nyuuzyou/wikis huggingface dataset:
Equivalence of Definitions of Well-Ordering/Definition 1 implies Definition 2

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set such that:
:$\forall T \subseteq S: \exists a \in T: \forall x \in T: a \preceq x$
That is, such that $\preceq$ is a well-ordering by definition 1.
Then $\preceq$ is a total ordering.
== Proof ==
Consider $X = \set {a, b}$ where $a, b \in S$.
We have that $X$ has a smallest element.
So either $\min X = a$ or $\min X = b$.
If $\min X = a$, then $a \preceq b$.
If $\min X = b$, then $b \preceq a$.
So either $a \preceq b$ or $b \preceq a$.
That is, $a$ and $b$ are comparable.
As this applies to all $a, b \in S$, the ordering $\preceq$ is total.
== Sources ==
* : $\S 17$: Well Ordering
* : $\S 7$: Exercise $4$
Category:Equivalence of Definitions of Well-Ordering
Equivalence of Forms of Axiom of Countable Choice

== Theorem ==
The following forms of the Axiom of Countable Choice are equivalent in $\mathrm{ZF}^-$:
=== Form 1 ===
=== Form 2 ===
== Proof ==
=== Form 1 implies Form 2 ===
Suppose that Axiom of Countable Choice: Form 1 holds.
Let $S$ be a countable set of non-empty sets.
Then $S$ is either finite or countably infinite.
If $S$ is finite, then it has a choice function by the Principle of Finite Choice.
Suppose instead that $S$ is countably infinite.
Then there exists a bijection $f: \N \to S$.
Thus by Form 1, there exists a mapping $g: \N \to \bigcup S$ such that:
:$\forall n \in \N: g \left({n}\right) \in f \left({n}\right)$
Then $g \circ f^{-1}: S \to \bigcup S$ is a choice function for $S$.
As every countable set has a choice function, Axiom of Countable Choice: Form 2 holds.
=== Form 2 implies Form 1 ===
Suppose that Axiom of Countable Choice: Form 2 holds.
Let $\langle x_n \rangle_{n \mathop \in \N}$ be a sequence of non-empty sets.
By the Axiom of Replacement:
:$I = \left\{{x_n: n \in \N}\right\}$
is a set.
By Surjection from Natural Numbers iff Countable, $I$ is countable.
By Axiom of Countable Choice: Form 2, $I$ has a choice function $h: I \to \bigcup I$.
Define a sequence in $\bigcup I$ by letting $y_n = h \left({x_n}\right)$.
Then by the definition of choice function:
:$\forall n \in \N: y_n \in x_n$
As this holds for all such sequences, Axiom of Countable Choice: Form 1 holds.
Category:Axiom of Countable Choice
Equivalence of Formulations of Axiom of Choice

== Theorem ==
The following formulations of the Axiom of Choice are equivalent:
=== Formulation 1 ===
=== Formulation 2 ===
=== Formulation 3 ===
=== Formulation 4 ===
== Proof ==
=== Formulation 1 implies Formulation 2 ===
=== Formulation 2 implies Formulation 1 ===
=== Formulation 1 implies Formulation 3 ===
=== Formulation 3 implies Formulation 1 ===
=== Formulation 1 iff Formulation 4 ===
== Sources ==
* : Introduction to Set Theory: $3$. The Axiom of Choice and Its Equivalents
Category:Axiom of Choice
Category:Equivalence of Formulations of Axiom of Choice
Equivalence of Formulations of Axiom of Choice/Formulation 1 iff Formulation 4

== Theorem ==
The following formulation of the Axiom of Choice:
=== Formulation 1 ===
implies the following formulation of the Axiom of Choice:
=== Formulation 4 ===
== 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
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 1$ Introduction to well ordering: Exercise $1.6$
Category:Equivalence of Formulations of Axiom of Choice
Equivalence of Formulations of Axiom of Choice/Formulation 1 implies Formulation 2

== Theorem ==
The following formulation of the Axiom of Choice:
=== Formulation 1 ===
implies the following formulation of the Axiom of Choice:
=== Formulation 2 ===
== Proof ==
Let $\family {X_i}_{i \mathop \in I}$ be an indexed family of sets all of which are non-empty, indexed by $I$ which is also non-empty.
By hypothesis, Formulation 1 holds.
That is, there exists a choice function on every set of non-empty sets.
Let $f$ be a choice function on $\set{X_i}$.
Let $x_i = \map f {X_i}$.
By definition of choice function, each $x_i \in X_i$.
Therefore, $\family {x_i}_{i \mathop \in I}$ satisfies Formulation 2.
Category:Equivalence of Formulations of Axiom of Choice
Equivalence of Formulations of Axiom of Choice/Formulation 1 implies Formulation 3

== Theorem ==
The following formulation of the Axiom of Choice:
=== Formulation 1 ===
implies the following formulation of the Axiom of Choice:
=== Formulation 3 ===
== Proof ==
Let $\SS$ be the set:
:$\SS = \set {s: \O \notin s \land \forall t, u \in s: t = u \lor t \cap u = \O}$
Let $c$ be a choice function on $\SS$ and consider the image set $c \sqbrk \SS$:
:$c \sqbrk \SS = \set {\map c s: \O \notin s \land \forall t, u \in s: t = u \lor t \cap u = \O}$
By the definition of choice function:
:$\map c s \in s$
By construction of $\SS$, for any $s \in \SS$:
:$s \cap c \sqbrk \SS = \set {\map c s}$
== Sources ==
* : (P.5)
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 1$ Introduction to well ordering: Exercise $1.5$
Category:Equivalence of Formulations of Axiom of Choice
Equivalence of Formulations of Axiom of Choice/Formulation 2 implies Formulation 1

== Theorem ==
The following formulation of the Axiom of Choice:
=== Formulation 2 ===
implies the following formulation of the Axiom of Choice:
=== Formulation 1 ===
== Proof ==
Suppose that Formulation 2 holds.
That is, the Cartesian product of a non-empty family of non-empty sets is non-empty.
Let $\CC$ be a non-empty set of non-empty sets.
$\CC$ may be converted into an indexed set by using $\CC$ itself as the indexing set and using the identity mapping on $\CC$ to do the indexing.
Then the Cartesian product of all the sets in $\CC$ has at least one element.
An element of such a Cartesian product is a mapping, that is a family whose domain is the indexing set which in this context is $\CC$.
The value of this mapping at each index is an element of the set which bears that index.
So we have a mapping $f$ whose domain is $\CC$ such that:
:$A \in \CC \implies \map f A \in A$
Now let $\CC$ be the set of all non-empty subsets of some set $X$.
Then the assertion means that there exists a mapping $f$ whose domain is $\powerset X \setminus \set \O$ such that:
:$A \in \powerset X \setminus \set \O \implies \map f A \in A$
That is, Formulation 1 holds.
== Sources ==
* : $\S 15$: The Axiom of Choice
Category:Equivalence of Formulations of Axiom of Choice
Equivalence of Formulations of Axiom of Choice/Formulation 3 implies Formulation 1

== Theorem ==
The following formulation of the Axiom of Choice:
=== Formulation 3 ===
implies the following formulation of the Axiom of Choice:
=== Formulation 1 ===
== Proof ==
Let $\BB$ be a non-empty indexed family of non-empty sets indexed by $\II$.
Consider sets of the following form:
:$\CC = \set {\tuple {B_i, x}: i \in \II, B_i \in \BB, x \in B_i}$
That is, it is the set of ordered pairs of which the first coordinate is a set $B_i \in \BB$ and the second coordinate is an element of $B_i$.
This is, by construction, a subset of the cartesian product:
:$\CC \subseteq \ds \BB \times \bigcup_{i \mathop \in \II} B_i$
By hypothesis all sets $B$ are non-empty.
Thus there exists at least one $\tuple {B_i, x}$ element in $\CC$ for each $B_i \in \CC$.
By Equality of Ordered Pairs, if $B_j \ne B_k$ in $\BB$, then $\tuple {B_i, x} \ne \tuple {B_j, x}$ for all such pairs in $\CC$.
So any two sets $\set {\tuple {B_j, x}: x \in B_j}$ and $\set {\tuple {B_k, x} :x \in B_k}$ are disjoint, by the inequality of their first coordinates.
Then $\CC$ is an indexed family of disjoint non-empty sets in $\BB \times \bigcup_{i \mathop \in \II} B_i$.
Hence $\CC$ satisfies the hypotheses of the third formulation of the Axiom of Choice.
Let $c$ be a set satisfying:
:$\forall r \in \CC: c \cap r = \set t$
All elements of $c$ are ordered pairs:
:the first coordinate of which is a set in $B_i \in \BB$
and:
:the second coordinate of which is an element $x \in B_i$.
Viewed as a relation, each set in $\BB$ bears $c$ to an element in that set.
Every set in $\BB$ bears $c$ to exactly one element inside that set.
So $c$ is in fact a mapping and satisfies the criteria of a choice function as expounded in formulation $(1)$.
== Sources ==
* : $\S 9.2$
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 1$ Introduction to well ordering: Exercise $1.5$
Category:Equivalence of Formulations of Axiom of Choice
Equivalence of Formulations of Axiom of Empty Set

== Theorem ==
The following formulations of the are equivalent:
=== Formulation 1 ===
=== Formulation 2 ===
== Proof ==
=== Formulation 1 implies Formulation 2 ===
=== Formulation 2 implies Formulation 1 ===
== Sources ==
* : $\S 0.2$. Sets
Category:Empty Set
Axiom of the Empty Set
Equivalence of Formulations of Axiom of Empty Set for Classes

== Theorem ==
In the context of class theory, the following formulations of the are equivalent:
=== Formulation 1 ===
=== Formulation 2 ===
== Proof ==
It is assumed throughout that the and the both hold.
=== Formulation $1$ implies Formulation $2$ ===
Let formulation $1$ be axiomatic:
There exists a set that has no elements:
From the it follows that the $x$ so created is a class, and we label it $\O$.
But this $\O$ is a set.
Hence it follows that the truth of formulation $2$ follows from acceptance of the truth of formulation $1$.
=== Formulation $2$ implies Formulation $1$ ===
Let formulation $2$ be axiomatic:
By the , we can define the class $\O$ as:
:$\O := \set {y: \lnot {y \ne y} }$
which is a class with no elements.
We have that $\O$ is a set .
By the we have that $\O$ is unique.
From Equivalence of Formulations of Axiom of Empty Set, we can express $\O$ as:
:$\O := \set {y: \lnot {y \in \O} }$
That is, the truth of formulation $1$ follows from acceptance of the truth of formulation $2$.
Category:Axiom of the Empty Set
Axiom of the Empty Set
Equivalence of Formulations of Axiom of Infinity for Zermelo Universe

== Theorem ==
The following formulations of the in the context of a Zermelo universe are equivalent:
=== Formulation 1 ===
=== Formulation 2 ===
=== Formulation 3 ===
== Proof ==
Let the , in each of its formulations, be applied to a basic universe $V$ separately, as follows.
=== Formulation $1$ implies Formulation $2$ ===
Let formulation 1 be taken as an axiom.
Then the class $\omega$ as so defined is a set.
But, by definition, $\omega$ is itself an inductive set.
Hence an inductive set exists.
Thus formulation 2 holds.
=== Formulation $2$ implies Formulation $1$ ===
Let formulation 2 be taken as an axiom.
That is, there exists an inductive set, which we shall call $b$.
By the inductive set definition of the natural numbers:
:$\forall n \in \omega: n \in b$
where $\omega$ denotes the set of natural numbers.
That is:
:$\omega \subseteq b$
We have that:
:$b$ is a set
and
:$\omega$ is a subclass of $b$.
By the , $\omega$ is a set.
Thus formulation 1 holds.
=== Formulation $1$ implies Formulation $3$ ===
Let formulation 1 be taken as an axiom.
In the basic universe $V$, we are able to create the set:
:$S = \set {\O, \set \O}$
which is the natural number $2$.
By the , its power set is also a set:
:$\powerset S = \set {\O, \set \O, \set {\set \O}, \set {\set \O, \set \O} }$
But $\powerset S$ is not a natural number.
Thus formulation 3 holds.
=== Formulation $3$ implies Formulation $2$ ===
Let formulation 3 be taken as an axiom.
 formulation 2 did not hold.
Then there would be no inductive sets.
But then vacuously every set is an element of all inductive sets.
Thus, by the inductive set definition of the natural numbers:
:Every set is a natural number.
Thus it is seen that each of these formulations of the directly or indirectly implies each of the other two.
Hence the result.
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 2$ Definition of the Natural Numbers
Category:Axiom of Infinity
Axiom of Infinity
Equivalence of Formulations of Axiom of Pairing

== Theorem ==
The following formulations of the in the context of axiomatic set theory are equivalent:
=== Strong Form ===
=== Weak Form ===
== Proof ==
=== Strong Form implies Weak Form ===
Let the strong form of the Axiom of Pairing be assumed:
By definition of the biconditional, this can be expressed as:
:$\forall a: \forall b: \exists c: \forall z: \paren {\paren {z = a \lor z = b \implies z \in c} \land \paren {z \in c \implies z = a \lor z = b} }$
from which, by the Rule of Simplification:
:$\forall a: \forall b: \exists c: \forall z: \paren {z = a \lor z = b \implies z \in c}$
Thus the weak form of the Axiom of Pairing is seen to hold.
=== Weak Form implies Strong Form ===
Let the weak form of the Axiom of Pairing be assumed:
By the , let us create the set $c'$ as:
:$c' = \set {z: \paren {z \in c} \land \paren {z = a \lor z = b} }$
By the , it follows that:
:$c' = \set {a, b}$
Thus we have:
:$\forall a: \forall b: \exists c': \forall z: \paren {z = a \lor z = b \implies z \in c'}$
and:
:$\forall a: \forall b: \exists c': \forall z: \paren {z \in c' \implies z = a \lor z = b}$
That is:
:$\forall a: \forall b: \exists c': \forall z: \paren {z = a \lor z = b \iff z \in c'}$
Thus the strong form of the Axiom of Pairing is seen to hold.
Category:Axiom of Pairing
Axiom of Pairing
Equivalence of Formulations of Axiom of Pairing for Classes

== Theorem ==
The following formulations of the in the context of class theory are equivalent:
=== Formulation 1 ===
=== Formulation 2 ===
== Proof ==
It is assumed that all classes are subclasses of a basic universe $V$.
=== $(1)$ implies $(2)$ ===
Let formulation $1$ of the Axiom of Pairing be assumed:
Thus we have that $c = \set {a, b}$ is a set such that both $a \in c$ and $b \in c$.
Thus formulation $2$ of the Axiom of Pairing is seen to hold.
=== $(2)$ implies $(1)$ ===
Let formulation $2$ of the axiom of pairing be assumed:
Then the class $\set {a, b}$ is a subclass of $c$.
We have that $c$ is a subclass of a basic universe $V$.
Hence by the , every subclass of $c$ is a set.
That is, $\set {a, b}$ is a set.
Thus formulation $1$ of the Axiom of Pairing is seen to hold.
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 4$ The pairing axiom: Note $4$
Category:Axiom of Pairing
Axiom of Pairing
Equivalence of Formulations of Axiom of Powers

== Theorem ==
In the context of class theory, the following formulations of the are equivalent:
=== Formulation 1 ===
=== Formulation 2 ===
== Proof ==
It is assumed throughout that the and the both hold.
=== Formulation $1$ implies Formulation $2$ ===
Let formulation $1$ be axiomatic:
Thus it is posited that for a given set $x$ the power set of $x$ exists:
:$y := \powerset x := \set {z: z \subseteq x}$
and that this is a set.
Formulation $2$ asserts that given the existence of $\powerset x$, it is axiomatic that $\powerset x$ is itself a set.
Hence it follows that the truth of formulation $2$ follows from acceptance of the truth of formulation $1$.
From Power Set Exists and is Unique, which depends on:
:the 
:the 
$\powerset x$ is unique for a given $x$.
=== Formulation $2$ implies Formulation $1$ ===
Let formulation $2$ be axiomatic:
Let $x$ be a set.
Let us create the power set $y$ of x:
:$\powerset x := \set {z: z \subseteq x}$
From Power Set Exists and is Unique, which depends on:
:the 
:the 
$\powerset x$ exists and is unique for a given $x$.
We have asserted the truth of formulation $2$.
That is, $\powerset x$ is a set.
As $x$ is arbitrary, it follows that $\powerset x$ exists and is unique for all sets $x$.
That is, the truth of formulation $1$ follows from acceptance of the truth of formulation $2$.
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 6$ The power axiom: Remarks $(1)$
Category:Axiom of Powers
Axiom of Powers
Equivalence of Formulations of Axiom of Replacement

== Theorem ==
In the context of class theory, the following formulations of the Axiom of Replacement are equivalent:
=== Formulation 1 ===
=== Formulation 2 ===
=== Formulation 3 ===
== Proof ==
== Sources ==
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 3$ The axiom of substitution: Exercise $3.1$
Axiom of Replacement
Equivalence of Formulations of Axiom of Unions

== Theorem ==
In the context of class theory, the following formulations of the are equivalent:
=== Formulation 1 ===
=== Formulation 2 ===
== Proof ==
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 5$ The union axiom: Note $1$
Category:Axiom of Unions
Axiom of Unions
Category:Equivalence of Formulations of Axiom of Unions
Equivalence of Formulations of Axiom of Unions/Proof

== Theorem ==
== Proof ==
It is assumed throughout that the and the both hold.
=== Formulation $1$ implies Formulation $2$ ===
Let formulation $1$ be axiomatic:
Thus it is posited that for a given set of sets $A$ the union of $A$ exists:
:$x := \bigcup A := \set {y: \exists z \in A: y \in z}$
and that this is a set.
Formulation $2$ asserts that given the existence of $\bigcup A$, it is axiomatic that $\bigcup A$ is itself a set.
Hence it follows that the truth of formulation $2$ follows from acceptance of the truth of formulation $1$.
From Class Union Exists and is Unique, which depends on:
:the 
:the 
$\bigcup A$ is unique for a given $A$.
=== Formulation $2$ implies Formulation $1$ ===
Let formulation $2$ be axiomatic:
Let $A$ be a set of sets.
Let us create the class:
:$x := \bigcup A := \set {y: \exists z \in A: y \in z}$
From Class Union Exists and is Unique, which depends on:
:the 
:the 
$\bigcup A$ exists and is unique for a given $A$.
We have asserted the truth of formulation $2$.
That is, $\bigcup A$ is a set.
As $A$ is arbitrary, it follows that $\bigcup A$ exists and is unique for all sets of sets $A$.
That is, the truth of formulation $1$ follows from acceptance of the truth of formulation $2$.
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 5$ The union axiom: Exercise $5.1$
Category:Equivalence of Formulations of Axiom of Unions
Equivalence of Formulations of Pasch's Axiom

== Theorem ==
The two forms of Pasch's Axiom in Tarski's Geometry are consistent.
That is, the expressions:
:$(1): \quad \forall a, b, c, p, q: \exists x: \mathsf B a p c \land \mathsf B b q c \implies \mathsf B p x b \land \mathsf B q x a$
and:
:$(2): \quad \forall a, b, c, p, q: \exists x: \mathsf B a p c \land \mathsf B q c b \implies \mathsf B a x q \land \mathsf B b p x$
are logically equivalent.
== Proof ==
== Sources ==
Category:Tarski's Geometry
Equivalence of Formulations of Peano's Axioms

== Theorem ==
Let $P$ be a set.
Let $s: P \to P$ be a mapping.
Let $0 \in P$ be a distinguished element.
=== Formulation 1 ===
=== Formulation 2 ===
== Proof ==
=== Formulation 1 implies Formulation 2 ===
For $(\text P 3)$, there is nothing to prove.
Next, axiom $(\text P 4)$ of Formulation 2.
Recall the definition of the image of a mapping as applicable to $s$:
:$\Img s = \set {n \in P: \exists m \in P: \map s m = n}$
From this definition, it follows that $0 \notin \Img s$, and hence:
:$\Img s \ne P$
Lastly, axiom $(\text P 5)$ of Formulation 2.
By Non-Successor Element of Peano Structure is Unique, we know that there is exactly one element of $P$ that is not in the image of $s$.
By axiom $(\text P 3)$ of Formulation 1, we know this element is $0$.
Therefore, the premises:
:$0 \in A$
and:
:$\exists x \in A: \neg \exists y \in P: x = \map s y$
are logically equivalent, and hence so are both forms of axiom $(\text P 5)$.
=== Formulation 2 implies Formulation 1 ===
By Non-Successor Element of Peano Structure is Unique, we know that there is exactly one element of $P$ that is not in the image of $s$.
We will identify this element with the distinguished element $0$.
This implies that axioms $(\text P 3)$ and $(\text P 4)$ of Formulation 1 are satisfied.
Lastly, axiom $(\text P 5)$ of Formulation 1.
Since $0$ satisfies the premise:
:$\neg \exists y \in P: 0 = \map s y$
we conclude that axiom $(\text P 5)$ of Formulation 1 follows a fortiori from axiom $(\text P 5)$ of Formulation 2.
== Sources ==
* : Introduction $\S 4$ (implicit)
Category:Peano's Axioms
Equivalence of Local Uniform Convergence and Compact Convergence

== Theorem ==
Let $U \subseteq \C$ be an open set of the complex plane.
Let $f_n: U \to \C$ be a sequence of functions which converges pointwise to $f: U \to \C$.
Then $f_n$ converges to $f$ uniformly on all compact subsets of $U$ $f_n$ converges locally uniformly on $U$.
== Proof ==
=== Sufficient Condition ===
Suppose that $f_n$ converges to $f$ uniformly on all compact subsets of $U$.
Let $z \in U$.
Since $U$ is open, there is an open ball $\map {B_\epsilon} z$, around $z$, of radius $\epsilon$, with the closure of the ball, $\overline{\map {B_\epsilon} z}$, contained in $U$.
Since $\overline {\map {B_\epsilon} z}$ is closed and bounded, it is compact, by the Heine-Borel Theorem. 
We then have by assumption that $f_n$ converges uniformly on $\overline {\map {B_\epsilon} z}$.
Thus $f_n$ converges locally uniformly on $U$.
=== Necessary Condition ===
Now suppose that $f_n$ converges to $f$ locally uniformly.
That is, for every $z \in U$, there is an $r \in \R_{>0}$ such that $\map {B_r} z \subset U$ and $f_n$ converges uniformly on $\map {B_r} z$.
Let $K \subset U$ be an arbitrary compact subset.
Let $\set {\map {B_{r_z} } z}_{z \mathop \in K}$ be an open cover of $K$ by open balls, one centered at each $z \in K$, such that $f_n$ converges uniformly on each ball.
Since $K$ is compact, for every open cover of the form $\set {\map {B_{r_z} } z}_{z \mathop \in K}$, there exists a subcover:
:$\set {\map {B_{r_i} } {z_i} }_{i \mathop = 1}^n \subseteq \set {\map {B_{r_z} } z}_{z \mathop \in K}$
which is finite.
Let $\epsilon > 0$, and $1 \le i \le n$. 
Since the convergence is uniform, there is an $N_i$ large enough so that:
:$\ds \sup_{z \mathop \in \map {B_{r_i} } {z_i} } \size {\map {f_n} z - \map f z} < \epsilon$
for all $n > N_i$.
Let $N = \ds \max_{1 \mathop \le i \mathop \le n} \set {N_i}$.
Let $B = \ds \bigcup_{i \mathop = 1}^n \map {B_{r_i} } {z_i}$.
Then:
:$\ds \sup_{z \mathop \in B} \size {\map {f_n} z - \map f z} < \epsilon$
for all $n > N$.
Thus $f_n$ converges uniformly on $B$, and so also on $K$.
Category:Complex Analysis
Category:Uniform Convergence
Equivalence of Logical Implication and Conditional

== Theorem ==
:$\paren {p \implies q} \dashv \vdash \paren {p \vdash q}$
That is, the conditional is logically equivalent to logical implication.
== Proof ==
This directly follows from:
* The Modus Ponendo Ponens: $p \implies q, p \vdash q$
* The Rule of Implication: $\left({p \vdash q}\right) \vdash p \implies q$.
== Caution ==
This is not to say that the conditional and the logical implication are the same thing.
If $p \not \vdash q$ it does not mean that $\neg \left({p \implies q}\right)$.
The latter statement is true only when $p$ is true and $q$ is false.
The former statement just says that it is not always true that when $p$ is true then $q$ is true.
Category:Propositional Logic
Category:Conditional