nyuuzyou/wikis huggingface dataset:
Equivalence of Definitions of Meromorphic Function

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== $(1)$ implies $(2)$ ===
Let $f$ be a meromorphic function by definition $1$.
By Meromorphic Function is Quotient of Holomorphic Functions:
:$f$ can be expressed as the ratio of two holomorphic functions.
Thus $f$ is a meromorphic function by definition $2$.
=== $(2)$ implies $(1)$ ===
Let $f$ be a meromorphic function by definition $2$.
Thus $f$ is a meromorphic function by definition $1$.
== Sources ==
* 
Category:Meromorphic Functions
Equivalence of Definitions of Metric Space Continuity at Point

== Theorem ==
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $f: A_1 \to A_2$ be a mapping from $A_1$ to $A_2$.
Let $a \in A_1$ be a point in $A_1$.
=== $\epsilon$-$\delta$ Definition ===
=== Definition by Limits ===
=== $\epsilon$-Ball Definition ===
=== Definition by Neighborhood ===
== Proof ==
=== $\epsilon$-$\delta$ Definition iff Definition by Limits ===
This is proved in Metric Space Continuity by Epsilon-Delta.
=== $\epsilon$-$\delta$ Definition iff $\epsilon$-Ball Definition ===
This is proved in Metric Space Continuity by Open Ball.
=== $\epsilon$-Ball Definition iff Definition by Neighborhood ===
This is proved in Metric Space Continuity by Neighborhood.
Category:Continuous Mappings on Metric Spaces
Equivalence of Definitions of Metrizable Topology

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $d$ be a metric on $S$ such that $\tau$ is the topology induced by $d$.
From Identity Mapping is Homeomorphism:
:$T$ is homeomorphic to a topological space with a topology induced by a metric.
=== Definition 2 implies Definition 1 ===
Let $M = \struct{A, d}$ be a metric space such that $T$ is homeomorphic to $\struct{A,\tau_d}$ where $\tau_d$ is the topology induced by $d$.
Let $\phi : \struct{S, \tau} \to \struct{A, \tau_d}$ be a homeomorphism.
Let $d_\phi : S \times S \to \R_{\ge 0}$ be the mapping defined by:
:$\forall s,t \in S: \map {d_\phi} {s,t} = \map d {\map \phi s, \map \phi t}$
==== Lemma 1 ====
==== Lemma 2 ====
It remains to show that $\tau$ is the topology induced by the metric $d_\phi$.
We have:
Hence $\tau$ is a topology induced by a metric by definition.
Category:Metrizable Topologies
Category:Equivalence of Definitions of Metrizable Topology
Equivalence of Definitions of Metrizable Topology/Lemma 1

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $M = \struct {A, d}$ be a metric space.
Let $\tau_d$ be the topology induced by $d$ on $A$.
Let $\phi: \struct {S, \tau} \to \struct {A, \tau_d}$ be a homeomorphism between $\struct {S, \tau}$ and $\struct {A, \tau_d}$.
Let $d_\phi: S \times S \to \R_{\ge 0}$ be the mapping defined by:
:$\forall s, t \in S: \map {d_\phi} {s,t} = \map d {\map \phi s, \map \phi t}$
Then:
:$d_\phi$ is a metric on $S$.
== Proof ==
We note that by definition $d: M \to \R$ is a metric on $M$.
Hence $d$ satisfies all the metric space axioms.
=== $d_\phi$ satisfies ===
We have:
Hence $d_\phi$ satisfies .
=== $d_\phi$ satisfies ===
We have:
Hence $d_\phi$ satisfies .
=== $d_\phi$ satisfies ===
We have:
Hence $d_\phi$ satisfies .
=== $d_\phi$ satisfies ===
By definition of homeomorphism:
:$\phi$ is an injection
By definition of injection:
:$\forall s, t \in S : s \ne t \implies \map \phi s \ne \map \phi t$
We have:
Hence $d_\phi$ satisfies .
It follows that $d_\phi$ is a metric on $S$ by definition.
Category:Equivalence of Definitions of Metrizable Topology
Equivalence of Definitions of Metrizable Topology/Lemma 2

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $M = \struct {A, d}$ be a metric space.
Let $\tau_d$ be the topology induced by $d$ on $A$.
Let $\phi : \struct{S, \tau} \to \struct{A, \tau_d}$ be a homeomorphism between topological spaces $\struct{S, \tau}$ and $\struct{A, \tau_d}$.
Let $d_\phi : S \times S \to \R_{\ge 0}$ be the metric defined by:
:$\forall s,t \in S: \map {d_\phi} {s,t} = \map d {\map \phi s, \map \phi t}$
Then:
:$\forall U \subseteq S : U$ is open in $\struct{S, d_\phi}$ $\phi \sqbrk U$ is open in $\struct{A, d}$
== Proof ==
=== Lemma 3 ===
Let $U \subseteq S$.
=== Necessary Condition ===
Let $U$ be open in $\struct{S, d_\phi}$.
By definition of open set:
:$\forall s \in U: \exists \epsilon_s : \map {B_{\epsilon_s}} s \subseteq U$
From Lemma 3:
:$\forall s \in U: \exists \epsilon_s : \map {B_{\epsilon_s}} {\map \phi s} = \phi \sqbrk {\map {B_{\epsilon_s}} s} \subseteq \phi \sqbrk U$
By definition of open set:
:$\phi \sqbrk U$ is open in $\struct{A, d}$.
=== Sufficient Condition ===
Let $\phi \sqbrk U$ be open in $\struct{A, d}$.
By definition of open set:
:$\forall s \in U: \exists \epsilon_s : \map {B_{\epsilon_s}} {\map \phi s} \subseteq \phi \sqbrk U$
From Lemma 3:
:$\forall s \in U: \exists \epsilon_s : \phi \sqbrk {\map {B_{\epsilon_s}} s} = \map {B_{\epsilon_s}} {\map \phi s} \subseteq \phi \sqbrk U$
From Preimage of Subset is Subset of Preimage:
:$\forall s \in U: \exists \epsilon_s : \phi^{-1} \sqbrk {\phi \sqbrk {\map {B_{\epsilon_s}} s}} \subseteq \phi^{-1} \sqbrk {\phi \sqbrk {U}}$
By definition of homeomorphism:
:$\phi$ is an injection
From Preimage of Image of Subset under Injection equals Subset:
::$\forall s \in U: \exists \epsilon_s : \map {B_{\epsilon_s}} s = \phi^{-1} \sqbrk {\phi \sqbrk {\map {B_{\epsilon_s}} s}} \subseteq \phi^{-1} \sqbrk {\phi \sqbrk {U}} = U$
By definition of open set:
:$U$ is open in $\struct{S, d_\phi}$.
Category:Equivalence of Definitions of Metrizable Topology
Equivalence of Definitions of Metrizable Topology/Lemma 3

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $M = \struct {A, d}$ be a metric space.
Let $\tau_d$ be the topology induced by $d$ on $A$.
Let $\phi : \struct{S, \tau} \to \struct{A, \tau_d}$ be a homeomorphism between topological spaces $\struct{S, \tau}$ and $\struct{A, \tau_d}$.
Let $d_\phi : S \times S \to \R_{\ge 0}$ be the metric defined by:
:$\forall s,t \in S: \map {d_\phi} {s,t} = \map d {\map \phi s, \map \phi t}$
Then:
:$\forall s \in S, \epsilon \in \R_{\ge 0} : \phi \sqbrk {\map {B_\epsilon} s} = \map {B_\epsilon} {\map \phi s}$
where 
:$(1) \quad \map {B_\epsilon} s$ is the open ball in $\struct{S, d_\phi}$ with center $s$ and radius $\epsilon$
:$(2) \quad \map {B_\epsilon} {\map \phi s}$ is the open ball in $\struct{A, d}$ with center $\map \phi s$ and radius $\epsilon$
== Proof ==
Let $s \in S$.
Let $\epsilon \in \R_{\ge 0}$.
We have:
By set equality:
:$\map {B_\epsilon} s = \phi^{-1} \sqbrk {\map {B_\epsilon} {\map \phi s}}$
Hence:
:$\phi \sqbrk {\map {B_\epsilon} s} = \phi \sqbrk {\phi^{-1} \sqbrk {\map {B_\epsilon} {\map \phi s}}}$
By definition of homeomorphism:
:$\phi$ is a surjection
From Image of Preimage of Subset under Surjection equals Subset:
:$\phi \sqbrk {\phi^{-1} \sqbrk {\map {B_\epsilon} {\map \phi s}}} = \map {B_\epsilon} {\map \phi s}$
The result follows.
Category:Equivalence of Definitions of Metrizable Topology
Equivalence of Definitions of Minimal Element

== Theorem ==
Let $\struct {S, \RR}$ be a relational structure.
Let $T \subseteq S$ be a subset of $S$.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $x$ be an minimal element by definition 1.
That is:
:$(1): \quad \forall y \in T: y \mathrel \RR x \implies x = y$
:
:$\exists y \in T: y \mathrel {\RR^\ne} x$
Then by definition:
:$y \mathrel \RR x \land x \ne y$
which contradicts $(1)$.
Thus by Proof by Contradiction:
:$\nexists y \in T: y \mathrel {\RR^\ne} x$
That is $x$ is a minimal element by definition 2.
=== Definition 2 implies Definition 1 ===
Let $x$ be a minimal element by definition 2.
That is:
:$(2): \quad \nexists y \in T: y \mathrel {\RR^\ne} x$
:
:$\exists y \in T: y \mathrel \RR x: x \ne y$
That is:
:$\exists y \in T: y \mathrel {\RR^\ne} x$
which contradicts $(2)$.
Thus:
:$\forall y \in T: y \mathrel \RR x \implies x = y$
Thus $x$ is a minimal element by definition 1.
== Sources ==
* : $\S 14$: Order
Category:Order Theory
Equivalence of Definitions of Minimal Polynomial

== Theorem ==
Let $L / K$ be a field extension.
Let $\alpha \in L$ be algebraic over $K$.
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== 1 equals 2 ===
By Minimal Polynomial is Irreducible, it follows that the two are equal.
=== 1 equals 3 ===
Let $f \in K \sqbrk x$ be the monic polynomial of smallest degree such that $\map f \alpha = 0$.
Let $g \in K \sqbrk x$ be a polynomial.
If $f \divides g$, then $g = q f$ for some $q \in K \sqbrk x$.
Thus $\map g \alpha = \map f \alpha \, \map q \alpha = 0$.
Conversely, suppose $\map g \alpha = 0$.
By the Division Theorem for Polynomial Forms over Field, there exists $q, r \in K \sqbrk x$ such that:
:$g = q f + r$
and:
:$r = 0$ or $\deg r < \deg f$.
Evaluating this expression at $\alpha$ we find that:
:$\map g \alpha = \map q \alpha \, \map f \alpha + \map r \alpha \implies \map r \alpha = 0$
since $\map f \alpha = \map g \alpha = 0$.
But $f$ has minimal degree among the non-zero polynomials that are zero at $\alpha$.
Therefore $r = 0$.
Therefore:
:$g = q f$
That is, $f$ divides $g$.
Category:Minimal Polynomials
Equivalence of Definitions of Minimally Closed Class

== Theorem ==
Let $A$ be a class.
Let $g$ be a mapping on $A$.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
Let it be given that $A$ is closed under $g$.
=== $(1)$ implies $(2)$ ===
Let $A$ be a minimally closed class under $g$ by definition 1.
Then by definition:
:there exists $b \in A$ such that $A$ has no proper subclass $B$ such that:
:$b \in B$
:$B$ is closed under $g$.
Let $A$ have a subclass $C$ which is closed under $g$ such that $b \in C$.
Then by definition, $C$ is not a proper subclass of $A$.
Thus by definition of proper subclass:
:$C = A$
By definition of subclass:
:$A \subseteq C$
and:
:$C \subseteq A$
That is:
:$\forall x \in A: x \in C$
and:
:$\forall x \in C: x \in A$
That is:
$C$ contains all elements of $A$.
Thus $A$ is a minimally closed class under $g$ by definition 2.
=== $(2)$ implies $(1)$ ===
Let $A$ be a minimally closed class under $g$ by definition 2.
Then by definition:
:Every subclass of $A$ containing $b$ which is closed under $g$ contains all the elements of $A$.
Let $C$ be a subclass of $A$ which is closed under $g$ such that $b \in C$.
By definition of subclass:
:$C \subseteq A$
But :
:$\forall x \in A: x \in C$
That is:
:$A \subseteq C$
By definition of equality of classes it follows that:
:$A = C$
and so by definition $C$ cannot be a proper subclass of $A$.
Thus $A$ is a minimally closed class under $g$ by definition 1.
Category:Minimally Closed Classes
Equivalence of Definitions of Minimally Inductive Class

== Theorem ==
Let $A$ be a class.
Let $g$ be a mapping on $A$.
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== $(1)$ implies $(2)$ ===
Let it be given that $A$ is inductive under $g$.
Let $A$ be a minimally inductive class under $g$ by definition 1.
Then by definition:
:$A$ has no proper subclass $B$ such that $B$ is inductive under $g$.
Let $A$ have a subclass $C$ which is inductive under $g$.
Then by definition, $C$ is not a proper subclass of $A$.
Thus by definition of proper subclass:
:$C = A$
By definition of subclass:
:$A \subseteq C$
and:
:$C \subseteq A$
That is:
:$\forall x \in A: x \in C$
and:
:$\forall x \in C: x \in A$
That is:
$C$ contains all elements of $A$.
Thus $A$ is a minimally inductive class under $g$ by definition 2.
=== $(2)$ implies $(1)$ ===
Let it be given that $A$ is inductive under $g$.
Let $A$ be a minimally inductive class under $g$ by definition 2.
Then by definition:
:Every subclass of $A$ which is inductive under $g$ contains all the elements of $A$.
Let $C$ be a subclass of $A$ which is inductive under $g$.
By definition of subclass:
:$C \subseteq A$
But by hypothesis:
:$\forall x \in A: x \in C$
That is:
:$A \subseteq C$
By definition of equality of classes it follows that:
:$A = C$
and so by definition $C$ cannot be a proper subclass of $A$.
Thus $A$ is a minimally inductive class under $g$ by definition 1.
=== $(1)$ implies $(3)$ ===
Let $A$ be a minimally inductive class under $g$ by definition 1.
Then by definition:
By definition of inductive class:
:$\O \in A$
:$A$ is closed under $g$.
Putting these definitions together, we have that:
:No proper subclass of $A$ such that $\O \in A$ is closed under $g$.
That is, 
$A$ is minimally closed under $g$ with respect to $\O$
Thus $A$ is a minimally inductive class under $g$ by definition 3.
=== $(3)$ implies $(1)$ ===
Let $A$ be a minimally inductive class under $g$ by definition 3.
That is, $A$ is minimally closed under $g$ with respect to $\O$.
That is:
:$(1): \quad \O \in A$
:$(2): \quad$ No proper subclass of $A$ such that $\O \in A$ is closed under $g$.
Hence by definition of inductive class:
Thus $A$ is a minimally inductive class under $g$ by definition 1.
Category:Minimally Inductive Classes
Equivalence of Definitions of Modulo Multiplication

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== $(1)$ implies $(2)$ ===
Let $\times_m$ be the operation of multiplication modulo $m$ by definition $1$.
Then $\times_m$ is defined as:
:$\eqclass a m \times_m \eqclass b m = \eqclass {a b} m$
By the Division Theorem:
Hence $p \in \eqclass a m$
Thus:
Thus:
:$p q \in \eqclass {a b} m$
By the Division Theorem:
:$p q = s + k m$
for some $k \in \Z$ and $0 \le s < m$
Then:
:$s \in \eqclass {p q} m$
while $s$ equals the remainder after $p q$ has been divided by $m$.
That is:
:$p \times_m q$ equals the remainder after $p q$ has been divided by $m$.
Thus $\times_m$ is the operation of multiplication modulo $m$ by definition $2$.
=== $(2)$ implies $(1)$ ===
Let $\times_m$ be the operation of multiplication modulo $m$ by definition $2$.
Then by definition:
:$\forall a, b \in \Z_m: a \times_m b$ equals the remainder after $a b$ has been divided by $m$
where:
:$\Z_m = \set {0, 1, \ldots, m - 1}$
Thus by definition of residue class:
Let $s$ equal the remainder after $a b$ has been divided by $m$:
:$a b = s + k m$
where $k \in \Z$.
Hence:
:$s \in \eqclass {a b} m$
So as Modulo Multiplication is Well-Defined:
:$\eqclass a m \times_m \eqclass b m = \eqclass {a b} m$
Thus $\times_m$ is the operation of multiplication modulo $m$ by definition $1$.
=== $(2)$ implies $(3)$ ===
Let $\times_m$ be the operation of multiplication modulo $m$ by definition $2$.
Then by definition:
:$\forall a, b \in \Z_m: a \times_m b$ equals the remainder after $a b$ has been divided by $m$
where:
:$\Z_m = \set {0, 1, \ldots, m - 1}$
Let $s$ equal the remainder after $a b$ has been divided by $m$.
That is, there exists $k \in \Z$ such that:
:$a b = s + k m$
where $0 \le s < m$.
It follows that:
:$k m \le a b$
but:
:$\paren {k + 1} m > a b$
Thus we have:
where $k$ is the largest integer such that $k m \le a b$.
Thus $\times_m$ is the operation of multiplication modulo $m$ by definition $3$.
=== $(3)$ implies $(2)$ ===
Let $\times_m$ be the operation of multiplication modulo $m$ by definition $3$.
Then by definition:
:$a \times_m b := a b - k m$
where $k$ is the largest integer such that $k m \le a b$.
Let $s$ be defined as:
:$s = a b - k m$
By definition:
:$0 \le s < m$
That is, $s$ equals the remainder after $a b$ has been divided by $m$.
Thus $\times_m$ is the operation of multiplication modulo $m$ by definition $2$.
Category:Modulo Multiplication
Equivalence of Definitions of Nilradical of Ring

== Theorem ==
Let $A$ be a commutative ring.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
By Nilpotent Element is Contained in Prime Ideals, $\Nil A$ is contained in the intersection of all prime ideals.
It remains to prove the other inclusion.
Let $f \in A$ be not nilpotent.
Let $S$ be the set of ideals of $A$ that are disjoint from $\set {f^n: n \in \N}$.
By Zorn's Lemma, $S$ has a maximal element $P$.
In particular, $f \notin P$.
We want to show that $P$ is prime.
Let $a, b \in A$ with $a, b \notin P$.
Then the sums of ideals $\ideal a + P$ and $\ideal b + P$ contain $P$ strictly.
By the maximality of $P$, there exist $n, m \in \N$ with $f^n \in \ideal a + P$ and $f^m \in \ideal b + P$.
Then:
:$f^{m + n} \in \ideal {a b} + P$
Thus:
:$\ideal {a b} + P \notin S$
In particular:
:$a b \notin P$
Thus $P$ is prime, and so:
:$\ds f \notin \bigcap_{\mathfrak p \mathop \in \Spec A} \mathfrak p$
== Sources ==
* : Chapter $1$: Rings and Ideals: $\S$ Nilradical and Jacobson Radical
Category:Nilradicals of Rings
Equivalence of Definitions of Noetherian Module

== Theorem ==
Let $A$ be a commutative ring with unity.
Let $M$ be an $A$-module.
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Assume $N_1 \subseteq N_2 \subseteq N_3 \subseteq \cdots$ is an increasing sequence of submodules of $M$.
By Definition 1 $N := \bigcup_{i \mathop = 1}^\infty N_i$ is a finitely generated submodule of $M$.
Suppose that $N$ is generated by $a_1, \dotsc, a_k \in N$.
For all $i \in \set {1, \dots, k}$, there is some $j_i \in \N$, such that $a_i \in N_{j_i}$.
For $j := \max \set {j_1, \dots, j_k}$, we have:
:$a_1, \dotsc, a_k \in N_j$
Hence $N_j = N$.
This shows, that the increasing sequence stabilizes, as desired.
=== Definition 2 implies Definition 1 ===
Let $N$ be a submodule of $M$.
 $N$ is not finitely generated.
Any finitely generated submodule of $N$ is not equal to $N$.
So we can inductively choose a sequence:
:$a_i \in N \setminus \sequence {a_1, \dots, a_{i - 1} }$.
The chain:
:$\sequence {a_1} \subsetneq \sequence {a_1, a_2} \subsetneq \sequence {a_1, a_2, a_3} \subsetneq \dotsb$
is strictly increasing.
This contradicts Definition 2.
Thus $N$ is finitely generated.
=== Definition 2 iff Definition 3 ===
This follows by Increasing Sequence in Ordered Set Terminates iff Maximal Element.
== Also see ==
* Equivalence of Definitions of Noetherian Ring
Category:Noetherian Modules
Equivalence of Definitions of Noetherian Ring

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
=== Definition 4 ===
== Proof ==
=== Definition 2 iff Definition 3 ===
This follows by Increasing Sequence in Ordered Set Terminates iff Maximal Element.
=== Definition 2 implies Definition 1 ===
Assume there is an ideal $I$ which is not finitely generated.
For any finite set $\set {a_1, \dotsc, a_n}$ where $n \in \N$, the generated ideal is not equal to $I$.
Consider the chain:
:$\sequence {a_1} \subset \sequence {a_1, a_2} \subset \cdots$
This chain does not satisfy the ascending chain condition (note that $I$ has infinitely many elements by assumption). 
=== Definition 1 implies Definition 2 ===
Let there be a chain of ideals $I_1 \subset I_2 \subset \cdots$.
Then $\ds J = \bigcup_{n \mathop \ge 1} I_n$ is an ideal.
Let $J$ be finitely generated, by $\set {b_1, \dotsc b_m}$ for some $m \in \N$.
As the chain is ascending, there exists an ideal such that:
:$\set {b_1, \dotsc b_m} \subset I_k$
for some $k \in \N$.
It follows that:
:$I_k = \ideal {b_1, \dotsc, b_m}$
Hence:
:$\forall l \ge k: I_l = I_k$
=== Definition 1 iff Definition 4 ===
Let $A$ is a ring.
For any subset $B\subseteq A$, we have:
:$B$ is an ideal $B$ is a submodule.
:$B$ is a finitely generated ideal $B$ is a finitely generated module. 
The claim follows from these observations.
== Also see ==
* Equivalence of Definitions of Noetherian Module
Category:Noetherian Rings
Equivalence of Definitions of Noetherian Topological Space

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
=== Definition 4 ===
== Proof ==
=== Definition 2 implies Definition 4 ===
Let $A \subseteq \tau$ be non-empty.
 
 $A$ has no maximal elements.
That is:
:$\forall X \in A : \set {Y \in A : X \subsetneq Y} \ne \O$
Thus by axiom of choice there is a mapping:
:$f : A \to A$
such that:
:$\forall X \in A : \map f X \subsetneq X$
Choose $X_1 \in A$ arbitrarily.
Recursively, for $i = 1,2,\ldots$ let:
:$X_{i + 1} := \map f {X_i}$
so that:
:$X_1 \subsetneq X_2 \subsetneq X_3 \subsetneq \cdots$
This contradicts the ascending chain condition.
Therefore $A$ has a maximal element.
Category:Noetherian Topological Spaces
Equivalence of Definitions of Non-Archimedean Division Ring Norm

== Theorem ==
Let $\struct {R, +, \circ}$ be a division ring whose zero is denoted $0_R$.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
 
Let $\norm {\,\cdot\,} : R \to \R_{\ge 0}$ be a norm on a division ring satisfying:
It remains only to show that $\norm {\,\cdot\,}$ satisfies $(\text N 1)$ and $(\text N 2)$.
This follows from the definition of a norm on a division ring.
=== Definition 2 implies Definition 1 ===
Let $\norm{\,\cdot\,} : R \to \R_{\ge 0}$ satisfy the non-Archimedean norm axioms: $(\text N 1)$, $(\text N 2)$ and $(\text N 4)$.
To show that $\norm{\,\cdot\,}$ is a norm on a division ring satisfying $(\text N 4)$, it remains to show that $\norm{\,\cdot\,}$ satisfies:
Let $x, y \in R$.
, suppose $\norm x \le \norm y$.
From :
:$0 \le \norm x$
Then:
The result follows.
== Sources ==
* : $\S 2.1$: Absolute Values on a Field: Definition $2.1.1 \, (iv)$
* : $\S 1.2$ Normed fields: Remark $1.13$ 
Category:Norm Theory
Equivalence of Definitions of Non-Archimedean Vector Space Norm

== Theorem ==
Let $\struct {R, +, \circ}$ be a division ring with norm $\norm {\,\cdot\,}_R$.
Let $X$ be a vector space over $R$, with zero $0_X$.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
 
Let $\norm {\,\cdot\,} : X \to \R_{\ge 0}$ be a norm on a division ring satisfying:
It remains only to show that $\norm {\,\cdot\,}$ satisfies $(\text N 1)$ and $(\text N 2)$.
This follows from the definition of a norm on a division ring.
=== Definition 2 implies Definition 1 ===
Let $\norm{\,\cdot\,} : X \to \R_{\ge 0}$ satisfy the non-Archimedean norm axioms: $(\text N 1)$, $(\text N 2)$ and $(\text N 4)$.
To show that $\norm{\,\cdot\,}$ is a norm on a division ring satisfying $(\text N 4)$, it remains to show that $\norm{\,\cdot\,}$ satisfies:
Let $x, y \in X$.
, suppose $\norm x \le \norm y$.
From non-Archimedean norm axiom $(\text N 1)$ : Positive Definiteness:
:$0 \le \norm x$
Then:
The result follows.
Category:Norm Theory