd s$
{{explain|Definition of $\paren {T x}$}}
Then a fixed point of $T$ in $\XX$ is a solution to the above ODE.

Proof:
Let $\map y t$ be a fixed point of the map $T$.
That is:
:$\ds \map y t = x_0 + \int_{t_0}^t \map f {s, \map y s} \rd s$
Then:
:$\ds \map y {t_0} = x_0 + \int_{t_0}^{t_0} \map f {s, \map y s} \rd s = x_0$
By the fundamental theorem of calculus we have that $y$ is differentiable, and for $t \in \closedint a b$:
{{begin-eqn}}
{{eqn | l = \map {y'} t 
      | r = \frac {\d} {\d t} \int_{t_0}^t \map f {s, \map y s} \rd s
      | c = Derivative of Constant
}}
{{eqn | r = \map f {t, \map y t}
      | c = Fundamental Theorem of Calculus
}}
{{end-eqn}}
This shows that $y$ is a solution to the ODE as claimed.
{{Qed}}
Category:Ordinary Differential Equations
6r44x14mstc6xr1q04weqsng3a8j8b6
Theorem: Cardinality of Infinite Union of Infinite Sets

Let $\kappa$ be an infinite cardinal.
Let $X_i$ be sets of cardinality at most $\kappa$ indexed by a set $I$ of cardinality at most $\kappa$.
Then their union $\ds \bigcup_{i \mathop \in I} X_i$ has cardinality at most $\kappa$.
Furthermore, if at least one of the $X_i$ is size $\kappa$, then the union has cardinality $\kappa$.

Proof:
We can assume that $I$ has cardinality $\kappa$ and that the sets $X_i$ are disjoint and all of size $\kappa$.
The more general case follows since these other possibilities can only decrease the cardinality of the union.
When at least one of the $X_i$ is size $\kappa$, then the union must be at least size $\kappa$ since it contains a copy of $X_i$.
The idea is to prove by transfinite induction that for limit ordinals $\omega_\alpha$, the order type of $\omega_\alpha \times \omega_\alpha$ is at most $\omega_\alpha$.
From this we can conclude that the cardinality of $\aleph_\alpha \times \aleph_\alpha$ is still $\aleph_\alpha$.
We can then identify $\ds \bigcup_{i \mathop \in I} X_i$ with some $\aleph_\alpha \times \aleph_\alpha$ to yield the theorem.
We begin by defining a well-ordering $\prec$ for each $\omega_\alpha \times \omega_\alpha$.
Let $\tuple {\beta_1, \beta_2} \prec \tuple {\gamma_1, \gamma_2}$ {{iff}} one of the following holds:
:$(1): \quad \max \beta_i < \max \gamma_i$
or:
:$(2): \quad \max \beta_i = \max \gamma_i$ and $\beta_1 < \gamma_1$
or:
:$(3): \quad \max \beta_i = \max \gamma_i$ and $\beta_1 = \gamma_1$ and $\beta_2 < \gamma_2$.
Informally rephrased, this says that one pair is less than another pair if its max is smaller or if their max is the same but it is lexicographically smaller.
Theorem: Zermelo's Well-Ordering Theorem

Let the Axiom of Choice be accepted.
Then every set is well-orderable.

Proof:
Let $S$ be an arbitrary set.
By assumption $S$ is well-orderable.
From Well-Orderable Set has Choice Function, $S$ has a choice function.
As $S$ is arbitrary, the result follows.
{{qed}}
Theorem: Zermelo's Well-Ordering Theorem

Let the Axiom of Choice be accepted.
Then every set is well-orderable.

Proof:
Let $\FF$ be an arbitrary collection of sets.
By assumption all sets can be well-ordered.
Hence the set $\bigcup \FF$ of all elements of sets contained in $\FF$ is well-ordered by some ordering $\preceq$.
By definition then, every subset of $\ds \bigcup \FF$ has a smallest element under $\preceq$.
Also, note that each set in $\FF$ is a subset of $\bigcup \FF$.
Thus, we may define the choice function $C$:
:$C: \FF \to \bigcup \FF, \map C X = \min X$
where $\min X$ is the smallest element of $X$ under $\preceq$.
{{qed}}
Theorem: Zermelo's Well-Ordering Theorem

Let the Axiom of Choice be accepted.
Then every set is well-orderable.

Proof:
Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
By the {{Axiom-link|Choice}}, there exists a choice function on $S$.
The result follows from Set with Choice Function is Well-Orderable.
{{qed}}
Theorem: Zermelo's Well-Ordering Theorem

Let the Axiom of Choice be accepted.
Then every set is well-orderable.

Proof:
Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
By the Axiom of Choice, there is a choice function $c$ defined on $\powerset S \setminus \set \O$.
We will use $c$ and the Principle of Transfinite Induction to define a bijection between $S$ and some ordinal.
Intuitively, we start by pairing $\map c S$ with $0$, and then keep extending the bijection by pairing $\map c {S \setminus X}$ with $\alpha$, where $X$ is the set of elements already dealt with.
=== Basis for the Induction ===
$\alpha = 0$
Let $s_0 = \map c S$.
=== Inductive Step ===
Suppose $s_\beta$ has been defined for all $\beta < \alpha$.
If $S \setminus \set {s_\beta: \beta < \alpha}$ is empty, we stop.
Otherwise, define:
:$s_\alpha := \map c {S \setminus \set {s_\beta: \beta < \alpha} }$
The process eventually stops, else we have defined bijections between subsets of $S$ and arbitrarily large ordinals.
{{handwaving|I think this can be fixed with Hartogs' Lemma (Set Theory)}}
Now, we can impose a well-ordering on $S$ by embedding it via $s_\alpha \to \alpha$ into the ordinal $\beta = \ds {\bigcup_{s_\alpha \mathop \in S} \alpha}$ and using the well-ordering of $\beta$.
{{qed}}
Theorem: Zermelo's Well-Ordering Theorem

Let the Axiom of Choice be accepted.
Then every set is well-orderable.

Proof:
Let $S$ be a non-empty set.
Let $C$ be a choice function for $S$.
Let $A$ be an arbitrary set.
Let the mapping $h$ be defined as:
:$\map h A = \begin {cases} \map C {S \setminus A} & : A \subsetneqq S \\ x & : \text {otherwise} \end {cases}$
where $x$ is an arbitrary element such that $x \notin S$.
The latter is known to exist from Exists Element Not in Set.
From the Transfinite Recursion Theorem: Formulation $5$, there exists a mapping $F$ on the class of all ordinals $\On$ such that:
:$\forall \alpha \in \On: \map F \alpha = \map h {F \sqbrk \alpha}$
It remains to show the following:
:$(1): \quad \exists \delta \in \On: \map F \delta \notin S$
:$(2): \quad$ Let $\beta$ be the smallest ordinal such that $\map F \beta \notin S$. Then:
::::$F \sqbrk \beta = S$
:::and:
::::$F \restriction \beta$ is injective
:::and so $\beta$ can be put into a one-to-one correspondence with $S$.
{{finish|Proof needed for both $1$ and $2$ above}}
Hence $S$ can be well-ordered.
Theorem: Zermelo's Well-Ordering Theorem is Equivalent to Axiom of Choice

Zermelo's Well-Ordering Theorem holds {{iff}} the Axiom of Choice holds.
That is, every set is well-orderable {{iff}} every collection of sets has a choice function.

Proof:
=== Necessary Condition ===
Suppose the Axiom of Choice holds.
Then Zermelo's Well-Ordering Theorem holds by Zermelo's Well-Ordering Theorem itself.
That is, every set is well-orderable.
{{qed|lemma}}
Theorem: Product Topology is Coarsest Topology such that Projections are Continuous

Let $\mathbb X = \family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be an indexed family of topological spaces where $I$ is an arbitrary index set.
Let $X$ be the cartesian product of $\mathbb X$:
:$\ds X := \prod_{i \mathop \in I} X_i$
Let $\tau$ be the product topology on $X$.
For each $i \in I$, let $\pr_i : X \to X_i$ be the corresponding projection which maps each ordered tuple in $X$ to the corresponding element in $X_i$:
:$\forall x \in X: \map {\pr_i} x = x_i$
Then $\tau$ is the coarsest topology on $X$ such that all the $\pr_i$ are continuous.

Proof:
The result follows from the definition of the product topology and Equivalence of Definitions of Initial Topology.
{{qed}}
Theorem: Directed Hamilton Cycle Problem is NP-complete

Both versions of the Directed Hamilton Cycle problem are NP-complete.

Proof:
=== Function Version is Reducible to Decision Version ===
The algorithm described below solves the function version of the problem with $O \left({n^2}\right)$ calls of the decision version of the problem.
;Input: The directed graph $G$
;Output: Either:
:A Hamilton cycle in $G$ if one exists
or:
:''no solution'' if not.
;Auxiliary Function: $f \left({G}\right) \to \left\{{0, 1}\right\}$ defined as:
:$f \left({G}\right) = 1$ {{iff}} $G$ has a Hamilton cycle
:$f \left({G}\right) = 0$ {{iff}} $G$ has no Hamilton cycle
If $f \left({G}\right) = 0$ output ''no solution''.
Pick a starting vertex.
Call it $v_0$.
The solution path starts at $v_0$.
While $G$ has more then one vertex:
: Pick one of the edges going out of $v_0$.
: Call that edge $\left({v_0, u}\right)$.
: If $f \left({G - \left({v_0, u} \right)}\right) = 1$ remove $\left({v_0, u}\right)$ from $G$.
: Otherwise:
::add $u$ to the end of the solution set
::remove the vertices $v_0$ and $u$ from $G$
::replace them with a new $v_0$ where:
:::$\left({v_0, w}\right)$ is in the new graph if $\left({u, w}\right)$ is in the old graph
::and:
:::$\left({w, v_0}\right)$ is in the new graph if $\left({w, v_0}\right)$ is in the old graph.
Finally add $v_0$ to the end of the solution path and we have our solution.
Every time $f \left({G}\right)$ is calculated at least one edge is removed.
Thus $f \left({G}\right)$ is called $O \left({n^2}\right)$ times.
This algorithm shows the functional problem polynomially reduces to the decision problem.
If the output of the functional problem is anything other than ''no solution'' then $G$ has a Hamilton cycle.
Thus the functional problem reduces to the the decision problem.
{{explain|The below does not follow from the above -- where is the "mutually polynomially reducible"? We have two statements that FP reduces to DP but no indication that DP reduces to FP. Is the second of the above statements backwards?}}
Because they are mutually polynomially reducible we can show that both are NP or NP-hard by showing that either of them are NP or NP-hard.
{{qed|lemma}}
Theorem: Condition for Closed Extension Space to be T0 Space

Let $T = \struct {S, \tau}$ be a topological space.
Let $T^*_p = \struct {S^*_p, \tau^*_p}$ be the closed extension space of $T$.
Then $T^*_p$ is a $T_0$ space {{iff}} $T$ is.

Proof:
By definition:
:$\tau^*_p = \set {U \cup \set p: U \in \tau} \cup \set \O$
Let $T = \struct {S, \tau}$ be a $T_0$ space.
Then:
:$\forall x, y \in S$ such that $x \ne y$, either:
::$\exists U \in \tau: x \in U, y \notin U$
:or:
::$\exists U \in \tau: y \in U, x \notin U$
Let $x, y \in S^*_p$ such that $x \ne y, x \ne p, y \ne p$.
Then:
:$\exists U \in \tau: x \in U, y \notin U$
or:
:$\exists U \in \tau: y \in U, x \notin U$
Now let $U^* = U \cup \set p$.
It follows that:
:$x \in U^*, y \notin U^*$
or:
:$y \in U^*, x \notin U^*$
Now consider $x, p \in S^*_p$ such that $x \ne p$.
We have that:
:$\set p \in \tau^*_p: p \in \set p, x \notin \set p$
So we have shown that:
:$\forall x, y \in S^*_p$ such that $x \ne y$, either:
::$\exists U \in \tau^*_p: x \in U, y \notin U$
:or:
::$\exists U \in \tau^*_p: y \in U, x \notin U$
and so $T^*_p$ is a $T_0$ space.
Now suppose that $T = \struct {S, \tau}$ is '''not''' a $T_0$ space.
Let $x, y \in S$ such that:
:$\forall U \in \tau: x \in U \implies y \in U$
:$\forall U \in \tau: y \in U \implies x \in U$
Now consider $U^* = U \cup \set p$ for any $U \in \tau$.
Then:
:As $x \in U \implies y \in U$ it follows that $x \in U^* \implies y \in U^*$.
:As $y \in U \implies x \in U$ it follows that $y \in U^* \implies x \in U^*$.
Hence $T^*_p$ is not a $T_0$ space.
{{qed}}
Theorem: Excluded Point Space is T0

Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.
Then $T$ is a $T_0$ space.

Proof:
Let $T$ be a trivial space.
That is, let $S = \set p$.
Then the result holds vacuously, as there are no two distinct points in $T$.
Now suppose $T$ is not trivial.
Then $\exists x \in S: x \ne p$.
Now we have that $\set x \subseteq T$ is open in $T$ such that $p \notin \set x$ but $x \in \set x$.
Finally, suppose that $x, y \in S: x \ne y, x \ne p, y \ne p$.
Then we have that (for example) $\set x \subseteq T$ is open in $T$ such that $x \in \set x$ but $y \notin \set x$
Hence the result.
{{qed}}
Theorem: Excluded Point Space is T0

Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.
Then $T$ is a $T_0$ space.

Proof:
We have:
:Excluded Point Topology is Open Extension Topology of Discrete Topology
:Discrete Space is $T_0$
Then by Condition for Open Extension Space to be $T_0$, as a discrete space is $T_0$ then so is its open extension.
{{qed}}
Theorem: Open Extension Topology is not T1

Let $T = \struct {S, \tau}$ be a topological space.
Let $T^*_{\bar p} = \struct {S^*_p, \tau^*_{\bar p} }$ be the open extension space of $T$.
Then $T^*_{\bar p}$ is not a $T_1$ space.

Proof:
By definition:
:$\tau^*_{\bar p} = \set {U: U \in \tau} \cup \set {S^*_p}$
Let $x \in S^*_p, x \ne p$.
Let $U = \set x$.
Then $U \in \tau^*_p$ such that $x \in U, p \notin U$.
But the only $v \in \tau^*_p$ such that $p \in V$ is the set $S^*_p$, and we have that $x \in S^*_p$.
So there is no $V \in \tau^*_p$ such that $x \notin V, p \in V$, by definition of the open extension topology.
Hence $T^*_{\bar p}$ can not be a $T_1$ space.
{{qed}}
Theorem: Extreme Value Theorem

Let $X$ be a compact metric space and $Y$ a normed vector space.
Let $f: X \to Y$ be a continuous mapping.
Then $f$ is bounded, and there exist $x, y \in X$ such that:
:$\forall z \in X: \norm {\map f x} \le \norm {\map f z} \le \norm {\map f y}$
where $\norm {\map f x}$ denotes the norm of $\map f x$.
Moreover, $\norm f$ attains its minimum and maximum.

Proof:
{{AxiomReview|ZF version of this proof (use Continuous Image of Compact Space is Compact/Corollary 3)}}
{{explain|What should be reviewed??}}
By Continuous Image of Compact Space is Compact, $f \sqbrk X \subseteq Y$ is compact.
Therefore, by Compact Subspace of Metric Space is Bounded, $f$ is bounded.
Let $\ds A = \inf_{x \mathop \in X} \norm {\map f x}$.
It follows from the definition of infimum that there exists a sequence $\sequence {y_n}$ in $X$ such that:
:$\ds \lim_{n \mathop \to \infty} \norm {\map f {y_n} } = A$
By Sequence of Implications of Metric Space Compactness Properties, $X$ is sequentially compact.
So there exists a convergent subsequence $\sequence {x_n}$ of $\sequence {y_n}$.
Let $\ds x = \lim_{n \mathop \to \infty} x_n$.
Since $f$ is continuous and a norm is continuous, it follows by Composite of Continuous Mappings at Point is Continuous that:
:$\ds \norm {\map f x} = \norm {\map f {\lim_{n \mathop \to \infty} x_n} } = \norm {\lim_{n \mathop \to \infty} \map f {x_n} } = \lim_{n \mathop \to \infty} \norm {\map f {x_n} } = A$
So $\norm f$ attains its minimum at $x$.
By replacing the infimum with the supremum in the definition of $A$, we also see that $\norm f$ attains its maximum by the same reasoning.
{{qed}}
Theorem: Continuous Functions on Compact Space form Banach Space

Let $X$ be a compact Hausdorff space.
Let $Y$ be a Banach space.
Let $\CC = \CC \struct {X; Y}$ be the space of continuous functions on $X$ valued in $Y$. 
Let $\norm {\,\cdot\,}_\infty$ be the supremum norm on $\CC$.
Then $\struct {\CC, \norm {\,\cdot\,}_\infty}$ is a Banach space.

Proof:
From Continuous Function on Compact Space is Bounded, every continuous mapping $f : X \to Y$ is bounded.
Hence $\map \CC {X; Y} = \map {\CC_b} {X; Y}$, where $\map {\CC_b} {X; Y}$ is the space of bounded continuous functions on $X$ valued in $Y$.
From Bounded Continuous Functions on Topological Space form Banach Space, we have that $\struct {\map {\CC_b} {X; Y}, \norm {\, \cdot \,}_\infty} = \struct {\CC, \norm {\,\cdot\,}_\infty}$ is a Banach space.
{{qed}}
Category:Functional Analysis
iqu7vyqu1qnmq8vsh6nt0cgvhsb9lgw
Theorem: Supremum Norm is Norm

Let $S$ be a set.
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space over $K \in \set {\R, \C}$.
Let $\BB$ be the set of bounded mappings $S \to X$.
Let $\norm {\, \cdot \,}_\infty$ be the supremum norm on $\BB$.
Then $\norm {\, \cdot \,}_\infty$ is a norm on $\BB$.
{{MissingLinks|Add a link that establishes that $\BB$ is a vector space}}

Proof:
First:
{{begin-eqn}}
{{eqn | l = \norm f_\infty
      | r = 0
}}
{{eqn | ll= \leadstoandfrom
      | l = \sup_{x \mathop \in S} \norm {\map f x}
      | r = 0
}}
{{eqn | ll= \leadstoandfrom
      | q = \forall x \in S
      | l = \norm {\map f x}
      | r = 0
      | c = since $\norm {\, \cdot \,}$ is a norm, and hence non-negative
}}
{{eqn | ll= \leadstoandfrom
      | q = \forall x \in S
      | l = \map f x
      | r = 0
      | c = since $\norm x = 0 \iff x = 0$
}}
{{eqn | ll= \leadstoandfrom
      | l = f
      | r = 0
      | c = 
}}
{{end-eqn}}
Now let $\lambda \in K, f \in \BB$
We have:
{{begin-eqn}}
{{eqn | l = \norm {\lambda f}_\infty
      | r = \sup_{x \mathop \in S} \norm {\lambda \map f x}
      | c = 
}}
{{eqn | r = \size \lambda_K \sup_{x \mathop \in S} \norm {\map f x}
      | c = Multiple of Supremum, and because $\norm {\, \cdot \,}$ is a norm
}}
{{eqn | r = \size \lambda_K \, \norm f_\infty
      | c = 
}}
{{end-eqn}}
Finally let $f, g \in \BB$.
We have:
{{begin-eqn}}
{{eqn | l = \norm {f + g}_\infty
      | r = \sup_{x \mathop \in S} \norm {\map f x + \map g x}
      | c = 
}}
{{eqn | o = \le
      | r = \sup_{x \mathop \in S} \paren {\norm {\lambda \map f x} + \norm {\lambda \map g x} }
      | c = because $\norm {\, \cdot \,}$ is a norm
}}
{{eqn | o = \le
      | r = \sup_{x \mathop \in S} \norm {\lambda \map f x} + \sup_{x \mathop \in S} \norm {\lambda \map g x}
      | c = Supremum of Sum
}}
{{eqn | r = \norm f_\infty + \norm g_\infty
      | c = 
}}
{{end-eqn}}
Thus $\norm {\, \cdot \,}_\infty$ has the defining properties of a norm.
{{qed}}
Category:Examples of Norms
Category:Supremum Norm
Category:Supremum Norm is Norm
oij2vnoc9jg4367n6hxy9mmzjut0vl3
Theorem: One-Step Vector Subspace Test

Let $V$ be a vector space over a division ring $K$ whose unity is $1_K$.
Let $U \subseteq V$ be a non-empty subset of $V$ such that:
:$\forall u, v \in U: \forall \lambda \in K: u + \lambda v \in U$
Then $U$ is a subspace of $V$.

Proof:
We start with observing that the properties:
:{{Vector-space-axiom|1}}
:{{Vector-space-axiom|2}}
:{{Vector-space-axiom|3}}
:{{Vector-space-axiom|4}}
are true for all elements of $V$.
Hence, since $U \subseteq V$, they hold for all elements of $U$ as well.
The same holds for the axioms:
:{{Abelian-group-axiom|C||underlying}}
:{{Abelian-group-axiom|1||underlying}}
From Vector Inverse is Negative Vector, we have for all $u \in U$:
:$u + \paren {-1_K} u = 0_V$
which {{hypothesis}} is an element of $U$.
Since $U$ is non-empty, this means $0_V \in U$.
Hence it is seen that axioms:
:{{Abelian-group-axiom|2||underlying}}
:{{Abelian-group-axiom|3||underlying}}
are satisfied.
The last axiom that remains is {{Abelian-group-axiom|0||underlying}}.
From {{Vector-space-axiom|4}} and {{hypothesis}}:
:$\forall u, v \in U: u + v = u + 1_K v \in U$
Having verified all the vector space axioms, we conclude that $U$ is a subspace of $V$.
{{qed}}
Theorem: Ax-Grothendieck Theorem

Let $f: \C^n \to \C^n$ be a polynomial map.
Let $f$ be injective.
Then $f$ is surjective.

Proof:
The proof proceeds as follows:
:$(1): \quad$ showing that the theorem can be captured using a first-order sentences in the language of rings
:$(2): \quad$ showing that the theorem is true for at least one field of every characteristic $p > 0$
:$(3): \quad$ applying the Lefschetz Principle (First-Order).