nyuuzyou/wikis huggingface dataset:
Union of Indexed Family of Sets Equal to Union of Disjoint Sets/General Result

== Theorem ==
Let $I$ be a set which can be well-ordered by a well-ordering $\preccurlyeq$.
Let $\family {E_\alpha}_{\alpha \mathop \in I}$ be a countable indexed family of sets indexed by $I$ where at least two $E_\alpha$ are distinct.
Then there exists a countable indexed family of disjoint sets $\family {F_\alpha}_{\alpha \mathop \in I}$ defined by:
:$\ds F_\beta = E_\beta \setminus \paren {\bigcup_{\alpha \mathop \prec \beta} E_\alpha}$
satisfying:
:$\ds \bigsqcup_{\alpha \mathop \in I} F_n = \bigcup_{\alpha \mathop \in I} E_n$
where:
:$\bigsqcup$ denotes disjoint union.
:$\alpha \prec \beta$ denotes that $\alpha \preccurlyeq \beta$ and $\alpha \ne \beta$.
== Proof ==
Denote:
where:
:$\ds F_\beta = E_\beta \setminus \paren {\bigcup_{\alpha \mathop \prec \beta} E_\alpha}$
We first show that $E = F$.
That $x \in E \implies x \in F$ follows from the construction of $F$ from subsets of $E$.
Thus $E \subseteq F$.
Then:
so $F \subseteq E$.
Thus $E = F$ by definition of set equality.
To show that the sets in $F$ are (pairwise) disjoint, consider an arbitrary $x \in F$. 
Then $x \in F_\beta$ for some $F_\beta$.
By the Well-Ordering Principle, there is a smallest such $\beta$ with respect to $\preccurlyeq$.
Then:
:$\forall \gamma \prec \beta: x \notin F_\gamma$
Choose any distinct $\eta, \zeta \in I$. 
We have:
If $\eta \prec \zeta$, then:
If $\zeta < \eta$, then:
So the sets $F_\eta, F_\zeta$ are disjoint. 
Thus $F$ is the disjoint union of sets equal to $E$:
:$\ds \bigcup_{\alpha \mathop \in I} E_\alpha = \bigsqcup_{\alpha \mathop \in I} F_\alpha$
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $13$
Category:Union of Indexed Family of Sets Equal to Union of Disjoint Sets
Union of Initial Segments is Initial Segment or All of Woset

== Theorem ==
Let $\struct {X, \preccurlyeq}$ be a well-ordered non-empty set.
Let $A \subseteq X$.
Let:
:$\ds J = \bigcup_{x \mathop \in A} S_x$
be a union of initial segments defined by the elements of $A$.
Then either:
:$J = X$
or:
:$J$ is an initial segment of $X$.
== Proof 1 ==
== Proof 2 ==
Category:Well-Orderings
Category:Union of Initial Segments is Initial Segment or All of Woset
Union of Initial Segments is Initial Segment or All of Woset/Proof 1

== Theorem ==
== Proof ==
Suppose the hypotheses of the theorem hold.
If $J = X$ then the theorem is satisfied.
Assume $J \ne X$.
Then $X \setminus J$ is non-empty.
By Subset of Well-Ordered Set is Well-Ordered, $X \setminus J$ is itself well-ordered.
Thus $X \setminus J$ has a smallest element; call it $b$.
We claim that $b \preccurlyeq y$ for every $y \in J$.
 there exists a $y \in J$ with $b \preccurlyeq y$.
Then there exists some $x_0 \in A$ with $y$ in the initial segment $S_{x_0}$.
Thus $b \in S_{x_0}$ and so $b \in J$.
This contradicts the fact that $b \in X \setminus J$.
Thus there cannot exist a $y \in J$ with $b \preccurlyeq y$.
Any element strictly preceding $b$ is in $J$, by choice of $b$.
Also, $b \notin J$, because $b \in X \setminus J$.
From above, any element strictly succeeding $b$ is not in $J$.
Thus $J = S_b$ by definition of initial segment.
== Sources ==
* $\S P.16$
Category:Union of Initial Segments is Initial Segment or All of Woset
Union of Initial Segments is Initial Segment or All of Woset/Proof 2

== Theorem ==
== Proof ==
Suppose the hypotheses of the theorem hold.
For every $x \in A$, we have $S_x \subseteq X$ by the definition of initial segment.
By Union of Family of Subsets is Subset, we have $J \subseteq X$.
Consider $X \setminus J$.
If $X \setminus J$ is empty, then $X \subseteq J$ by Set Difference with Superset is Empty Set, and hence $J = X$ by Subset Relation is Antisymmetric, so the theorem is satisfied.
So assume $X \setminus J$ is non-empty.
By Subset of Well-Ordered Set is Well-Ordered, $X \setminus J$ is itself well-ordered.
Thus $X \setminus J$ has a smallest element; call it $a$.
We will show that $J = S_a$, thus showing the veracity of the theorem.
We need to show that:
:$J = \set {b \in S: b \prec a}$
which by definition of set equality is equivalent to the claim that:
:$\forall z \in X: \in J \iff z \prec a$
For the forward direction, suppose $z \in J$.
We need to show that $z \prec a$.
By definition of union, there is $x \in A$ such that $z \in S_x$, that is:
:$z \prec x$
 $a \preceq z$.
Then $a \prec x$, so $a \in S_x$ by the definition of initial segment
Hence $a \in J$ by definition of union.
This contradicts the fact that $a \in X \setminus J$.
Thus $z \prec a$.
For the backward direction, suppose $z \prec a$ where $z \in X$. We need to show that $z \in J$
 $z \notin J$.
Then $z \in X \setminus J$.
Hence $a \preceq z$ by definition of smallest element.
This contradicts the fact that $z \prec a$.
Thus $z \in J$.
Since $z$ was arbitrary, we conclude that $J = S_a$ as required.
Category:Union of Initial Segments is Initial Segment or All of Woset
Union of Interiors and Boundary Equals Whole Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $A$ be a subset of $T$.
Then:
:$S = \Int A \cup \partial A \cup \Int {A'}$
where:
:$A' = S \setminus A$ denotes the complement of $A$ relative to $S$
:$\Int A$ denotes the interior of $A$
:$\partial A$ denotes the boundary of $A$.
== Proof ==
== Sources ==
* 
* 
Category:Set Boundaries
Category:Set Interiors
Union of Interiors is Subset of Interior of Union

== Theorem ==
Let $T$ be a topological space.
Let $\H$ be a set of subsets of $T$.
That is, let $\H \subseteq \powerset T$ where $\powerset T$ is the power set of $T$.
Then the union of the interiors of the elements of $\H$ is a subset of the interior of the union of $\H$:
:$\ds \bigcup_{H \mathop \in \H} H^\circ \subseteq \paren {\bigcup_{H \mathop \in \H} H}^\circ $
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Interior of Union is not necessarily Union of Interiors
Category:Union of Interiors is Subset of Interior of Union
Category:Set Interiors
Category:Set Union
Union of Interiors is Subset of Interior of Union/Proof 1

== Theorem ==
== Proof ==
In the following, $H^-$ denotes the closure of the set $H$.
At this point we note that:
:$(1): \quad \ds \paren {\bigcap_{H \mathop \in \mathbb H} \paren {T \setminus H} }^- \subseteq \bigcap_{H \mathop \in \mathbb H} \paren {T \setminus H}^-$
from Closure of Intersection is Subset of Intersection of Closures.
Then we note that:
:$\ds T \setminus \paren {\bigcap_{H \mathop \in \mathbb H} \paren {T \setminus H}^-} \subseteq T \setminus \paren {\paren {\bigcap_{H \mathop \in \mathbb H} \paren {T \setminus H} }^-} $
from $(1)$ and Set Complement inverts Subsets.
Then we continue:
Category:Union of Interiors is Subset of Interior of Union
Union of Interiors is Subset of Interior of Union/Proof 2

== Theorem ==
== Proof ==
Let $\mathbb U$ be the set of all open subsets of $\bigcup \H$.
Then by definition of interior:
:$\ds \paren {\bigcup_{H \mathop \in \H} H}^\circ = \paren {\bigcup \H}^\circ = \bigcup \mathbb U$
As $\mathbb U$ contains all open subsets of $\bigcup \H$, and $H^\circ$ is open for any $H \in \H$:
:$\ds \set {H^\circ : H \in \H} \subseteq \mathbb U$
Thus:
:$\ds \bigcup_{H \mathop \in \H} H^\circ = \bigcup \set {H^\circ : H \in \H} \subseteq \bigcup \mathbb U$
Category:Union of Interiors is Subset of Interior of Union
Union of Intersections/Examples/4 Arbitrarily Chosen Sets of Complex Numbers

== Example of Union of Intersections ==
Let:
Then:
:$\paren {A \cap C} \cup \paren {B \cap D} = \set {1, i, -i}$
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Point Sets: $121 \ \text{(b)}$
Category:Union of Intersections
Union of Intersections of 2 from 3 equals Intersection of Unions of 2 from 3

== Theorem ==
Let $A$, $B$ and $C$ be sets.
Then:
:$\paren {A \cap B} \cup \paren {B \cap C} \cup \paren {C \cap A} = \paren {A \cup B} \cap \paren {B \cup C} \cap \paren {C \cup A}$
== Proof ==
== Sources ==
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: Exercise $13$
Category:Set Intersection
Category:Set Union
Union of Inverse of Relations is Inverse of their Union

== Theorem ==
For $i \in \set {1, 2}$, let $\RR_i \subseteq S_i \times T_i$ be relations on $S_i \times T_i$.
Let ${\RR_i}^{-1} \subseteq T_i \times S_i$ be the inverse of $\RR_i$.
Then:
:${\RR_1}^{-1} \cup {\RR_2}^{-1} = \paren {\RR_1 \cup \RR_2}^{-1}$
== Proof ==
Let $\tuple {t, s} \in {\RR_1}^{-1} \cup {\RR_2}^{-1}$.
By definition of union:
:$\tuple {t, s} \in {\RR_1}^{-1} \lor \tuple {t, s} \in {\RR_2}^{-1}$.
For $i \in \set {1, 2}$, let $\tuple {t, s} \in {\RR_i}^{-1}$.
By definition of inverse:
:$\tuple {s, t} \in \RR_i$
That is:
:$\tuple {s, t} \in \RR_1 \lor \tuple {s, t} \in \RR_2$
By definition of union:
:$\tuple {s, t} \in \RR_1 \lor \tuple {s, t} \in \RR_2 \iff \tuple {s, t} \in \RR_1 \cup \RR_2$
By the definition of inverse:
:$\tuple {t, s} \in \paren {\RR_1 \cup \RR_2}^{-1}$
Category:Inverse Relations
Category:Set Union
Union of Inverses of Mappings is Inverse of Union of Mappings

== Theorem ==
Let $I$ be an indexing set.
Let $\family {f_i: i \in I}$ be an indexed family of mappings.
For each $i \in I$, let $f^{-1}$ denote the inverse of $f$.
Then the inverse of the union of $\family {f_i: i \in I}$ is the union of the inverses of $f_i, i \in I$.
That is:
:$\ds \paren {\bigcup \family {f_i: i \in I} }^{-1} = \bigcup \family {f_i^{-1}: i \in I}$
== Proof ==
Category:Inverse Relations
Category:Set Union
Category:Indexed Families
Union of Left-Total Relations is Left-Total

== Theorem ==
Let $S_1, S_2, T_1, T_2$ be sets or classes.
Let $\RR_1 \subseteq S_1 \times T_1$ and $\RR_2 \subseteq S_2 \times T_2$ be left-total relations.
Then $\RR_1 \cup \RR_2$ is left-total.
== Proof ==
Let both $\RR_1$ and $\RR_2$ be left-total.
Let $\RR = \RR_1 \cup \RR_2$.
Let $s \in S_1 \cup S_2$.
By the definition of union:
:$s \in S_1 \lor s \in S_2$
Thus $s \in S_i$ for $i \in \set {1, 2}$.
By definition of left-total relation, there is a $t \in T_i$ such that $\tuple {s, t} \in \RR_i$.
We have that $\RR$ is a superset of $\RR_i$.
Hence from Union is Smallest Superset:
:$\tuple {s, t} \in \RR_i \subseteq \RR \implies \tuple {s, t} \in \RR$
== Also see ==
* Union of Right-Total Relations is Right-Total
Category:Relation Theory
Union of Local Bases is Basis

== Theorem ==
Let $T = \struct {X, \tau}$ be a topological space.
For each $x \in X$, let $\BB_x$ be a local basis at $x$ which consists entirely of open sets.
Then $\ds \BB = \bigcup_{x \mathop \in X} \BB_x$ is a basis for the topology $\tau$.
== Proof ==
Let $U \in \tau$ be any open set of $T$.
Consider any $x \in U$.
Then, by definition of local basis, there exists a $B_x \in \BB_x$ such that $B_x \subseteq U$.
Also by definition of local basis:
:$x \in B_x$
So by Set Union Preserves Subsets:
:$U = \ds \bigcup_{x \mathop \in U} \set x \subseteq \bigcup_{x \mathop \in U} B_x$
Also, since $B_x \subseteq U$ for all $x$, by Union is Smallest Superset: General Result:
:$\ds \bigcup_{x \mathop \in U} B_x \subseteq U$
Thus by definition of set equality:
:$\ds U = \bigcup_{x \mathop \in U} B_x$
So, by definition, $\ds \BB = \bigcup_{x \mathop \in X} \BB_x$ is a basis for the topology $\tau$.
== Also see ==
* Basis induces Local Basis
Category:Topological Bases
Category:Local Bases
Union of Many-to-One Relations with Disjoint Domains is Many-to-One

== Theorem ==
Let $S_1, S_2, T_1, T_2$ be sets or classes.
Let $\RR_1$ be a many-to-one relation on $S_1 \times T_1$.
Let $\RR_2$ be a many-to-one relation on $S_2 \times T_2$.
Suppose that the domains of $\RR_1$ and $\RR_2$ are disjoint.
Then $\RR_1 \cup \RR_2$ is a many-to-one relation on $\paren {S_1 \cup S_2} \times \paren {T_1 \cup T_2}$.
== Proof ==
Let $\RR = \RR_1 \cup \RR_2$.
Let $\tuple {x, y_1}, \tuple {x, y_2} \in \RR$.
By the definition of union, $\tuple {x, y_1}$ and $\tuple {x, y_2}$ are each in $\RR_1$ or $\RR_2$.
Suppose that both are in $\RR_1$.
Then since $\RR_1$ is a many-to-one relation, $y_1 = y_2$.
Suppose that $\tuple {x, y_1} \in \RR_1$ and $\tuple {x, y_2} \in \RR_2$.
Then $x$ is in the domain of $\RR_1$ and that of $\RR_2$, contradicting the premise, so this cannot occur.
The other two cases are precisely similar.
Thus in all cases $y_1 = y_2$.
As this holds for all such pairs, $\RR$ is many-to-one.
Category:Relation Theory
Union of Mappings which Agree is Mapping

== Theorem ==
Let $A, B, Y$ be sets.
Let $f: A \to Y$ and $g: B \to Y$ be mappings.
Let $X = A \cup B$.
Let $f$ and $g$ agree on $A \cap B$.
Then $f \cup g: X \to Y$ is a mapping.
=== Family of Sets ===
== Proof ==
By definition, $f \cup g$ is a relation whose domain is $X = A \cup B$.
Let $\tuple {x, y_1} \in f \cup g$ and $\tuple {x, y_2} \in f \cup g$.
At least one of the following must be true:
:$(1): \quad \tuple {x, y_1} \in f, \tuple {x, y_2} \in f$
:$(2): \quad \tuple {x, y_1} \in g, \tuple {x, y_2} \in g$
:$(3): \quad \tuple {x, y_1} \in f, \tuple {x, y_2} \in g$
:$(4): \quad \tuple {x, y_1} \in g, \tuple {x, y_2} \in f$
Because $f$ and $g$ are mappings, $(1)$ and $(2)$ imply that $y_1 = y_2$.
If $(3)$ holds, then $y_1 = \map f x$ and $y_2 = \map g x$.
But then $x \in A \cap B$.
So by hypothesis:
:$y_1 = \map f x = \map g x = y_2$
Similarly if $(4)$ holds.
Thus in all cases:
:$\tuple {x, y_1}, \tuple {x, y_2} \in f \cup g \implies y_1 = y_2$
and so by definition $f \cup g: X \to Y$ is a mapping.
== Sources ==
* : $\S 1.13$: The Restriction of a Function: Theorem $13.2$
Category:Mapping Theory
Category:Set Union
Union of Mappings which Agree is Mapping/Family of Mappings

== Theorem ==
Let $Y$ be a set.
Let $\family {A_i}_{i \mathop \in I}$ be a family of sets indexed by $I$.
Let $\family {f_i: A_i \to Y}$ be a family of mappings indexed by $I$.
Let $X = \ds \bigcup_{i \mathop \in I} A_i$.
Let $f = \ds \bigcup_{i \mathop \in I} f_i : X \to Y$ where $\ds \bigcup_{i \mathop \in I} f_i$ is the union of relations.
Let for all $i, j \in I$, $f_i$ and $f_j$ agree on $A_i \cap A_j$.
Then:
:$f : X \to Y$ is a mapping
== Proof ==
By Union of Relations is Relation, $f = \ds \bigcup_{i \mathop \in I} f_i$ is a relation whose domain is $X = \ds \bigcup_{i \mathop \in I} A_i$.
Let $\tuple {x, y_1}, \tuple {x, y_2} \in f$.
Then there exists $i, j \in I$ such that:
:$\tuple {x, y_1} \in f_i$ and $\tuple {x, y_2} \in f_j$
=== Case 1 : $i = j$ ===
Let $i = j$.
Then $\tuple {x, y_1}, \tuple {x, y_2} \in f_i$.
By definition of a mapping:
:$y_1 = y_2$.
=== Case 2 : $i \ne j$ ===
Let $i \ne j$.
Then $y_1 = \map {f_i} x$ and $y_2 = \map {f_i} x$.
But then $x \in A_i \cap A_j$.
By definition of agreement of mappings:
:$y_1 = \map {f_i} x = \map {f_j} x = y_2$
Thus in both cases:
:$\tuple {x, y_1}, \tuple {x, y_2} \in f \implies y_1 = y_2$
It follows that $f : X \to Y$ is a mapping by definition.
Category:Mapping Theory
Category:Set Union
Category:Indexed Families
Union of Mappings with Disjoint Domains is Mapping

== Theorem ==
Let $S_1, S_2, T_1, T_2$ be sets.
Let $f: S_1 \to T_1$ and $g: S_2 \to T_2$ be mappings.
Let $h = f \cup g$ be their union.
If $S_1 \cap S_2 = \O$, then $h: S_1 \cup S_2 \to T_1 \cup T_2$ is a mapping whose domain is $S_1 \cup S_2$.
== Proof ==
From the definition of mapping, it is clear that $h$ is a relation.
Suppose $\tuple {x, y_1}, \tuple {x, y_2} \in h$.
Clearly $x \in S_1$ or $x \in S_2$ but as $S_1 \cap S_2 = \O$ it is not in both.
If $x \in S_1$ then $y_1 = y_2$ as $\tuple {x, y_1}, \tuple {x, y_2} \in f$, and $f$ is a mapping.
Similarly, if $x \in S_2$ then $y_1 = y_2$ as $\tuple {x, y_1}, \tuple {x, y_2} \in g$, and $g$ is a mapping.
So $\tuple {x, y_1}, \tuple {x, y_2} \in h \implies y_1 = y_2$.
Now suppose $x \in \Dom h$.
Either $x \in S_1$ or $x \in S_2$.
As both $f$ and $g$ are mappings it follows that either $\exists y \in T_1: \tuple {x, y} \in f$ or $\exists y \in T_2: \tuple {x, y} \in g$.
In either case, $\exists y \in T_1 \cup T_2: \tuple {x, y} \in h$.
So $h$ is a mapping whose domain is $S_1 \cup S_2$, as we were to show.
== Warning ==
If $S_1 \cap S_2 \ne \O$ then there may exist $\tuple {x, y_1} \in f$ and $\tuple {x, y_2} \in g$ such that $y_1 \ne y_2$.
In such a case $h = f \cup g$ is not a mapping.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 3$: Unions and Intersections of Sets: Exercise $3.8$
Category:Mapping Theory
Category:Set Union
Union of Matroid Base with Element of Complement is Dependent

== Theorem ==
Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $B \subseteq S$ be a base of $M$.
Let $x \in S \setminus B$.
Then:
:$B \cup \set x$ is a dependent superset of $B$
== Proof ==
From Set is Subset of Union:
:$B \subseteq B \cup \set x$
Because $x \in B \cup \set x$ and $x \notin B$:
:$B \ne B \cup \set x$
Hence:
:$B \subsetneq B \cup \set x$
By definition of base:
:$B$ is a maximal independent subset
Hence:
:$B \cup \set x \notin \mathscr I$
Category:Matroid Bases
Category:Matroid Dependent Subsets
Union of Meager Sets is Meager Set

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $A$ and $B$ be meager in $T$.
Then $A \cup B$ is meager in $T$.
== Proof ==
Since $A$ is meager in $T$:
:there exists a countable collection of sets $\set {U_n: n \in \N}$ nowhere dense in $T$ such that $\ds A = \bigcup_{n \mathop \in \N} U_n$.
Since $B$ is meager in $T$:
:there exists a countable collection of sets $\set {V_m: m \in \N}$ nowhere dense in $T$ such that $\ds B = \bigcup_{m \mathop \in \N} V_m$.
Then:
:$\ds A \cup B = \paren {\bigcup_{n \mathop \in \N} U_n} \cup \paren {\bigcup_{m \mathop \in \N} V_m}$
The is a countable union of sets nowhere dense in $T$, so:
:$A \cup B$ is meager in $T$.
Category:Meager Spaces
Category:Set Union
Union of Nest of Injections is Injection

== Theorem ==
Let $N$ be a nest of mappings which are all injections.
Then:
:$\bigcup N$ is an injection
where $\bigcup N$ denotes the union of $N$.
== Proof ==
== Sources ==
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 1$ A few preliminaries: Proposition $1.4$
Category:Union of Nest of Injections is Injection
Category:Nests of Mappings
Union of Nest of Injections is Injection/Proof

== Theorem ==
== Proof ==
From Union of Nest of Mappings is Mapping we have that $\bigcup N$ is a mapping.
:
:$\exists x_1, x_2: \tuple {x_1, y} \in \bigcup N \land \tuple {x_2, y} \in \bigcup N$
such that:
:$x_1 \ne x_2$
Then:
:$\exists f \subseteq \bigcup N: \tuple {x_1, y} \in f$
and:
:$\exists g \subseteq \bigcup N: \tuple {x_1, y} \in g$
But because $N$ is a nest of mappings, either:
:$f \subseteq g$
or:
:$g \subseteq f$
That means either:
:$\tuple {x_1, y} \in g$
or:
:$\tuple {x_2, y} \in f$
Hence either $f$ or $g$ has both $\tuple {x_1, y}$ and $\tuple {x_2, y}$ in it.
That is, either $f$ or $g$ is not an injection.
From this contradiction it follows that if: 
:$\exists x_1, x_2: \tuple {x_1, y} \in \bigcup N \land \tuple {x_2, y} \in \bigcup N$
then it has to be the case that:
:$x_1 = x_2$
It follows that $\bigcup N$ is a an injection.
== Sources ==
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 1$ A few preliminaries: Exercise $1.1$
Category:Union of Nest of Injections is Injection
Union of Nest of Mappings is Mapping

== Theorem ==
Let $N$ be a nest of mappings.
Then:
:$\bigcup N$ is a mapping
where $\bigcup N$ denotes the union of $N$.
== Proof ==
== Sources ==
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 1$ A few preliminaries: Proposition $1.4$
Category:Union of Nest of Mappings is Mapping
Category:Nests of Mappings
Union of Nest of Mappings is Mapping/Proof

== Theorem ==
== Proof ==
By definition of union, the elements of $\bigcup N$ are elements of elements of $N$.
That is, the elements of $\bigcup N$ are ordered pairs of sets.
Let $x \in \bigcup N$.
Then:
:$\exists f \subseteq \bigcup N: \exists y: \tuple {x, y} \in f$
:
:$\exists y_1, y_2: \tuple {x, y_1} \in \bigcup N \land \tuple {x, y_2} \in \bigcup N$
such that:
:$y_1 \ne y_2$
Then:
:$\exists f \subseteq \bigcup N: \tuple {x, y_1} \in f$
and:
:$\exists g \subseteq \bigcup N: \tuple {x, y_2} \in g$
But because $N$ is a nest of mappings, either:
:$f \subseteq g$
or:
:$g \subseteq f$
That means either:
:$\tuple {x, y_1} \in g$
or:
:$\tuple {x, y_2} \in f$
Hence either $f$ or $g$ has both $\tuple {x, y_1}$ and $\tuple {x, y_2}$ in it.
That is, either $f$ or $g$ is not a mapping.
From this contradiction it follows that if: 
:$\exists y_1, y_2: \tuple {x, y_1} \in \bigcup N \land \tuple {x, y_2} \in \bigcup N$
then it has to be the case that:
:$y_1 = y_2$
It follows that $\bigcup N$ is a mapping.
== Sources ==
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 1$ A few preliminaries: Exercise $1.1$
Category:Union of Nest of Mappings is Mapping
Union of Nest of Ordinal Sequences which is Proper Class

== Theorem ==
Let $N$ be a nest of ordinal sequences such that $N$ is a proper class.
Let $\bigcup N$ denote the union of $N$.
Then $\bigcup N$ is a mapping whose domain is the class of all ordinals $\On$.
== Proof ==
== Sources ==
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 5$ Transfinite recursion theorems: Lemma $5.2 \ (2)$
Category:Ordinal Sequences
Category:Nests
Category:Class of All Ordinals
Union of Non-Disjoint Bounded Subsets of Metric Space is Bounded

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $B$ and $C$ be bounded subsets of $M$ such that $B \cap C \ne \O$.
Let $\map \diam B$ and $\map \diam C$ denote the diameters of $B$ and $C$.
Then $B \cup C$ is a bounded subset of $M$ such that:
:$\map \diam {B \cup C} \le \map \diam B + \map \diam C$
== Proof ==
 there exists $x, y \in B \cup C$ such that:
:$\map d {x, y} > \map \diam B + \map \diam C$
$x$ and $y$ cannot both be in $B$ or $C$, otherwise either $\map d {x, y} \le \map \diam B$ or $\map d {x, y} \le \map \diam C$.
, let $x \in B$ and $y \in C$.
Let $z \in B \cap C$.
This is possible because $B \cap C \ne \O$ .
Then:
This contradicts our supposition that $\map d {x, y} > \map \diam B + \map \diam C$.
Hence:
== Sources ==
* : $2$: Continuity generalized: metric spaces: Exercise $2.6: 5$
Category:Bounded Metric Spaces
Union of Non-Disjoint Convex Sets is Convex Set

== Theorem ==
Let $\struct {S, \preccurlyeq}$ be an ordered set.
Let $\CC$ be a set of convex sets of $S$ such that their intersection is non-empty:
:$\ds \bigcap \CC \ne \O$
Then the union $\ds \bigcup \CC$ is also convex.
== Proof ==
Let $x, y, z \in S$ be arbitrary elements of $S$ such that $x \prec y \prec z$.
Let $x, z \in \ds \bigcup \CC$.
First let $x, z \in C$ where $C \in \CC$.
Then as $C$ is convex, $y \in C$.
Hence, by definition of union, $y \in \ds \bigcup \CC$.
Now let $x \in C_1, z \in C_2$ where $C_1, C_2 \in \CC$.
We have that $\ds \bigcap \CC \ne \O$.
Thus $C_1 \cap C_2 \ne \O$.
Then $\exists a \in C_1 \cap C_2: x < a < z$.
Hence one of the following cases holds:
:$(1): \quad x < y < a < z$, whence $y \in C_1$, by convexity of $C_1$
:$(2): \quad x < a < y < z$, whence $y \in C_2$, by convexity of $C_2$
:$(3): \quad y = a$, whence $y \in C_1$ and $y \in C_2$, by definition of $a$.
Thus in all cases $y \in \ds \bigcup \CC$.
Thus $\ds \bigcup \CC$ is convex by definition.
== Sources ==
* : Part $\text {II}$: Counterexamples: $39$. Order Topology: $2$
Category:Convex Sets (Order Theory)
Category:Set Union