nyuuzyou/wikis huggingface dataset:
Union of Total Ordering with Lower Sections is Total Ordering

== Theorem ==
Let $\struct {Y, \preceq}$ be a totally ordered set.
Let $X$ be the disjoint union of $Y$ with the set of lower sections of $Y$.
Define a relation $\preceq'$ on $X$ extending $\preceq$ by letting:
:$y_1 \preceq' y_2 \iff y_1 \preceq y_2$
:$y \preceq' L \iff y \in L$
:$L_1 \preceq' L_2 \iff L_1 \subseteq L_2$
:$L \preceq' y \iff y \in Y \setminus L$
Then $\preceq'$ is a total ordering.
== Proof ==
First note that by Lower Sections in Totally Ordered Set form Chain:
:$\subseteq$ is a total ordering on the set of lower sections.
Also note that by Complement of Lower Section is Upper Section, the complement of each $\preceq$-lower section is a $\preceq$-upper section.
=== Reflexivity ===
This follows immediately from the fact that $\preceq$ and $\subseteq$ are reflexive.
Thus $\preceq'$ is reflexive.
=== Transitivity ===
There are eight possibilities to consider.
If $y_1 \preceq' y_2$ and $y_2 \preceq' y_3$, then $y_1 \preceq' y_3$ because $\preceq$ is transitive.
If $L_1 \preceq' L_2$ and $L_2 \preceq' L_3$, then $L_1 \preceq' L_3$ because $\subseteq$ is transitive.
If $y_1 \preceq' y_2$ and $y_2 \preceq' L$, then:
:$y_1 \preceq y_2$ and $y_2 \in L$
Since $L$ is a lower section in $Y$:
:$y_1 \in L$
so:
:$y_1 \preceq' L$
If $L \preceq' y_1$ and $y_1 \preceq' y_2$, then:
:$y_1 \in Y \setminus L$ and $y_1 \preceq y_2$
Since $Y \setminus L$ is an upper section in $Y$:
:$y_2 \in Y \setminus L$
so:
:$L \preceq' y_2$
If $y \preceq' L_1$ and $L_1 \preceq' L_2$, then:
:$y \in L_1$ and $L_1 \subseteq L_2$
By the definition of subset:
:$y \in L_2$
so:
:$y \preceq' L_2$
If $L_1 \preceq' L_2$ and $L_2 \preceq' y$, then:
:$y \in Y \setminus L_2$ and $L_2 \supseteq L_1$
so:
:$y \in Y \setminus L_1$
so:
:$L_1 \preceq' y$
If $y_1 \preceq' L$ and $L \preceq' y_2$ then:
:$y_1 \in L$ and $y_2 \in Y \setminus L$
Since $L$ is a lower section:
:$y_2 \npreceq y_1$
Since $\preceq$ is a total ordering:
:$y_1 \preceq y_2$
so:
:$y_1 \preceq' y_2$
If $L_1 \preceq' y$ and $y \preceq' L_2$, then:
:$y \in L_2$ but $y \notin L_1$
Thus:
:$L_2 \nsubseteq L_1$
Since $\subseteq$ is a total ordering on the lower sections:
:$L_1 \subseteq L_2$
so:
:$L_1 \preceq' L_2$
Thus $\preceq'$ is transitive.
=== Antisymmetry ===
There are three cases.
If $y_1 \preceq' y_2$ and $y_2 \preceq' y_1$ then:
:$y_1 \preceq y_2$ and $y_2 \preceq y_1$
Since $\preceq$ is antisymmetric:
:$y_1 = y_2$
If $L_1 \preceq' L_2$ and $L_2 \preceq' L_1$ then:
:$L_1 \subseteq L_2$ and $L_2 \subseteq L_1$
Since $\subseteq$ is antisymmetric:
:$L_1 = L_2$
By the definition of $\preceq'$, it is impossible for $y \preceq' L$ and $L \preceq' y$, so the third case cannot occur.
Thus$\preceq'$ is antisymmetric.
Since $\preceq'$ is reflexive, transitive, and antisymmetric, it is an ordering.
$\preceq'$ is a total ordering of $X$ because:
:$\preceq$ is a total ordering
:the set of lower sections is a chain
and:
:for any $y$ and $L$ either $y \in L$ or $y \in Y \setminus L$.
Category:Lower Sections
Category:Total Orderings
Union of Transitive Class is Subclass

== Theorem ==
Let $A$ be a transitive class.
Let $\bigcup A$ denote the union of $A$.
Then:
:$\bigcup A \subseteq A$
== Proof ==
Let $A$ be transitive.
Let $x \in \bigcup A$.
Then by definition:
:$\exists y \in A: x \in y$
By definition of transitive class:
:$x \in y \land y \in A \implies x \in A$
and so:
:$x \in A$
Hence the result by definition of subclass.
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 5$ The union axiom: Exercise $5.3. \ \text {(a)}$
Category:Class is Transitive iff Union is Subclass
Union of Transitive Class is Transitive

== Theorem ==
Let $A$ be a class.
Let $\bigcup A$ denote the union of $A$.
Let $A$ be transitive.
Then $\bigcup A$ is also transitive.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 5$ The union axiom: Exercise $5.3. \ \text {(c)}$
Category:Class Union
Category:Transitive Classes
Category:Union of Transitive Class is Transitive
Union of Transitive Class is Transitive/Proof 1

== Theorem ==
== Proof ==
Let $A$ be transitive.
By Class is Transitive iff Union is Subclass:
:$\bigcup A \subseteq A$
By Union of Subclass is Subclass of Union of Class:
:$\map \bigcup {\bigcup A} \subseteq \bigcup A$
Then by Class is Transitive iff Union is Subclass:
:$\bigcup A$ is transitive.
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 10$ Some useful facts about transitivity: Theorem $10.2$
Category:Union of Transitive Class is Transitive
Union of Transitive Class is Transitive/Proof 2

== Theorem ==
== Proof ==
Let $A$ be transitive.
Let $x \in \bigcup A$.
By Class is Transitive iff Union is Subclass we have that:
:$\bigcup A \subseteq A$
Thus by definition of subclass:
:$x \in A$
As $A$ is transitive:
:$x \subseteq A$
Let $z \in x$.
As $x \subseteq A$, it follows by definition of subclass that:
:$z \in A$
Thus we have that:
:$\exists x \in A: z \in x$
and so by definition of union of class:
:$z \in \bigcup A$
Thus we have that:
:$z \in x \implies z \in \bigcup A$
and so by definition of subclass:
:$x \subseteq \bigcup A$
Thus we have that:
:$x \in \bigcup A \implies x \subseteq \bigcup A$
Hence $\bigcup A$ is a transitive class by definition.
Category:Union of Transitive Class is Transitive
Union of Transitive Relations Not Always Transitive

== Theorem ==
The union of transitive relations is not necessarily itself transitive.
== Proof ==
Proof by counterexample:
Let $S = \set {a, b, c, d}$.
Let $\RR_1$ be the transitive relation $\set {\tuple {a, b}, \tuple {b, c}, \tuple {a, c} }$.
Let $\RR_2$ be the transitive relation $\set {\tuple {b, c}, \tuple {c, d}, \tuple {b, d} }$.
Then we have that $\tuple {a, b} \in \RR_1 \cup \RR_2$ and $\tuple {b, d} \in \RR_1 \cup \RR_2$.
However, $\tuple {a, d} \notin \RR_1 \cup \RR_2$, and so $\RR_1 \cup \RR_2$ is not transitive.
Category:Transitive Relations
Category:Set Union
Union of Turing Machines

== Theorem ==
Let $T_1, T_2$ be Turing machines.
Let $\Sigma_1, \Sigma_2$ be the input symbols of $T_1, T_2$, respectively.
Let $L_1, L_2$ be the languages accepted by $T_1, T_2$, respectively.
There exists a Turing machine $T$ such that:
* The input symbols of $T$ are $\Sigma_1 \cup \Sigma_2$.
* The language accepted by $T$ is $L_1 \cup L_2$.
* $T$ halts on some input it is in the accepted language, or both $T_1$ and $T_2$ halt on it.
== Proof ==
Let:
:$T_1 = \tuple {Q_1, \Sigma_1, \Gamma_1, \delta_1, q_1, B, F_1}$
:$T_2 = \tuple {Q_2, \Sigma_2, \Gamma_2, \delta_2, q_2, B, F_2}$
As implied by the notation, the blank symbols of the two machines will be identified with each other, and denoted as $B$.
Construct the $2$-tape Turing machine $T' = \tuple{Q, \Sigma, \Gamma, \delta, q_0, B, F}$ as follows:
:$Q = \set {q_0, q_L} \cup \paren {Q_1 \cup \set {q_H} } \times \paren {Q_2 \cup \set {q_H} }$
:$\Sigma = \Sigma_1 \cup \Sigma_2$
:$\Gamma = \Gamma_1 \cup \Gamma_2$
:$F = F_1 \times Q_2 \cup Q_1 \times F_2$
:$\map \delta {q, s, t} = \begin{cases}
\tuple {q_0, s, s, R, R} & : q = q_0 \land s \ne B \\
\tuple {q_L, B, B, L, L} & : q = q_0 \land s = B \\
\tuple {q_L, s, t, L, L} & : q = q_L \land s \ne B \\
\tuple {\tuple {q_1, q_2}, B, B, R, R} & : q = q_L \land s = B \\
\tuple {\tuple {a', b'}, s', t', d_1, d_2} & : q = \tuple {a, b}
\end{cases}$
::where $\tuple {a', s', d_1} = \map {\delta_1} {a, s}$ if $\map {\delta_1} {a, s}$ is defined, and $\tuple {a, s, S}$ otherwise; and likewise for $\delta_2$.
The following special cases will apply to $\delta$:
* $\map {\delta} {\tuple {q_H, q_H}, s, t}$ is undefined for any $s, t$
* $\map {\delta} {\tuple {a, b}, s, t}$ is undefined if at least one of $a \in F_1$ or $b \in F_2$ is true.
The state $q_0$ copies the input string from tape $1$ to tape $2$, until it encounters a blank.
State $q_L$ moves both tapes back to the start of the input string.
After this, $T_1$ and $T_2$ are represented on the two tapes of $T'$, in the natural way.
$T'$ will accept and halt when either machine enters one of its accepting states, as required.
When either machine halts, that is when $\map {\delta_1} {a, s}$ is undefined, $T'$ changes the respective state to $q_H$, and does not alter its tape or state after that.
If both machines halt without accepting, $T'$ does the same.
The result follows from Multitape Turing Machine Reduces to Turing Machine.
Category:Turing Machines
Union of Two Compact Sets is Compact

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological spaces.
Let $H$ and $K$ be compact subsets of $T$.
Then $H \cup K$ is compact in $T$.
== Proof ==
Let $\CC$ be an open cover of $H \cup K$.
Then $\CC$ is an open cover of both $H$ and $K$.
As $H$ and $K$ are both compact in $T$:
:$H$ has a finite subcover $C_H$ of $\CC$
:$K$ has a finite subcover $C_K$ of $\CC$.
Their union $C_H \cup C_K$ is a finite subcover of $\CC$ for $H \cup K$.
From Union of Finite Sets is Finite it follows that $C_H \cup C_K$ is finite.
As $\CC$ is arbitrary, it follows by definition that $H \cup K$ is compact in $T$.
== Sources ==
* : $5$: Compact spaces: Exercise $5.10: 3$
Category:Compact Spaces
Category:Set Union
Union of Union of Cartesian Product

== Theorem ==
Let $A$ and $B$ be sets such that $A \ne \O$ and $B \ne \O$.
Let the ordered pair $\tuple {a, b}$ be defined using the Kuratowski formalization:
:$\tuple {a, b} := \set {\set a, \set {a, b} }$
Then:
:$\ds \bigcup \bigcup \paren {A \times B} = A \cup B$
where:
:$\cup$ denotes union
:$\times$ denotes Cartesian product.
== Proof ==
== Also see ==
* Union of Union of Cartesian Product with Empty Factor
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $1$: Pairs, Relations, and Functions: Exercise $4 \ \text {(a)}$
Category:Set Union
Category:Cartesian Product
Union of Union of Cartesian Product with Empty Factor

== Theorem ==
Let $A$ and $B$ be sets such that either $A = \O$ or $B = \O$.
Let the ordered pair $\tuple {a, b}$ be defined using the Kuratowski formalization:
:$\tuple {a, b} := \set {\set a, \set {a, b} }$
Then:
:$\ds \bigcup \bigcup \paren {A \times B} = A \cup B \iff A = B = \O$
where:
:$\cup$ denotes union
:$\times$ denotes Cartesian product.
That is, if either $A$ or $B$ is empty:
:$\ds \bigcup \bigcup \paren {A \times B} = A \cup B$
holds they are both empty
== Proof ==
Let $A = \O$ or $B = \O$.
From Cartesian Product is Empty iff Factor is Empty:
:$A \times B = \O$
Hence from Union of Empty Set:
:$\ds \bigcup \bigcup \paren {A \times B} = \O$
However, from Union is Empty iff Sets are Empty:
:$A \cup B = \O \iff A = \O \text { and } B = \O$
The result follows.
== Also see ==
* Union of Union of Cartesian Product
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $1$: Pairs, Relations, and Functions: Exercise $4 \ \text {(b)}$
Category:Set Union
Category:Cartesian Product
Union of Union of Relation is Union of Domain with Image

== Theorem ==
Let $V$ be a basic universe.
Let $\RR \subseteq V \times V$ be a relation.
Let $\Dom \RR$ denote the domain of $\RR$.
Then:
:$\map \bigcup {\bigcup \RR} = \Dom \RR \cup \Img \RR$
where:
:$\bigcup \RR$ denotes the union of $\RR$
:$\Dom \RR$ denotes the domain of $\RR$
:$\Img \RR$ denotes the image of $\RR$.
== Proof ==
Category:Class Union
Category:Relation Theory
Union of Unordered Tuples

== Theorem ==
Let $x_1, \dots, x_n, x_{n+1}, \dots, x_m$ be arbitrary.
Then
:$\set {x_1, \dots, x_n} \cup \set {x_{n + 1}, \dots, x_m} = \set {x_1, \dots, x_n, x_{n + 1}, \dots, x_m}$
== Proof ==
Let $a$ be arbitrary.
:$a \in \set {x_1, \dots, x_n} \cup \set {x_{n + 1}, \dots, x_m}$
:$a \in \set {x_1, \dots, x_n}$ or $a \in \set {x_{n + 1}, \dots, x_m}$ by definition of union
:$a = x_1 \lor \dots \lor a = x_n$ or $a = x_{n + 1} \lor \dots \lor a = x_m$ by definition of unordered tuple
:$a = x_1 \lor \dots \lor a = x_n \lor a = x_{n + 1} \lor \dots \lor a = x_m$
:$a \in \set {x_1, \dots, x_n, x_{n + 1}, \dots, x_m}$ by definition of unordered tuple
Thus result follows by definition of set equality.
Category:Set Theory
Union of Upper Sections is Upper

== Theorem ==
Let $\struct {S, \preceq}$ be a preordered set.
Let $A$ be a set of subsets of $S$ such that:
:$\forall X \in A: X$ is an upper section.
Then:
:$\bigcup A$ is also an upper section.
== Proof ==
Let $x \in \bigcup A, y \in S$ such that:
:$x \preceq y$
By definition of union:
:$\exists Y \in A: x \in Y$
By assumption:
:$Y$ is an upper section.
By definition of upper section:
:$y \in Y$
Thus by definition of union:
:$y \in \bigcup A$
Hence
:$\bigcup A$ is an upper section.
== Sources ==
* 
* 
Category:Upper Sections
Category:Set Union
Union of Vertical Sections is Vertical Section of Union

== Theorem ==
Let $X$ and $Y$ be sets.
Let $\set {E_\alpha : \alpha \in A}$ be a set of subsets of $X \times Y$. 
Let $x \in X$. 
Then: 
:$\ds \paren {\bigcup_{\alpha \in A} E_\alpha}_x = \bigcup_{\alpha \in A} \paren {E_\alpha}_x$
where:
:$\ds \paren {\bigcup_{\alpha \in A} E_\alpha}_x$ is the $x$-vertical section of $\ds \bigcup_{\alpha \in A} E_\alpha$
:$\paren {E_\alpha}_x$ is the $x$-vertical section of $E_\alpha$.
 
== Proof ==
Note that:
:$\ds y \in \bigcup_{\alpha \in A} \paren {E_\alpha}_x$
: 
:$y \in \paren {E_\alpha}_x$ for some $\alpha \in A$.
From the definition of the $x$-vertical section, this is equivalent to:
:$\tuple {x, y} \in E_\alpha$ for some $\alpha \in A$. 
This in turn is equivalent to: 
:$\ds \tuple {x, y} \in \bigcup_{\alpha \in A} E_\alpha$
Again applying the definition of the $x$-vertical section:
:$\ds y \in \paren {\bigcup_{\alpha \in A} E_\alpha}_x$
So:
:$\ds y \in \bigcup_{\alpha \in A} \paren {E_\alpha}_x$ $\ds y \in \paren {\bigcup_{\alpha \in A} E_\alpha}_x$
giving:
:$\ds \paren {\bigcup_{\alpha \in A} E_\alpha}_x = \bigcup_{\alpha \in A} \paren {E_\alpha}_x$
Category:Set Union
Category:Vertical Section of Sets
Union of g-Tower is Greatest Element and Unique Fixed Point

== Theorem ==
Let $M$ be a set.
Let $g: M \to M$ be a progressing mapping on $M$.
Let $M$ be a $g$-tower.
Then:
:$\ds \bigcup M \in M$
:$\ds \bigcup M$ is the greatest element (under the subset relation) of $M$
:$\ds \bigcup M$ is the unique fixed point of $M$.
== Proof ==
Let the hypothesis be assumed.
By G-Tower is Nest, $M$ is a nest. By assumption, $M$ is a set.
Hence by definition $M$ is a chain.
By $g$-Tower is Closed under Chain Unions:
:$\ds \bigcup M \in M$
It follows by Draft:Union of Set of Sets is Greatest Element under Subset Relation that $\ds \bigcup M$ is the greatest element (under the subset relation) of $M$.
Hence $\ds \bigcup M$ cannot be a proper subset of $\ds \map g {\bigcup M}$.
Thus:
:$\ds \bigcup M = \map g {\bigcup M}$
and the result follows.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 2$ Superinduction and double superinduction: Theorem $2.6$
Category:G-Towers
Union of two Zero Loci is Zero Locus

== Theorem ==
Let $k$ be a field.
Let $n \in \N_{>0}$.
Let $A = k \sqbrk {X_1, \ldots, X_n}$ be the ring of polynomials in $n$ variables over $k$.
Let $S,T \subseteq A$.
Then:
:$\ds {\map V S} \cup \map V T = \map V {S \cdot T}$
where:
:$\map V \cdot$ denotes the zero locus
:$S \cdot T := \set { fg : f \in S, g \in T }$
== Proof ==
Let $x \in {\map V S} \cup \map V T$.
That is, either $x \in \map V S$ or $x \in \map V T$.
Then, if $x \in \map V S$:
:$\forall \paren {f,g} \in S \times T : \map f x \map g x = 0 \map g x = 0$
else:
:$\forall \paren {f,g} \in S \times T : \map f x \map g x = \map f x 0 = 0$
Therefore:
:$x \in \map V {S \cdot T}$
On the other hand:
Category:Algebraic Geometry
Union with Complement

== Theorem ==
The union of a set and its complement is the universal set:
:$S \cup \map \complement S = \mathbb U$
== Proof ==
Substitute $\mathbb U$ for $S$ and $S$ for $T$ in $T \cup \relcomp S T = S$ from Union with Relative Complement.
== Also see ==
* Intersection with Complement
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Exercise $\text{B v}$
* : $\S 1.6$: Set Identities and Other Set Relations: Exercise $2 \ \text{(j)}$
* : $\S 2$
* : $\text {(vii)}$
Category:Set Union
Category:Set Complement
Category:Universal Set
Union with Disjoint Singleton is Dependent if Element Depends on Subset

== Theorem ==
Let $M = \struct{S, \mathscr I}$ be a matroid.
Let $A \subseteq S$.
Let $x \in S : x \notin A$.
Let $x$ depend on $A$.
Then $A \cup \set x$ is a dependent subset of $M$.
== Proof ==
We proceed by Proof by Contraposition.
Let $A \cup \set x$ be independent.
By matroid axiom $(\text I 2)$:
:$A$ is independent
We have:
Then $x$ does not depend on $A$ by definition.
The theorem holds by the Rule of Transposition.
Category:Matroid Dependence
Category:Matroid Dependent Subsets
Union with Empty Set

== Theorem ==
The union of any set with the empty set is the set itself:
:$S \cup \O = S$
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Intersection with Empty Set
== Sources ==
* : $\S 4$: Unions and Intersections
* : Exercise $1.1: \ 8 \ \text{(b)}$
* : Introduction to Set Theory: $1$. Elementary Operations on Sets
* : Chapter $1$: Exercise $1 \ \text {(e)}$
* : Chapter $\text I$: Algebraic Structures: $\S 3$: Unions and Intersections of Sets: Theorem $3.1$
* : Chapter $\text{I}$: Sets and Functions: Exercise $\text{B ii}$
* : Chapter $1$: The Notation and Terminology of Set Theory: $\S 6 \ \text{(c)}$
* : $\S 1.6$: Set Identities and Other Set Relations: Exercise $2 \ \text{(a)}$
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems
* : $1$: Set Theory and Logic: $\S 1$: Fundamental Concepts
* : $\S 2$
* : Appendix $\text{A}.2$: Theorem $\text{A}.11$
* : $\text {(iii)}$
* : $\text {(i)}$
Category:Subsets
Category:Empty Set
Category:Set Union
Category:Union with Empty Set
Union with Intersection equals Intersection with Union iff Subset

== Theorem ==
Let $A$, $B$ and $C$ be sets.
Then:
:$\paren {A \cap B} \cup C = A \cap \paren {B \cup C} \iff C \subseteq A$
== Proof ==
=== Necessary Condition ===
Let $C \subseteq A$.
We have:
and another way:
Thus:
:$C \subseteq A \implies \paren {A \cap B} \cup C = A \cap \paren {B \cup C}$
=== Sufficient Condition ===
Let $\paren {A \cap B} \cup C = A \cap \paren {B \cup C}$.
 $\exists x \in C: x \notin A$.
From Set is Subset of Union:
:$C \subseteq \paren {A \cap B} \cup C$
and so by definition of subset:
:$x \in \paren {A \cap B} \cup C$
From Intersection is Subset:
:$A \cap \paren {B \cup C} \subseteq A$
and so (indirectly) by definition of subset:
:$x \notin A \cap \paren {B \cup C}$
Hence by definition of subset:
:$\paren {A \cap B} \cup C \nsubseteq A \cap \paren {B \cup C}$
that is:
:$\paren {A \cap B} \cup C \ne A \cap \paren {B \cup C}$
This contradicts our initial assertion.
Hence by Proof by Contradiction:
:$\neg \exists x \in C: x \notin A$
From Assertion of Universality:
:$\forall x \in C: x \in A$
and so $C \subseteq A$ by definition of subset.
== Sources ==
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: Exercise $11 \ \text{(d)}$
Category:Set Union
Category:Set Intersection
Category:Subsets
Union with Relative Complement

== Theorem ==
The union of a set $T$ and its relative complement in $S$ is the set $S$:
:$\relcomp S T \cup T = S$
== Proof ==
=== Step 1 ===
By the definition of relative complement, we have that:
:$\relcomp S T \subseteq S$
and:
:$T \subseteq S$
Hence by Union is Smallest Superset:
:$\relcomp S T \cup T\subseteq S$
=== Step 2 ===
Let $x \in S$.
By Law of Excluded Middle, one of the following two applies:
:$(1): \quad x \in T$
:$(2): \quad x \notin T$
If $(2)$, then by definition of relative complement:
:$x \in S \setminus T = \relcomp S T$
So:
:$x \in T \lor x \in \relcomp S T$
By definition of set union:
:$x \in \relcomp S T \cup T$
Thus:
:$x \in S \implies x \in \relcomp S T \cup T$
By definition of subset it follows that:
:$S \subseteq \relcomp S T \cup T$
=== Step 3 ===
From:
:$\relcomp S T \cup T\subseteq S$
and:
:$S \subseteq \relcomp S T \cup T$
it follows from definition of set equality that:
:$S = \relcomp S T \cup T$
== Also see ==
* Intersection with Relative Complement is Empty
== Sources ==
* : $\S 5$: Complements and Powers
* : Chapter $1$: Exercise $1 \ \text {(a)}$
* : Chapter $\text I$: Algebraic Structures: $\S 3$: Unions and Intersections of Sets: Theorem $3.2$
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: $\text{(h)}$
* : $\S 1$: Naive Set Theory: $\S 1.2$: Operations on Sets: Exercise $1.2.2 \ \text{(iii)}$
Category:Relative Complement
Category:Set Union
Union with Superclass is Superclass

== Theorem ==
Let $A$ and $B$ be classes.
Then:
:$A \subseteq B \iff A \cup B = B$
where:
:$A \subseteq B$ denotes that $A$ is a subclass of $B$
:$A \cup B$ denotes the union of $A$ and $B$.
== Proof ==
Let $A \cup B = B$.
Then by definition of class equality:
:$A \cup B \subseteq B$
Thus:
 
Now let $A \subseteq B$.
We have:
 
From Subset of Union, we have $S \cup T \supseteq T$.
We also have:
 
Thus:
By definition of class equality:
:$A \cup B = B$
== Also see ==
* Intersection with Subclass is Subclass
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 5$ The union axiom: Exercise $5.6. \ \text {(f)}$
Category:Subclasses
Category:Class Union
Union with Superset is Superset

== Theorem ==
:$S \subseteq T \iff S \cup T = T$
where:
:$S \subseteq T$ denotes that $S$ is a subset of $T$
:$S \cup T$ denotes the union of $S$ and $T$.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Intersection with Subset is Subset
== Sources ==
* : Chapter $0$: Subsets and Complements; Union and Intersection: Theorem $1$
* : $\S 4$: Unions and Intersections
* : Exercise $1.1: \ 9$
* : $\S 2.1$: Sets: Exercise $\text{B} \ 1 \ \text{(a), (c)}$
* : Chapter $1$: Exercise $1 \ \text {(c)}$
* : Chapter $\text I$: Algebraic Structures: $\S 3$: Unions and Intersections of Sets: Exercise $3.3 \ \text{(d)}$
* : Chapter $\text{I}$: Sets and Functions: Exercise $\text{B iv}$
* : Chapter $1$: The Notation and Terminology of Set Theory: $\S 6 \ \text{(d)}$
* : Chapter $1$: Theory of Sets: $\S 3$: Set Operations: Union, Intersection and Complement: Exercise $1 \ \text{(a)}$
* : $\S 7.3 \ \text{(i)}$: Unions and Intersections
* : $\S 1$: Naive Set Theory: $\S 1.2$: Operations on Sets: Exercise $1.2.1 \ \text{(vii)}$
* : Appendix $\text{A}.2$: Theorem $\text{A}.11$
Category:Union with Superset is Superset
Category:Subsets
Category:Set Union
Union with Superset is Superset/Proof 1

== Theorem ==
== Proof ==
Let $S \cup T = T$.
Then by definition of set equality:
:$S \cup T \subseteq T$
Thus:
Now let $S \subseteq T$.
From Subset of Union, we have $S \cup T \supseteq T$.
We also have:
Then:
By definition of set equality:
:$S \cup T = T$
So:
and so:
:$S \subseteq T \iff S \cup T = T$
from the definition of equivalence.
== Sources ==
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: $\text{(f)}$
* : Appendix $\text{A}.2$: Theorem $\text{A}.11$
Category:Union with Superset is Superset
Union with Universal Set

== Theorem ==
The union of a set with the universal set is the universal set:
:$\mathbb U \cup S = \mathbb U$
== Proof ==
== Also see ==
* Intersection with Universal Set
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Exercise $\text{B ii}$
* : $\S 2$
* : $\text {(iv)}$
Category:Universal Set
Category:Set Union