nyuuzyou/wikis huggingface dataset:
Condition for Agreement of Family of Mappings

== Theorem ==
Let $\family {A_i}_{i \mathop \in I}, \family {B_i}_{i \mathop \in I}$ be families of non empty sets.
Let $\family {f_i}_{i \mathop \in I}$ be a family of mappings such that:
:$\forall i \in I: f_i \in \map \FF {A_i, B_i}$
We have that:
:$\ds \bigcup_{i \mathop \in I} f_i \in \map \FF {\bigcup_{i \mathop \in I} A_i, \bigcup_{i \mathop \in I} B_i}$
:
:$\ds \forall i, j \in I: \Dom {f_i} \cap \Dom {f_j} \ne \O \implies \paren {\forall a \in \paren {\Dom {f_i} \cap \Dom {f_j} }, \tuple {a, b} \in f_i \implies \tuple {a, b} \in f_j}$
== Proof ==
=== Sufficient Condition ===
Let:
:$\ds \bigcup_{i \mathop \in I} f_i \in \map \FF {\bigcup_{i \mathop \in I} A_i, \bigcup_{i \mathop \in I} B_i}$
Let $i, j \in I$ be such that:
:$\Dom {f_i} \cap \Dom {f_j} \ne \O$
Let $a \in \paren {\Dom {f_i} \cap \Dom {f_j} }$.
Let $\ds b \in \bigcup_{i \mathop \in I} B_i$ be such that:
:$\tuple {a, b} \in f_i$
:
:$\tuple {a, b} \notin f_j$
As $a \in \paren {\Dom {f_i} \cap \Dom {f_j} }$:
:$\ds \exists c \in \bigcup_{i \mathop \in I} B_i: \tuple {a, c} \in f_j$ 
As $\tuple {a, b} \in f_i$:
:$\ds \tuple {a, b} \in \bigcup_{i \mathop \in I} f_i$
Thus:
:$\ds \tuple {a, b}, \tuple {a, c} \in \bigcup_{i \mathop \in I} f_i$
such that $b \ne c$ and $\ds \bigcup_{i \mathop \in I} f_i$ is a mapping.
This is a contradiction.
Thus the supposition that the fact $\tuple {a, b} \notin f_j$ was false.
So:
:$\tuple {a, b} \in f_j$
=== Necessary Condition ===
Let:
:$\forall i, j \in I: \Dom {f_i} \cap \Dom {f_j} \ne \O \implies \paren {\forall a \in \paren {\Dom {f_i} \cap \Dom {f_j} }, \tuple {a, b} \in f_i \implies \tuple {a, b} \in f_j}$
Let $\ds a \in \bigcup_{i \mathop \in I} A_i$.
Hence:
:$\exists k \in I: a \in A_k$
Let $k \in I$.
Thus:
:$a \in \Dom f_k$
Let $l = \map {f_k} a$.
It follows that:
:$\tuple {a, l} \in f_k$
and so:
:$\ds \tuple {a, l} \in \bigcup_{i \mathop \in I} f_i$
:
:$\ds \exists m \in \bigcup_{i \mathop \in I} B_i: \paren {\tuple {a, m} \in \bigcup_{i \mathop \in I} f_i \land m \ne l}$
Let $\ds m \in \bigcup_{i \mathop \in I} B_i$.
Let $j \in I$ be such that:
:$\tuple {a, m} \in f_j$
We have:
:$a \in \paren {\Dom {f_k} \cap \Dom {f_j} }$
As $\tuple {a, l} \in f_k$:
:$\tuple {a, l} \in f_j$.
Therefore:
:$\tuple {a, m}, \tuple {a, l} \in f_j$
where $f_j \in \map \FF {A_j, B_j}$ and $m \ne l$.
This contradicts the definition of mapping.
So:
:$\ds \nexists m \in \bigcup_{i \mathop \in I} B_i: \paren {\tuple {a, m} \in \bigcup_{i \mathop \in I} f_i \land m \ne l}$
and so:
:$\ds \bigcup_{i \mathop \in I} f_i \in \map \FF {\bigcup_{i \mathop \in I} A_i, \bigcup_{i \mathop \in I} B_i}$
Category:Mapping Theory
Category:Set Union
Condition for Alexandroff Extension to be T1 Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a non-empty topological space.
Let $p$ be a new element not in $S$.
Let $S^* := S \cup \set p$.
Let $T^* = \struct {S^*, \tau^*}$ be the Alexandroff extension on $S$.
Then $T^*$ is a $T_1$ (Fréchet) space $T$ is a $T_1$ (Fréchet) space.
== Proof ==
=== Necessary Condition ===
Let $T = \struct {S, \tau}$ be a $T_1$ space.
By definition, $T$ is a $T_1$ space all points of $S$ are closed in $T$.
We have that $S$ is open in $T$ by definition of a topology.
Thus by definition of the Alexandroff extension, $S$ is open in $T^*$.
So as $S = S^* \setminus \set p$ is open in $T^*$, it follows that $\set p$ is closed in $T^*$.
That is, $p$ is a closed point of $T^*$.
Now let $x \in S^*$ such that $x \ne p$.
As $T$ is a $T_1$ space, $\set x$ is closed in $T$.
From Finite Topological Space is Compact, $\set x$ is a compact subset of $T$.
Thus by definition of the Alexandroff extension, $S^* \setminus \set x$ is open in $T^*$.
Thus by definition $\set x$ is closed in $T^*$.
That is, $x$ is a closed point of $T^*$.
Thus it has been shown that all points in $T^*$ are closed points of $T^*$.
Thus by definition $T^*$ is a $T_1$ space.
=== Sufficient Condition ===
Let $T^*$ be a $T_1$ space.
By definition of $T$ is a subspace of $T^*$.
By $T_1$ Property is Hereditary, it follows that $T$ is a $T_1$ space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $34$. One Point Compactification Topology: $2$
Category:Alexandroff Extensions
Category:T1 Spaces
Condition for Alexandroff Extension to be T2 Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a non-empty topological space.
Let $p$ be a new element not in $S$.
Let $S^* := S \cup \set p$.
Let $T^* =\struct {S^*, \tau^*}$ be the Alexandroff extension on $S$.
Then $T^*$ is a $T_2$ (Hausdorff) space $T$ is a locally compact Hausdorff space.
== Proof ==
=== Necessary Condition ===
Let $T = \struct {S, \tau}$ be a locally compact Hausdorff space.
Let $x, y \in S$.
Then as $T$ is Hausdorff, there exist two disjoint open sets $U, V \in \tau$ containing $x$ and $y$ respectively.
But by definition of the Alexandroff extension on $S$, $U$ and $V$ are also open sets of $T^*$.
That covers the case where both $x \ne p$ and $y \ne p$.
, let $y = p$.
As $T$ is a locally compact Hausdorff space, $x$ has a compact neighborhood $N_x$.
By definition of neighborhood, there exists an open set $U$ of $T$ containing $x$.
By definition of the Alexandroff extension, $U$ is an open set of $T^*$.
By Compact Subspace of Hausdorff Space is Closed, $N_x$ is closed in $T$.
Let $V = S^* \setminus N_x$.
By definition of $V$, $x \notin V$ and $p \in V$.
By definition of the Alexandroff extension, $V$ is an open set of $T^*$.
By Intersection with Complement is Empty iff Subset:
:$U \cup V = \O$
So we have two open sets $U, V$ of $T^*$ such that $x \in U, p \in V$ and $U \cup V = \O$.
Thus by definition $T^*$ is a $T_2$ space.
=== Sufficient Condition ===
Let $T^*$ be a $T_2$ space.
By definition of $T$ is a subspace of $T^*$.
By $T_2$ Property is Hereditary, it follows that $T$ is a $T_2$ space.
It remains to be shown that $T$ is also (weakly) locally compact.
Let $x \in S$.
As $T^*$ is a $T_2$ space, we can find.
:an open set $U$ of $T^*$ containing $x$
and:
:an open set $V$ of $T^*$ containing $p$
such that $U \cap V = \O$.
Then by definition of the Alexandroff extension, $S^* \setminus V$ is a compact (closed) neighborhood of $x$.
Thus by definition $S$ is a locally compact Hausdorff space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $34$. One Point Compactification Topology: $3$
Category:Alexandroff Extensions
Category:Hausdorff Spaces
Condition for Algebra Loop to be Abelian Group

== Theorem ==
Let $\struct {S, \odot}$ be an algebra loop.
Let $\struct {S \times S, \otimes}$ denote the external direct product of $\struct {S, \odot}$ with itself:
:$\forall \tuple {x_1, y_1}, \tuple {x_2, y_2} \in S \times S: \tuple {x_1, y_1} \otimes \tuple {x_2, y_2} = \tuple {x_1 \odot x_2, y_1 \odot y_2}$
Let $f$ and $g$ be mappings from $\struct {S \times S, \otimes}$ to $\struct {S, \odot}$.
Let $f \odot g$ denote the pointwise operation on $S^{S \times S}$ induced by $\odot$.
Let $\struct {S, \odot}$ satisfy the condition that:
If $f$ and $g$ are homomorphisms, then $f \odot g$ is also a homomorphism.
Then $\struct {S, \odot}$ is an abelian group.
== Proof ==
Recall that an algebra loop $\struct {S, \circ}$ is a quasigroup with an identity element:
:$\exists e \in S: \forall x \in S: x \circ e = x = e \circ x$
From Equivalent Conditions for Entropic Structure: Pointwise Operation of Homomorphisms from External Direct Product is Homomorphism:
:$\struct {S, \odot}$ is an entropic structure.
From Entropic Structure with Identity is Commutative Monoid, $\struct {S, \odot}$ is a commutative monoid.
Hence $\struct {S, \odot}$ is a commutative semigroup.
From Structure is Group iff Semigroup and Quasigroup, it follows that $\struct {S, \odot}$ is a group.
As $\odot$ is a commutative operation, it follows that $\struct {S, \odot}$ is an abelian group.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.13$
Category:Algebra Loops
Category:Abelian Groups
Condition for Algebraic Structure to be Self-Distributive Quasigroup

== Theorem ==
Let $\struct {S, \odot}$ be an algebraic structure.
Then:
:$\struct {S, \odot}$ is a self-distributive quasigroup
:
:for every $a \in S$, the left and right regular representations $\lambda_a$ and $\rho_a$ on $S$ are automorphisms of $\struct {S, \odot}$.
== Proof ==
=== Sufficient Condition ===
Let $\struct {S, \odot}$ be a self-distributive quasigroup.
By definition of quasigroup, we have that $\lambda_a$ and $\rho_a$ are both permutations on $S$.
It remains for the morphism property to be demonstrated.
Indeed:
and:
Thus $\lambda_a$ and $\rho_a$ have been shown to be automorphisms of $\struct {S, \odot}$.
=== Necessary Condition ===
Let $\odot$ be such that for every $a \in S$, the left and right regular representations $\lambda_a$ and $\rho_a$ on $S$ are automorphisms of $\struct {S, \odot}$.
We have that:
:$\lambda_a$ and $\rho_a$ are both permutations on $S$
:$\lambda_a$ and $\rho_a$ are both homomorphisms of $S$.
Hence by definition $\struct {S, \odot}$ is a quasigroup.
It remains to demonstrate self-distributivity.
Indeed:
and:
Thus $\struct {S, \odot}$ is a self-distributive quasigroup.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.21 \ \text{(b)}$
Category:Self-Distributive Operations
Category:Quasigroups
Condition for Bipartite Graph to be Hamiltonian

== Theorem ==
Let $G = \struct {A \mid B, E}$ be a bipartite graph.
Let $G$ be Hamiltonian.
Then $\card A = \card B$.
That is, there is the same number of vertices in $A$ as there are in $B$.
== Proof ==
Let $G = \struct {A \mid B, E}$ be a bipartite graph.
To be Hamiltonian, a graph $G$ needs to have a Hamilton cycle: that is, one which goes through all the vertices of $G$.
As each edge in $G$ connects a vertex in $A$ with a vertex in $B$, any cycle alternately passes through a vertex in $A$ then a vertex in $B$.
, suppose that $\card A > \card B$, that is, that there are more vertices in $A$ than in $B$.
Let $\card A = m, \card B = n$.
Suppose $G$ has a Hamilton cycle $C$.
Let that cycle start at $u \in B$.
After $2 n$ edges have been traversed, we will have arrived back at $u$ again, and all the vertices of $B$ will have been visited.
But there will still be $m - n$ vertices in $A$ which have not been visited.
Hence $C$ can not be a Hamilton cycle.
== Note ==
The implication does not go both ways.
This graph:
:
clearly fulfils the conditions (that is, it is a bipartite graph such that $\card A = \card B$) and is equally clearly not Hamiltonian.
== Sources ==
* : $\S 3.2$: The Salesman's Problem: An Introduction to Hamiltonian Graphs: Problem $25$
Category:Hamiltonian Graphs
Category:Bipartite Graphs
Condition for Cartesian Product Equivalent to Associated Cardinal Number

== Theorem ==
Let $S$ and $T$ be nonempty sets.
Let $\card S$ denote the cardinal number of $S$.
Then:
:$S \times T \sim \card {S \times T} \iff S \sim \card S \land T \sim \card T$
where $S \times T$ denotes the cartesian product of $S$ and $T$.
== Proof ==
=== Necessary Condition ===
If $S \times T \sim \card {S \times T}$, then there is a mapping $f$ such that:
:$f : S \times T \to \card {S \times T}$ is a bijection.
Since $f$ is a bijection, it follows that:
:$S$ is equivalent to the image of $S \times \set x$ under $f$ where $x \in T$.
This, in turn, is a subset of the ordinal $\card {S \times T}$.
$\card {S \times T}$ is an ordinal by Cardinal Number is Ordinal.
By Condition for Set Equivalent to Cardinal Number, it follows that $S \sim \card S$.
Similarly, $T \sim \card T$.
=== Sufficient Condition ===
Suppose $f: S \to \card S$ is a bijection and $g: T \to \card T$ is a bijection.
Let $\cdot$ denote ordinal multiplication, while $\times$ shall denote the Cartesian product.
Define the function $F$ to be:
:$\forall x \in S, y \in T: \map F {x, y} = \card S \cdot \map g y + \map f x$
It follows that $F: S \times T \to \card S \cdot \card T$ is a injection.
By Condition for Set Equivalent to Cardinal Number, it follows that $S \times T \sim \card {S \times T}$.
== Sources ==
* : $\S 10.15 \ (2)$
Category:Cardinals
Condition for Circles to be Orthogonal

== Theorem ==
Let $\CC_1$ and $\CC_2$ be circles embedded in a Cartesian plane.
Let $\CC_1$ and $\CC_2$ be described by Equation of Circle in Cartesian Plane as:
Then $\CC_1$ and $\CC_2$ are orthogonal :
:$2 \alpha_1 \alpha_2 + 2 \beta_1 \beta_2 = c_1 + c_2$
== Proof ==
When $\CC_1$ and $\CC_2$ are orthogonal, the distance between their centers forms the hypotenuse of a right triangle whose legs are equal to the radii.
From Equation of Circle in Cartesian Plane: Formulation 3, the radii $r_1$ and $r_2$ of $\CC_1$ and $\CC_2$ respectively are given by:
Hence we have:
== Sources ==
* : Chapter $\text {III}$. The Circle: $16$. condition that two circles should cut orthogonally
Category:Circles
Category:Orthogonal Circles
Condition for Closed Extension Space to be T0 Space

== Theorem ==
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 $T_0$ (Kolmogorov) space $T$ 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.
== Sources ==
* : Part $\text {II}$: Counterexamples: $12$. Closed Extension Topology: $21$
Category:T0 Spaces
Category:Closed Extension Topology
Condition for Closed Extension Space to be T4 Space

== Theorem ==
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 $T_4$ space $T$ is a $T_4$ space vacuously
and $T^*_p$ in this case is also a $T_4$ space vacuously.
== Proof ==
=== Sufficient Condition ===
Suppose $T^*_p$ is $T_4$.
Then for any two disjoint closed sets $A, B \subseteq S$ there exist disjoint open sets $U, V \in \tau^*_p$ containing $A$ and $B$ respectively.
However, for any non-empty set $U \in \tau^*_p$, $p \in U$.
Hence no non-empty open sets in $T^*_p$ are disjoint.
Therefore $T^*_p$ is a $T_4$ space vacuously: there do not exist two disjoint closed sets.
=== Necessary Condition ===
By Closed Sets of Closed Extension Topology, the closed sets of $T^*_p$ (apart from $S^*_p$) are the closed sets of $T$.
Therefore $T^*_p$ does not contain two disjoint closed sets $T$ does not contain two disjoint closed sets.
Hence $T^*_p$ is a $T_4$ space vacuously $T$ is a $T_4$ space vacuously.
== Sources ==
* : Part $\text {II}$: Counterexamples: $12$. Closed Extension Topology: $21$
Category:T4 Spaces
Category:Closed Extension Topology
Condition for Closed Extension Space to be T5 Space

== Theorem ==
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 $T_5$ space $T$ is a $T_5$ space vacuously
and $T^*_p$ in this case is also a $T_5$ space vacuously.
== Proof ==
=== Sufficient Condition ===
Let $T^*_p$ be a $T_5$ space.
Then for any two separated sets $A, B \subseteq S$ there exist disjoint open sets $U, V \in \tau^*_p$ containing $A$ and $B$ respectively.
However, for any non-empty set $U \in \tau^*_p$:
:$p \in U$
Hence no non-empty open sets in $T^*_p$ are separated.
Therefore $T^*_p$ is a $T_5$ space vacuously: there do not exist two separated sets.
=== Necessary Condition ===
Let $A, B \in \tau^*_p$.
By Closure of Open Set of Closed Extension Space, we have that:
:$\forall U \in \tau^*_p: U \ne \O \implies U^- = S$
where $U^-$ is the closure of $U$.
Thus for all $A, B \ne \O$, it cannot be the case that:
:$A^- \cap B = A \cap B^- = \O$
Therefore $T^*_p$ does not contain two separated sets $T$ does not contain two separated sets.
Hence $T^*_p$ is a $T_5$ space vacuously $T$ is a $T_5$ space vacuously.
== Sources ==
* : Part $\text {II}$: Counterexamples: $12$. Closed Extension Topology: $21$
Category:T5 Spaces
Category:Closed Extension Topology
Condition for Closure of Open Ball to be Closed Ball of Same Radius

== Theorem ==
Let $\struct {M, d}$ be a metric space.
:$(1) \quad$ for all $x \in X$ and $r > 0$ we have $\map \cl {\map {B_r} x} = \map { {B_r}^-} x$, where $\cl$ denotes closure
:$(2) \quad$ for each $\epsilon > 0$ and $x, y \in X$ with $x \ne y$, there exists $z \in X$ with $\map d {z, y} < \epsilon$ and $\map d {x, z} < \map d {x, y}$.
== Proof ==
=== $(1)$ implies $(2)$ ===
Suppose that $(1)$ holds. 
Let $\epsilon > 0$. 
Let $x, y \in X$ have $x \ne y$.
We aim to show that there exists $z \in X$ with $\map d {z, y} < \epsilon$ and $\map d {x, z} < \map d {x, y}$.
Let $r = \map d {x, y}$. 
From $(1)$, we have $\map \cl {\map {B_r} x} = \map { {B_r}^-} x$.
So $y \in \map \cl {\map {B_r} x}$. 
Hence by the definition of topological closure, for each $\epsilon > 0$ we have:
:$\map {B_\epsilon} y \cap \map {B_r} x \ne \O$
Let $z \in \map {B_\epsilon} y \cap \map {B_r} x$.
Then $\map d {y, z} < \epsilon$ and $\map d {x, z} < r = \map d {x, y}$.
=== $(2)$ implies $(1)$ ===
Suppose that $(2)$ holds.
Let $x \in X$ and $r > 0$.
From Closure of Open Ball in Metric Space, we have:
:$\map \cl {\map {B_r} x} \subseteq \map { {B_r}^-} x$
It remains to show that:
:$\map { {B_r}^-} x \subseteq \map \cl {\map {B_r} x}$
Since $\map {B_r} x \subseteq \map \cl {\map {B_r} x}$, we only need to show that if $y \in X$ has $\map d {x, y} = r$, then $y \in \map \cl {\map {B_r} x}$.
Let $y \in X$ have $\map d {x, y} = r$.
Let $\epsilon > 0$. 
Then there exists $z \in X$ such that $\map d {x, z} < \map d {x, y} = r$ and $\map d {z, y} < \epsilon$. 
So we have $z \in \map {B_\epsilon} y$ and $z \in \map {B_r} x$. 
So $\map {B_\epsilon} y \cap \map {B_r} x \ne \O$ for each $\epsilon > 0$.
Hence by the definition of closure, we have $y \in \map \cl {\map {B_r} x}$.
Hence we obtain $\map \cl {\map {B_r} x} = \map { {B_r}^-} x$.
== Sources ==
* 
Category:Closed Balls
Category:Open Balls
Condition for Codeword to be in Code

== Theorem ==
Let $C$ be a linear code over a Galois field $\GF$.
Let $\mathbf H$ be a parity check matrix for $C$.
A vector $\mathbf x$ is a codeword of $C$ :
:$\mathbf H \mathbf x^\intercal = \mathbf 0$
where:
:${}^\intercal$ denotes the transpose of a matrix
:$\mathbf 0$ denotes the zero vector.
== Proof ==
== Sources ==
* 
Category:Parity Check Matrices
Condition for Cofinal Nonlimit Ordinals

== Theorem ==
Let $x$ and $y$ be nonlimit ordinals.
Let $\operatorname{cof}$ denote the cofinal relation.
Let $\le$ denote the subset relation.
Furthermore, let $x$ and $y$ satisfy the condition:
:$0 < x \le y$
Then:
:$\map {\operatorname{cof}} { y,x }$
== Proof ==
Both $x$ and $y$ are non-empty, so by the definition of a limit ordinal:
:$x = z^+$ for some $z$.
:$y = w^+$ for some $w$.
:$\bigcup z^+ \le \bigcup w^+$ follows by Set Union Preserves Subsets/General Result.
:$z \le w$ follows by Union of Successor Ordinal.
Define the function $f : x \to y$ as follows:
:$\map f a = \begin{cases}
a &: a \ne z \\
w &: a = z
\end{cases}$
$a < b \le w$, so $f \left({a}\right) = a$.
Take any $a,b \in x$ such that $a < b$.
:$\map f a < \map f b$ shall be proven by cases:
=== Case 1: $b \ne z$ ===
If $b \ne z$:
:$\map f a < \map f b$ is simply a restatement of $a < b$.
=== Case 2: $b = z$ ===
If $b = z$, then $\map f b = w$ by the definition of $f$.
Since $a < z \le w$, $\map f a < \map f b$.
It follows that $f$ is strictly increasing.
Moreover, since $\bigcup y = w$ is the least upper bound, $f\left({z}\right) \ge a$ for all $a \in y$.
== Sources ==
* : $\S 10.53 \ (2)$
Category:Ordinals
Condition for Collinearity of Points in Complex Plane/Formulation 1

== Theorem ==
Let $z_1$, $z_2$ and $z_3$ be points in the complex plane.
Then $z_1$, $z_2$ and $z_3$ are collinear :
:$\dfrac {z_1 - z_3} {z_3 - z_2} = \lambda$
where $\lambda \in \R$ is a real number.
If this is the case, then $z_3$ divides the line segment in the ratio $\lambda$.
If $\lambda > 0$ then $z_3$ is between $z_1$ and $z_2$, and if $\lambda < 0$ then $z_3$ is outside the line segment joining $z_1$ to $z_2$.
== Proof ==
By Geometrical Interpretation of Complex Subtraction:
:$z_1 - z_3$ can be represented as the line segment from $z_3$ to $z_1$
:$z_3 - z_2$ can be represented as the line segment from $z_2$ to $z_3$.
Thus we have that $z_1$, $z_2$ and $z_3$ are collinear $z_1 - z_3$ is parallel to $z_3 - z_2$, when expressed as line segments.
Let $\dfrac {z_1 - z_3} {z_3 - z_2} = \lambda$ for $\lambda \in \C$.
That is:
:$z_1 - z_3 = \lambda \paren {z_3 - z_2}$
By Complex Multiplication as Geometrical Transformation‎:
:$\map \arg {z_1 - z_3} = \map \arg {z_3 - z_2} \iff \map \arg {z_3 - z_2} \arg \lambda \in \R_{>0}$
and:
:$\map \arg {z_1 - z_3} = -\map \arg {z_3 - z_2} \iff \map \arg {z_3 - z_2} \arg \lambda \in \R_{<0}$
where $\arg$ denotes the argument of a complex number.
Also by Complex Multiplication as Geometrical Transformation‎, $z_1 - z_3$ is $\lambda$ the length of $z_3 - z_2$.
The result follows.
== Sources ==
* : $\S 2$. Geometrical Representations: Example $2$.
Category:Condition for Collinearity of Points in Complex Plane
Condition for Collinearity of Points in Complex Plane/Formulation 2

== Theorem ==
Let $z_1, z_2, z_3$ be distinct complex numbers.
Then:
:$z_1, z_2, z_3$ are collinear in the complex plane
:
::$\exists \alpha, \beta, \gamma \in \R: \alpha z_1 + \beta z_2 + \gamma z_3 = 0$
:where:
::$\alpha + \beta + \gamma = 0$
::not all of $\alpha, \beta, \gamma$ are zero.
== Proof ==
=== Sufficient Condition ===
Let $z_1, z_2, z_3$ be collinear.
Then by Condition for Collinearity of Points in Complex Plane: Formulation 1 there exists a real number $b$ such that:
:$z_2 - z_1 = b \paren {z_3 - z_1}$
Then:
Setting $\alpha = b - 1, \beta = 1, \gamma = -b$ fits the bill, as $\paren {b - 1} + 1 + \paren {-b} = 0$.
=== Necessary Condition ===
Let $\alpha + \beta + \gamma = 0$ such that:
:$\alpha z_1 + \beta z_2 + \gamma z_3 = 0$
:at least one of $\alpha, \beta, \gamma$ is not zero.
 let $\alpha \ne 0$.
Then it follows that as $\alpha + \beta + \gamma = 0$, at least one of $\beta$ and $\gamma$ is also non-zero.
 let $\beta \ne 0$.
In the following it is immaterial whether $\gamma = 0$ or not.
We have:
Thus it is seen that:
:$z_2 - z_1 = b \paren {z_3 - z_1}$
for some $b \in \R$.
Hence by Condition for Collinearity of Points in Complex Plane: Formulation 1, $z_1$, $z_2$ and $z_3$ are collinear.
== Sources ==
* : $\S 2$. Geometrical Representations: Exercise $4$.
* : $1$: Complex Numbers: Supplementary Problems: Miscellaneous Problems: $134$
Category:Condition for Collinearity of Points in Complex Plane