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22973
|
\section{Uniformly Convergent iff Difference Under Supremum Metric Vanishes}
Tags: Uniform Convergence, Metric Spaces, Metric Spaces
\begin{theorem}
Let $X$ and $Y$ be metric spaces
Let $\sequence {f_n}$ be a sequence of mappings defined on $X$.
Let $f: X \to Y$ be a mapping.
Let $d_S: S \times S \to Y$ denote the supremum metric on $S \subseteq X$.
Then:
:$\sequence {f_n}$ converges uniformly to $f$ on $S$
{{iff}}:
:$\map {d_S} {f_n, f} \to 0$ as $n \to \infty$.
\end{theorem}
\begin{proof}
We have the following string of equivalences:
Start by {{Defof|Uniform Convergence}}:
:$\forall \epsilon \in \R_{>0}: \exists N \in \R: \forall n \ge N, \forall x \in S: \map d {\map {f_n} x, \map f x} < \epsilon$
By {{Defof|Upper Bound of Set}}, this is equivalent to:
:$\forall \epsilon \in \R_{>0}: \exists N \in \R: \forall n \ge N: \epsilon$ is an upper bound of $\set {\map d {\map {f_n} x, \map f x} : x \in X}$.
So, by {{Defof|Supremum of Set}}, this is equivalent to:
:$\forall \epsilon \in \R_{>0}: \exists N \in \R: \forall n \ge N: \sup_{x \mathop \in X} \map d {\map {f_n} x, \map f x} \le \epsilon$
By {{Defof|Supremum Metric}}, this is equivalent to:
:$\forall \epsilon \in \R_{>0}: \exists N \in \R: \forall n \ge N: \map {d_S} {f_n, f} \le \epsilon$
By {{Defof|Limit of Sequence in Metric Space}}, this is equivalent to:
:$\map {d_S} {f_n, f} \to 0$ as $n \to \infty$
{{qed}}
Category:Metric Spaces
Category:Uniform Convergence
\end{proof}
|
22974
|
\section{Union Distributes over Intersection/Family of Sets}
Tags: Intersection, Set Intersection, Families, Set Union, Indexed Families, Union, Families of Sets, Union Distributes over Intersection
\begin{theorem}
Let $I$ be an indexing set.
Let $\family {A_\alpha}_{\alpha \mathop \in I}$ be an indexed family of subsets of a set $S$.
Let $B \subseteq S$.
Then:
:$\ds \map {\bigcap_{\alpha \mathop \in I} } {A_\alpha \cup B} = \paren {\bigcap_{\alpha \mathop \in I} A_\alpha} \cup B$
where $\ds \bigcap_{\alpha \mathop \in I} A_\alpha$ denotes the intersection of $\family {A_\alpha}_{\alpha \mathop \in I}$.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \map {\bigcap_{\alpha \mathop \in I} } {A_\alpha \cup B}
| c =
}}
{{eqn | ll= \leadsto
| q = \forall \alpha \in I
| l = x
| o = \in
| r = A_\alpha \cup B
| c = Intersection is Subset
}}
{{eqn | ll= \leadsto
| q = \forall \alpha \in I
| l = x
| o = \in
| r = A_\alpha
| c = {{Defof|Set Union}}
}}
{{eqn | lo= \lor
| l = x
| o = \in
| r = B
| c =
}}
{{eqn | ll= \leadsto
| l = x
| o = \in
| r = \paren {\bigcap_{\alpha \mathop \in I} A_\alpha}
| c = {{Defof|Intersection of Family}}
}}
{{eqn | lo= \lor
| l = x
| o = \in
| r = B
| c =
}}
{{eqn | ll= \leadsto
| l = x
| o = \in
| r = \paren {\bigcap_{\alpha \mathop \in I} A_\alpha} \cup B
| c = {{Defof|Set Union}}
}}
{{end-eqn}}
By definition of subset:
:$\ds \map {\bigcap_{\alpha \mathop \in I} } {A_\alpha \cup B} \subseteq \paren {\bigcap_{\alpha \mathop \in I} A_\alpha} \cup B$
{{qed|lemma}}
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \paren {\bigcap_{\alpha \mathop \in I} A_\alpha} \cup B
| c =
}}
{{eqn | ll= \leadsto
| l = x
| o = \in
| r = \paren {\bigcap_{\alpha \mathop \in I} A_\alpha}
| c = {{Defof|Set Union}}
}}
{{eqn | lo= \lor
| l = x
| o = \in
| r = B
| c =
}}
{{eqn | ll= \leadsto
| q = \forall \alpha \in I
| l = x
| o = \in
| r = A_\alpha
| c = Intersection is Subset
}}
{{eqn | lo= \lor
| l = x
| o = \in
| r = B
| c =
}}
{{eqn | ll= \leadsto
| q = \forall \alpha \in I
| l = x
| o = \in
| r = A_\alpha \cup B
| c = {{Defof|Set Union}}
}}
{{eqn | ll= \leadsto
| l = x
| o = \in
| r = \map {\bigcap_{\alpha \mathop \in I} } {A_\alpha \cup B}
| c = {{Defof|Intersection of Family}}
}}
{{end-eqn}}
By definition of subset:
:$\ds \paren {\bigcap_{\alpha \mathop \in I} A_\alpha} \cup B \subseteq \map {\bigcap_{\alpha \mathop \in I} } {A_\alpha \cup B}$
{{qed|lemma}}
By definition of set equality:
:$\ds \map {\bigcap_{\alpha \mathop \in I} } {A_\alpha \cup B} = \paren {\bigcap_{\alpha \mathop \in I} A_\alpha} \cup B$
{{qed}}
\end{proof}
|
22975
|
\section{Union Distributes over Union}
Tags: Distributive Operations, Union Distributes over Union, Set Union, Union, Order Theory, Class Theory
\begin{theorem}
Set union is distributive over itself:
:$\forall A, B, C: \paren {A \cup B} \cup \paren {A \cup C} = A \cup B \cup C = \paren {A \cup C} \cup \paren {B \cup C}$
where $A, B, C$ are sets.
\end{theorem}
\begin{proof}
We have:
* Union is Associative
* Union is Commutative
* Union is Idempotent
The result follows from Associative Commutative Idempotent Operation is Distributive over Itself.
{{qed}}
Category:Set Union
Category:Distributive Operations
Category:Union Distributes over Union
\end{proof}
|
22976
|
\section{Union Distributes over Union/Families of Sets}
Tags: Union Distributes over Union, Families, Set Union, Indexed Families, Union, Families of Sets
\begin{theorem}
Let $I$ be an indexing set.
Let $\family {A_\alpha}_{\alpha \mathop \in I}$ and $\family {B_\alpha}_{\alpha \mathop \in I}$ be indexed families of subsets of a set $S$.
Then:
:$\ds \map {\bigcup_{\alpha \mathop \in I} } {A_\alpha \cup B_\alpha} = \paren {\bigcup_{\alpha \mathop \in I} A_\alpha} \cup \paren {\bigcup_{\alpha \mathop \in I} B_\alpha}$
where $\ds \bigcup_{\alpha \mathop \in I} A_\alpha$ denotes the union of $\family {A_\alpha}_{\alpha \mathop \in I}$.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \map {\bigcup_{\alpha \mathop \in I} } {A_\alpha \cup B_\alpha}
| c =
}}
{{eqn | ll= \leadsto
| q = \exists \beta \in I
| l = x
| o = \in
| r = A_\beta \cup B_\beta
| c = {{Defof|Union of Family}}
}}
{{eqn | ll= \leadsto
| l = x
| o = \in
| r = A_\beta
| c = {{Defof|Set Union}}
}}
{{eqn | lo= \lor
| l = x
| o = \in
| r = B_\beta
| c =
}}
{{eqn | ll= \leadsto
| l = x
| o = \in
| r = \bigcup_{\alpha \mathop \in I} A_\alpha
| c = Set is Subset of Union
}}
{{eqn | lo= \lor
| l = x
| o = \in
| r = \bigcup_{\alpha \mathop \in I} B_\alpha
| c = Set is Subset of Union
}}
{{eqn | ll= \leadsto
| l = x
| o = \in
| r = \paren {\bigcup_{\alpha \mathop \in I} A_\alpha} \cup \paren {\bigcup_{\alpha \mathop \in I} B_\alpha}
| c = {{Defof|Set Union}}
}}
{{end-eqn}}
Thus by definition of subset:
:$\ds \map {\bigcup_{\alpha \mathop \in I} } {A_\alpha \cup B_\alpha} \subseteq \paren {\bigcup_{\alpha \mathop \in I} A_\alpha} \cup \paren {\bigcup_{\alpha \mathop \in I} B_\alpha}$
{{qed|lemma}}
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \paren {\bigcup_{\alpha \mathop \in I} A_\alpha} \cup \paren {\bigcup_{\alpha \mathop \in I} B_\alpha}
| c =
}}
{{eqn | ll= \leadsto
| l = x
| o = \in
| r = \bigcup_{\alpha \mathop \in I} A_\alpha
| c = {{Defof|Set Union}}
}}
{{eqn | lo= \lor
| l = x
| o = \in
| r = \bigcup_{\alpha \mathop \in I} B_\alpha
| c =
}}
{{eqn | ll= \leadsto
| q = \exists \beta \in I
| l = x
| o = \in
| r = A_\beta
| c = {{Defof|Union of Family}}
}}
{{eqn | lo= \lor
| q = \exists \beta \in I
| l = x
| o = \in
| r = B_\beta
| c =
}}
{{eqn | ll= \leadsto
| q = \exists \beta \in I
| l = x
| o = \in
| r = A_\beta \cup B_\beta
| c = {{Defof|Union of Family}}
}}
{{eqn | ll= \leadsto
| l = x
| o = \in
| r = \map {\bigcup_{\alpha \mathop \in I} } {A_\alpha \cup B_\alpha}
| c =
}}
{{end-eqn}}
Thus by definition of subset:
:$\ds \paren {\bigcup_{\alpha \mathop \in I} A_\alpha} \cup \paren {\bigcup_{\alpha \mathop \in I} B_\alpha} \subseteq \map {\bigcup_{\alpha \mathop \in I} } {A_\alpha \cup B_\alpha}$
{{qed|lemma}}
By definition of set equality:
:$\ds \map {\bigcup_{\alpha \mathop \in I} } {A_\alpha \cup B_\alpha} = \paren {\bigcup_{\alpha \mathop \in I} A_\alpha} \cup \paren {\bigcup_{\alpha \mathop \in I} B_\alpha}$
{{qed}}
\end{proof}
|
22977
|
\section{Union Distributes over Union/General Result}
Tags: Set Union, Union Distributes over Union, Union
\begin{theorem}
Let $\family {\mathbb S_i}_{i \mathop \in I}$ be an $I$-indexed family of sets of sets.
Then:
:$\ds \bigcup_{i \mathop \in I} \bigcup \mathbb S_i = \bigcup \bigcup_{i \mathop \in I} \mathbb S_i$
\end{theorem}
\begin{proof}
By the definition of set union, we have:
:$\ds \forall i \in I: \forall S \in \mathbb S_i: S \in \bigcup_{j \mathop \in I} \mathbb S_j$
Hence, by Set is Subset of Union: General Result:
:$\ds \forall i \in I: \forall S \in \mathbb S_i: S \subseteq \bigcup \bigcup_{j \mathop \in I} \mathbb S_j$
By Union is Smallest Superset: General Result, it follows that:
:$\ds \forall i \in I: \bigcup \mathbb S_i \subseteq \bigcup \bigcup_{j \mathop \in I} \mathbb S_j$
Therefore, by Union is Smallest Superset: Family of Sets, we conclude that:
:$\ds \bigcup_{i \mathop \in I} \bigcup \mathbb S_i \subseteq \bigcup \bigcup_{j \mathop \in I} \mathbb S_j$
By Set is Subset of Union General Result and then Set is Subset of Union: Family of Sets, we have:
:$\ds \forall i \in I: \forall S \in \mathbb S_i: S \subseteq \bigcup \mathbb S_i \subseteq \bigcup_{j \mathop \in I} \bigcup \mathbb S_j$
Because $\subseteq$ is transitive, we can apply the definition set union to conclude that:
:$\ds \forall S \in \bigcup_{i \mathop \in I} \mathbb S_i: S \subseteq \bigcup_{j \mathop \in I} \bigcup \mathbb S_j$
Hence, by Union is Smallest Superset: General Result:
:$\ds \bigcup \bigcup_{i \mathop \in I} \mathbb S_i \subseteq \bigcup_{j \mathop \in I} \bigcup \mathbb S_j$
By definition of set equality, the result follows.
{{qed}}
Category:Union Distributes over Union
\end{proof}
|
22978
|
\section{Union Distributes over Union/Sets of Sets}
Tags: Set Union, Union Distributes over Union, Union
\begin{theorem}
Let $A$ and $B$ denote sets of sets.
Then:
:$\ds \bigcup \paren {A \cup B} = \paren {\bigcup A} \cup \paren {\bigcup B}$
where $\ds \bigcup A$ denotes the union of $A$.
\end{theorem}
\begin{proof}
Let $\ds s \in \bigcup \paren {A \cup B}$.
Then by definition of union of set of sets:
:$\exists X \in A \cup B: s \in X$
By definition of set union, either:
:$X \in A$
or:
:$X \in B$
If $X \in A$, then:
:$s \in \set {x: \exists X \in A: x \in X}$
If $X \in B$, then:
:$s \in \set {x: \exists X \in B: x \in X}$
Thus by definition of union of set of sets, either:
:$\ds s \in \bigcup A$
or:
:$\ds s \in \bigcup B$
So by definition of set union:
:$\ds s \in \paren {\bigcup A} \cup \paren {\bigcup B}$
So by definition of subset:
:$\ds \bigcup \paren {A \cup B} \subseteq \paren {\bigcup A} \cup \paren {\bigcup B}$
Now let $\ds s \in \paren {\bigcup A} \cup \paren {\bigcup B}$.
By definition of set union, either:
:$\ds s \in \bigcup A$
or:
:$\ds s \in \bigcup B$
That is, by definition of union of set of sets, either:
:$s \in \set {x: \exists X \in A: x \in X}$
or:
:$s \in \set {x: \exists X \in B: x \in X}$
{{WLOG}}, let $s \in X$ such that $X \in A$.
Then by Set is Subset of Union:
:$s \in X$ such that $X \in A \cup B$
That is:
:$\ds s \in \bigcup \paren {A \cup B}$
Similarly if $x \in X$ such that $X \in B$.
So by definition of subset:
:$\ds \paren {\bigcup A} \cup \paren {\bigcup B} \subseteq \bigcup \paren {A \cup B}$
Hence by definition of equality of sets:
:$\ds \bigcup \paren {A \cup B} = \paren {\bigcup A} \cup \paren {\bigcup B}$
{{qed}}
Category:Union Distributes over Union
\end{proof}
|
22979
|
\section{Union Operation on Supersets of Subset is Closed}
Tags: Set Union
\begin{theorem}
Let $S$ be a set.
Let $T \subseteq S$ be a given subset of $S$.
Let $\powerset S$ denote the power set of $S$
Let $\mathscr S$ be the subset of $\powerset S$ defined as:
:$\mathscr S = \set {Y \in \powerset S: T \subseteq Y}$
Then the algebraic structure $\struct {\mathscr S, \cup}$ is closed.
\end{theorem}
\begin{proof}
Let $A, B \in \mathscr S$.
We have that:
{{begin-eqn}}
{{eqn | l = T
| o = \subseteq
| r = A
| c = Definition of $\mathscr S$
}}
{{eqn | l = T
| o = \subseteq
| r = B
| c = Definition of $\mathscr S$
}}
{{eqn | n = 1
| ll= \leadsto
| l = T
| o = \subseteq
| r = A \cup B
| c = Set is Subset of Union
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = A
| o = \subseteq
| r = S
| c = {{Defof|Power Set}}
}}
{{eqn | l = B
| o = \subseteq
| r = S
| c = {{Defof|Power Set}}
}}
{{eqn | ll= \leadsto
| l = A \cup B
| o = \subseteq
| r = S
| c = Union is Smallest Superset
}}
{{eqn | n = 2
| ll= \leadsto
| l = A \cup B
| o = \in
| r = \powerset S
| c = {{Defof|Power Set}}
}}
{{end-eqn}}
Thus we have:
{{begin-eqn}}
{{eqn | l = T
| o = \subseteq
| r = A \cup B
| c = from $(1)$
}}
{{eqn | l = A \cup B
| o = \in
| r = \powerset S
| c = from $(2)$
}}
{{eqn | ll= \leadsto
| l = A \cup B
| o = \in
| r = \mathscr S
| c = Definition of $\mathscr S$
}}
{{end-eqn}}
Hence the result by definition of closed algebraic structure.
{{qed}}
\end{proof}
|
22980
|
\section{Union as Symmetric Difference with Intersection}
Tags: Symmetric Difference, Intersection, Set Intersection, Set Union, Union
\begin{theorem}
Let $A$ and $B$ be sets.
Then:
:$A \cup B = \paren {A \symdif B} \symdif \paren {A \cap B}$
where:
:$A \cup B$ denotes set union
:$A \cap B$ denotes set intersection
:$A \symdif B$ denotes set symmetric difference
\end{theorem}
\begin{proof}
From the definition of symmetric difference:
:$\paren {A \symdif B} \symdif \paren {A \cap B} = \paren {\paren {A \symdif B} \cup \paren {A \cap B} } \setminus \paren {\paren {A \symdif B} \cap \paren {A \cap B} }$
Also from the definition of symmetric difference:
{{explain}}
:$\paren {A \symdif B} \cap \paren {A \cap B} = \paren {\paren {A \cup B} \setminus \paren {A \cap B} } \cap \paren {A \cup B}$
From Set Difference Intersection with Second Set is Empty Set:
:$\paren {S \setminus T} \cap T = \O$
Hence:
:$\paren {\paren {A \cup B} \setminus \paren {A \cup B} } \cap \paren {A \cup B} = \O$
This leaves:
{{begin-eqn}}
{{eqn | o =
| r = \paren {\paren {A \symdif B} \cup \paren {A \cap B} } \setminus \paren {\paren {A \symdif B} \cap \paren {A \cap B} }
| c =
}}
{{eqn | r = \paren {\paren {A \symdif B} \cup \paren {A \cap B} } \setminus \O
| c =
}}
{{eqn | r = \paren {A \symdif B} \cup \paren {A \cap B}
| c = Set Difference with Empty Set is Self
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | o =
| r = \paren {A \symdif B} \cup \paren {A \cap B}
| c =
}}
{{eqn | r = \paren {A \setminus B} \cup \paren {B \setminus A} \cup \paren {A \cap B}
| c = {{Defof|Symmetric Difference|index = 1}}
}}
{{eqn | r = \paren {\paren {A \setminus B} \cup \paren {A \cap B} } \cup \paren {\paren {B \setminus A} \cup \paren {A \cap B} }
| c = Union is Idempotent, Union is Commutative and Union is Associative
}}
{{eqn | r = A \cup B
| c = Set Difference Union Intersection
}}
{{end-eqn}}
Hence the result.
{{qed}}
Category:Set Union
Category:Set Intersection
Category:Symmetric Difference
\end{proof}
|
22981
|
\section{Union equals Intersection iff Sets are Equal}
Tags: Set Union, Set Intersection, Union, Intersection
\begin{theorem}
Let $S$ and $T$ be sets.
Then:
:$\paren {S \cup T = S} \land \paren {S \cap T = S} \iff S = T$
where:
:$S \cup T$ denotes set union
:$S \cap T$ denotes set intersection.
\end{theorem}
\begin{proof}
From Intersection with Subset is Subset:
:$S \subseteq T \iff S \cap T = S$
From Union with Superset is Superset:
:$S \subseteq T \iff S \cup T = T$
That is:
:$T \subseteq S \iff S \cup T = S$
Thus:
:$\paren {S \cup T = S} \land \paren {S \cap T = S} \iff S \subseteq T \subseteq S$
By definition of set equality:
:$S = T \iff S \subseteq T \subseteq S$
Hence the result.
{{qed}}
\end{proof}
|
22982
|
\section{Union is Associative}
Tags: Set Theory, Set Union, Union is Associative, Union, Direct Proofs, Associativity, Examples of Associative Operations
\begin{theorem}
Set union is associative:
:$A \cup \paren {B \cup C} = \paren {A \cup B} \cup C$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | o =
| r = x \in A \cup \paren {B \cup C}
| c = {{Defof|Set Union}}
}}
{{eqn | o = \leadstoandfrom
| r = x \in A \lor \paren {x \in B \lor x \in C}
| c = {{Defof|Set Union}}
}}
{{eqn | o = \leadstoandfrom
| r = \paren {x \in A \lor x \in B} \lor x \in C
| c = Rule of Association: Disjunction
}}
{{eqn | o = \leadstoandfrom
| r = x \in \paren {A \cup B} \cup C
| c = {{Defof|Set Union}}
}}
{{end-eqn}}
Therefore:
:$x \in A \cup \paren {B \cup C}$ {{iff}} $x \in \paren {A \cup B} \cup C$
Thus it has been shown that:
: $A \cup \paren {B \cup C} = \paren {A \cup B} \cup C$
{{qed}}
\end{proof}
|
22983
|
\section{Union is Associative/Family of Sets}
Tags: Subsets, Set Union, Union is Associative, Union, Subset
\begin{theorem}
Let $\family {S_i}_{i \mathop \in I}$ and $\family {I_\lambda}_{\lambda \mathop \in \Lambda}$ be indexed families of sets.
Let $\ds I = \bigcup_{\lambda \mathop \in \Lambda} I_\lambda$ denote the union of $\family {I_\lambda}_{\lambda \mathop \in \Lambda}$.
Then:
:$\ds \bigcup_{i \mathop \in I} S_i = \bigcup_{\lambda \mathop \in \Lambda} \paren {\bigcup_{i \mathop \in I_\lambda} S_i}$
\end{theorem}
\begin{proof}
For every $\lambda \in \Lambda$, let $\ds T_\lambda = \bigcup_{i \mathop \in I_\lambda} S_i$.
Then:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \bigcup_{i \mathop \in I} S_i
| c =
}}
{{eqn | ll= \leadstoandfrom
| q = \exists i \in I
| l = x
| o = \in
| r = S_i
| c = {{Defof|Union of Family}}
}}
{{eqn | ll= \leadstoandfrom
| q = \exists \lambda \in \Lambda: \exists i \in I_\lambda
| l = x
| o = \in
| r = S_i
| c =
}}
{{eqn | ll= \leadstoandfrom
| q = \exists \lambda \in \Lambda
| l = x
| o = \in
| r = \bigcup_{i \mathop \in I_\lambda} S_i = T_\lambda
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = x
| o = \in
| r = \bigcup_{\lambda \mathop \in \Lambda} T_\lambda
| c =
}}
{{end-eqn}}
Thus:
:$\ds \bigcup_{i \mathop \in I} S_i = \bigcup_{\lambda \mathop \in \Lambda} T_\lambda = \bigcup_{\lambda \mathop \in \Lambda} \paren {\bigcup_{i \mathop \in I_\lambda} S_i}$
{{qed}}
\end{proof}
|
22984
|
\section{Union is Commutative/Family of Sets}
Tags: Set Union, Commutativity, Union is Commutative
\begin{theorem}
Let $\family {S_i}_{i \mathop \in I}$ be an indexed family of sets.
Let $\ds I = \bigcup_{i \mathop \in I} S_i$ denote the union of $\family {S_i}_{i \mathop \in I}$.
Let $J \subseteq I$ be a subset of $I$.
Then:
:$\ds \bigcup_{i \mathop \in I} S_i = \bigcup_{j \mathop \in J} S_j \cup \bigcup_{k \mathop \in \relcomp I J} S_k = \bigcup_{k \mathop \in \relcomp I J} S_k \cup \bigcup_{j \mathop \in J} S_j$
where $\relcomp I J$ denotes the complement of $J$ relative to $I$.
\end{theorem}
\begin{proof}
We have that both $\ds \bigcup_{j \mathop \in J} S_j$ and $\ds \bigcup_{k \mathop \in \relcomp I J} S_k$ are sets.
Hence by Union is Commutative we have:
:$\bigcup_{j \mathop \in J} S_j \cup \bigcup_{k \mathop \in \relcomp I J} S_k = \bigcup_{k \mathop \in \relcomp I J} S_k \cup \bigcup_{j \mathop \in J} S_j$
It remains to be demonstrated that $\ds \bigcup_{i \mathop \in I} S_i = \bigcup_{j \mathop \in J} S_j \cup \bigcup_{k \mathop \in \relcomp I J} S_k$.
So:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \bigcup_{i \mathop \in I} S_i
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = \exists i \in I: x
| o = \in
| r = S_i
| c = {{Defof|Union of Family}}
}}
{{eqn | ll= \leadstoandfrom
| l = \exists j \in J: x
| o = \in
| r = S_j
| c = {{Defof|Relative Complement}}
}}
{{eqn | lo= \lor
| l = \exists k \in \relcomp I J: x
| o = \in
| r = S_k
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = x
| o = \in
| r = \bigcup_{j \mathop \in J} S_j
| c = {{Defof|Union of Family}}
}}
{{eqn | lo= \lor
| l = x
| o = \in
| r = \bigcup_{k \mathop \in \relcomp I J} S_k
| c = {{Defof|Union of Family}}
}}
{{eqn | ll= \leadstoandfrom
| l = x
| o = \in
| r = \bigcup_{j \mathop \in J} S_j \cup \bigcup_{k \mathop \in \relcomp I J} S_k
| c = {{Defof|Set Union}}
}}
{{end-eqn}}
That is:
:$\ds x \in \bigcup_{i \mathop \in I} S_i \iff x \in \bigcup_{j \mathop \in J} S_j \cup \bigcup_{k \mathop \in \relcomp I J} S_k$
The result follows by definition of set equality.
{{qed}}
\end{proof}
|
22985
|
\section{Union is Dominated by Disjoint Union}
Tags: Set Theory
\begin{theorem}
Let $I$ be an indexing set.
For all $i \in I$, let $S_i$ be a set.
Then:
:$\ds \bigcup_{i \mathop \in I} S_i \preccurlyeq \bigsqcup_{i \mathop \in I} S_i$
where $\preccurlyeq$ denotes domination, $\bigcup$ denotes union, and $\bigsqcup$ denotes disjoint union.
\end{theorem}
\begin{proof}
For all $\ds x \in \bigcup_{i \mathop \in I} S_i$, there exists a $\map i x \in I$ such that $x \in S_{\map i x}$.
Thus the mapping $\ds \iota : \bigcup_{i \mathop \in I} S_i \to \bigsqcup_{i \mathop \in I} S_i$ defined by:
:$\map \iota x = \tuple {x, \map i x}$
is an injection.
{{handwaving}}
{{qed}}
Category:Set Theory
\end{proof}
|
22986
|
\section{Union is Empty iff Sets are Empty}
Tags: Empty Set, Set Union, Union is Empty iff Sets are Empty, Union
\begin{theorem}
If the union of two sets is the empty set, then both are themselves empty:
:$S \cup T = \O \iff S = \O \land T = \O$
\end{theorem}
\begin{proof}
{{begin-equation}}
{{equation | l=<math>S \cup T = \varnothing</math>
| o=<math>\iff</math>
| r=<math>\neg \exists x: x \in \left ({S \cup T}\right)</math>
| c=Definition of Empty Set
}}
{{equation | o=<math>\iff</math>
| r=<math>\forall x: \neg \left ({x \in \left ({S \cup T}\right)}\right)</math>
| c=De Morgan's Laws (Predicate Logic)
}}
{{equation | o=<math>\iff</math>
| r=<math>\forall x: \neg \left ({x \in S \or x \in T}\right)</math>
| c=Definition of Set Union
}}
{{equation | o=<math>\iff</math>
| r=<math>\forall x: x \notin S \and x \notin T</math>
| c=De Morgan's Laws
}}
{{equation | o=<math>\iff</math>
| r=<math>S = \varnothing \and T = \varnothing</math>
| c=Definition of Empty Set
}}
{{end-equation}}
{{qed}}
Category:Union
Category:Empty Set
45818
27146
2011-02-15T07:45:33Z
Prime.mover
59
45818
wikitext
text/x-wiki
\end{proof}
|
22987
|
\section{Union is Increasing}
Tags: Set Union, Union
\begin{theorem}
Let $U$ be a set.
Let $\FF$ and $\GG$ be sets of subsets of $U$.
Then $\FF \subseteq \GG \implies \bigcup \FF \subseteq \bigcup \GG$.
That is, $\bigcup$ is an increasing mapping from $\struct {\powerset {\powerset U}, \subseteq}$ to $\struct {\powerset U, \subseteq}$, where $\powerset U$ is the power set of $U$.
\end{theorem}
\begin{proof}
Let $\FF \subseteq \GG$.
Let $x \in \bigcup \FF$.
Then by the definition of union:
:$\exists S \in \FF: x \in S$
By the definition of subset:
:$S \in \GG$
Thus by the definition of union:
:$x \in \bigcup \GG$
Since this holds for all $x \in \bigcup \FF$:
:$\bigcup \FF \subseteq \bigcup \GG$
{{qed}}
Category:Set Union
\end{proof}
|
22988
|
\section{Union is Increasing Sequence of Sets}
Tags: Set Union
\begin{theorem}
Let $\sequence {D_n}_{n \mathop \in \N}$ be a sequence of sets.
Then:
:the sequence $\ds \sequence {\bigcup_{k \mathop = 1}^n D_k}_{n \mathop \in \N}$ is increasing.
\end{theorem}
\begin{proof}
We have:
:$\ds \bigcup_{k \mathop = 1}^{n + 1} D_k = D_n \cup \bigcup_{k \mathop = 1}^n D_k$
From Set is Subset of Union, we have:
:$\ds \bigcup_{k \mathop = 1}^n D_k \subseteq D_n \cup \bigcup_{k \mathop = 1}^n D_k$
so:
:$\ds \ds \bigcup_{k \mathop = 1}^n D_k \subseteq \bigcup_{k \mathop = 1}^{n + 1} D_k$
So:
:$\ds \sequence {\bigcup_{k \mathop = 1}^n D_k}_{n \mathop \in \N}$ is increasing.
{{qed}}
Category:Set Union
\end{proof}
|
22989
|
\section{Union is Smallest Superset/General Result}
Tags: Set Union, Subsets, Subset, Union
\begin{theorem}
Let $S$ and $T$ be sets.
Let $\powerset S$ denote the power set of $S$.
Let $\mathbb S$ be a subset of $\powerset S$.
Then:
:$\ds \paren {\forall X \in \mathbb S: X \subseteq T} \iff \bigcup \mathbb S \subseteq T$
\end{theorem}
\begin{proof}
Let $\mathbb S \subseteq \powerset S$.
By Union of Subsets is Subset: Subset of Power Set:
:$\ds \paren {\forall X \in \mathbb S: X \subseteq T} \implies \bigcup \mathbb S \subseteq T$
{{qed|lemma}}
Now suppose that $\ds \bigcup \mathbb S \subseteq T$.
Consider any $X \in \mathbb S$ and take any $x \in X$.
From Set is Subset of Union: General Result we have that:
:$\ds X \subseteq \bigcup \mathbb S$
Thus:
:$\ds x \in \bigcup \mathbb S$
But:
:$\ds \bigcup \mathbb S \subseteq T$
So it follows that:
:$X \subseteq T$
So:
:$\ds \bigcup \mathbb S \subseteq T \implies \paren {\forall X \in \mathbb S: X \subseteq T}$
{{qed|lemma}}
Hence:
:$\ds \paren {\forall X \in \mathbb S: X \subseteq T} \iff \bigcup \mathbb S \subseteq T$
{{Qed}}
\end{proof}
|
22990
|
\section{Union is Smallest Superset/Set of Sets}
Tags: Set Union, Subsets, Subset, Union
\begin{theorem}
Let $T$ be a set.
Let $\mathbb S$ be a set of sets.
Then:
:$\ds \paren {\forall X \in \mathbb S: X \subseteq T} \iff \bigcup \mathbb S \subseteq T$
\end{theorem}
\begin{proof}
By Union of Subsets is Subset: Set of Sets:
:$\ds \paren {\forall X \in \mathbb S: X \subseteq T} \implies \bigcup \mathbb S \subseteq T$
{{qed|lemma}}
For the converse implication, suppose that $\ds \bigcup \mathbb S \subseteq T$.
Consider any $X \in \mathbb S$ and take any $x \in X$.
From Set is Subset of Union: Set of Sets we have that $X \subseteq \bigcup \mathbb S$.
Thus $\ds x \in \bigcup \mathbb S$.
But $\ds \bigcup \mathbb S \subseteq T$.
So it follows that $X \subseteq T$.
So:
:$\ds \bigcup \mathbb S \subseteq T \implies \paren {\forall X \in \mathbb S: X \subseteq T}$
{{qed|lemma}}
Hence:
:$\ds \paren {\forall X \in \mathbb S: X \subseteq T} \iff \bigcup \mathbb S \subseteq T$
{{Qed}}
\end{proof}
|
22991
|
\section{Union of Bijections with Disjoint Domains and Codomains is Bijection}
Tags: Set Union, Mapping Theory, Bijections, Union of Bijections with Disjoint Domains and Codomains is Bijection
\begin{theorem}
Let $A$, $B$, $C$, and $D$ be sets or classes.
Let $A \cap B = C \cap D = \O$.
Let $f: A \to C$ and $g: B \to D$ be bijections.
Then $f \cup g: A \cup B \to C \cup D$ is also a bijection.
\end{theorem}
\begin{proof}
By the definition of bijection, $f$ and $g$ are many-to-one and one-to-many relations.
By Union of Many-to-One Relations with Disjoint Domains is Many-to-One and Union of One-to-Many Relations with Disjoint Images is One-to-Many:
:$f \cup g$ is many-to-one and one-to-many.
Thus to show $f \cup g$ is a bijection requires us only to demonstrate that it is both left-total and right-total.
We will show that $f \cup g$ is left-total.
Let $x \in A \cup B$.
Then $x \in A$ or $x \in B$.
If $x \in A$ then since $f$ is left-total there is a $y \in C$ such that $\tuple {x, y} \in f$.
By the definition of union, $\tuple {x, y} \in f \cup g$.
If $x \in B$ then since $g$ is left-total there is a $y \in D$ such that $\tuple {x, y} \in g$.
Then by the definition of union, $\tuple {x, y} \in f \cup g$.
As this holds for all $x$, $f \cup g$ is left-total.
The proof that $f \cup g$ is right-total is similar.
Thus it has been demonstrated that:
:$f \cup g$ is many-to-one
:$f \cup g$ is one-to-many
:$f \cup g$ is left-total
:$f \cup g$ is right-total
and therefore, by definition, a bijection.
{{qed}}
Category:Bijections
Category:Set Union
Category:Union of Bijections with Disjoint Domains and Codomains is Bijection
\end{proof}
|
22992
|
\section{Union of Bijections with Disjoint Domains and Codomains is Bijection/Corollary}
Tags: Bijections, Mapping Theory, Union of Bijections with Disjoint Domains and Codomains is Bijection
\begin{theorem}
Let $A$, $B$, $C$, and $D$ be sets or classes.
Let $A \cap B = C \cap D = \varnothing$.
Let $f: A \to C$ and $g: D \to B$ be bijections.
Then $f \cup g^{-1}: A \cup B \to C \cup D$ is also a bijection.
\end{theorem}
\begin{proof}
By definition of bijection, $g^{-1}: B \to D$ is a bijection.
Hence the result by Union of Bijections with Disjoint Domains and Codomains is Bijection.
{{qed}}
Category:Union of Bijections with Disjoint Domains and Codomains is Bijection
\end{proof}
|
22993
|
\section{Union of Blocks is Set of Points}
Tags: Design Theory, Union of Blocks is Set of Points
\begin{theorem}
Let $\struct {X, \BB}$ be a pairwise balanced design.
That is, let $\struct {X, \BB}$ be a design, with $\size X \ge 2$, and the number of occurrences of each pair of distinct points in $\BB$ be $\lambda$ for some $\lambda > 0$ constant.
Then the set union of all the subset elements in $\BB$ is precisely $X$.
\end{theorem}
\begin{proof}
Let $X = \set {x_1, x_2, \ldots, x_v}$.
Let $\BB = \multiset {y_1, y_2,\ldots, y_b}$, where the notation denotes a multiset.
Let $Y = \ds \bigcup_{i \mathop = 1}^b y_i$.
We shall show that $Y \subseteq X$ and $X \subseteq Y$.
\end{proof}
|
22994
|
\section{Union of Boundaries}
Tags: Boundaries, Set Boundaries
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space.
Let $A, B$ be subsets of $S$.
Then:
:$\partial A \cup \partial B = \map \partial {A \cup B} \cup \map \partial {A \cap B} \cup \paren {\partial A \cap \partial B}$
where $\partial A$ denotes the boundary of $A$.
\end{theorem}
\begin{proof}
First we will prove that
:$\partial A \subseteq \map \partial {A \cup B} \cup \map \partial {A \cap B} \cup \paren {\partial A \cap \partial B}$
Let $x \in \partial A$.
{{AimForCont}} that
:$x \notin \map \partial {A \cup B} \cup \map \partial {A \cap B} \cup \paren {\partial A \cap \partial B}$
Then by definition of union:
:$x \notin \map \partial {A \cup B} \land x \notin \map \partial {A \cap B} \land x \notin \paren {\partial A \cap \partial B}$
By Characterization of Boundary by Open Sets:
:$\exists Q \in \tau: x \in Q \land \paren {\paren {A \cup B} \cap Q = \O \lor \relcomp S {A \cup B} \cap Q = \O}$
By Intersection Distributes over Union. Complement of Union:
:$\paren {A \cap Q} \cup \paren {B \cap Q} = \O \lor \relcomp S A \cap \relcomp S B \cap Q = \O$
Because $x \notin \paren {\partial A \cap \partial B}$ therefore by definition of intersection:
:$x \notin \partial B$
By Characterization of Boundary by Open Sets:
:$\exists U \in \tau: x \in U \land \paren {B \cap U = \O \lor \relcomp S B \cap U = \O}$
As $x \in \partial A$ by Characterization of Boundary by Open Sets:
:$A \cap Q \ne \O$
Then by Union is Empty iff Sets are Empty:
:$\paren {A \cap Q} \cup \paren {B \cap Q} \ne \O$
Hence:
:$\relcomp S A \cap \relcomp S B \cap Q = \O$
We will show that:
:$B \cap U = \O \implies \relcomp S A \cap Q \cap U = \O$
Assume:
:$B \cap U = \O$
Then:
:$U \subseteq \relcomp S B$
By Intersection with Empty Set:
:$\relcomp S A \cap Q \cap \relcomp S B \cap U = \O \cap U = \O$
Thus by Intersection with Subset is Subset:
:$\relcomp S A \cap Q \cap U = \O$
By definition of intersection:
:$x \in Q \cap U$
By definition of topological space:
:$Q \cap U$ is open.
Then by Characterization of Boundary by Open Sets:
:$\relcomp S A \cap Q \cap U \ne \O$
Hence:
:$B \cap U \ne \O$
Then:
:$\relcomp S B \cap U = \O$
Therefore:
:$U \subseteq B$
Because $x \notin \map \partial {A \cap B}$ by Characterization of Boundary by Open Sets:
:$\exists V \in \tau: x \in V \land \paren {A \cap B \cap V = \O \lor \relcomp S {A \cap B} \cap V = \O}$
By Complement of Intersection:
:$A \cap B \cap V = \O \lor \paren {\relcomp S A \cup \relcomp S B} \cap V = \O$
By Intersection Distributes over Union:
:$A \cap V \cap B = \O \lor \paren {\relcomp S A \cap V} \cup \paren {\relcomp S B \cap V} = \O$
Because $x \in \partial A$ therefore by Characterization of Boundary by Open Sets:
:$\relcomp S A \cap V \ne \O$
Then by Union is Empty iff Sets are Empty:
:$\paren {\relcomp S A \cap V} \cup \paren {\relcomp S B \cap V} \ne \O$
Hence:
:$A \cap V \cap B = \O$
By Intersection with Empty Set:
:$A \cap V \cap B \cap U = \O \cap U = \O$
By Intersection with Subset is Subset:
:$A \cap V \cap U = \O$
By definition of intersection:
:$x \in V \cap U$
By definition of topological space:
:$V \cap U$ is open.
Then by Characterization of Boundary by Open Sets:
:$A \cap Q \cap V \ne \O$
This contradicts with $A \cap V \cap U = \O$
Thus the inclusion is proved.
{{qed|lemma}}
Analogically:
:$\partial B \subseteq \map \partial {A \cup B} \cup \map \partial {A \cap B} \cup \paren {\partial A \cap \partial B}$
By Union of Subsets is Subset:
:$\partial A \cup \partial B \subseteq \map \partial {A \cup B} \cup \map \partial {A \cap B} \cup \paren {\partial A \cap \partial B}$
By Boundary of Union is Subset of Union of Boundaries:
:$\map \partial {A \cup B} \subseteq \partial A \cup \partial B$
By Boundary of Intersection is Subset of Union of Boundaries:
:$\map \partial {A \cap B} \subseteq \partial A \cup \partial B$
By Intersection is Subset of Union:
:$\partial A \cap \partial B \subseteq \partial A \cup \partial B$
Hence by Union of Subsets is Subset:
:$\map \partial {A \cup B} \cup \map \partial {A \cap B} \cup \paren {\partial A \cap \partial B} \subseteq \partial A \cup \partial B$
Thus by definition of set equality the result follows:
:$\partial A \cup \partial B = \map \partial {A \cup B} \cup \map \partial {A \cap B} \cup \paren {\partial A \cap \partial B}$
{{qed}}
\end{proof}
|
22995
|
\section{Union of Bounded Above Real Subsets is Bounded Above}
Tags: Set Union, Boundedness
\begin{theorem}
Let $A$ and $B$ be sets of real numbers.
Let $A$ and $B$ be bounded above.
Then $A \cup B$ is also bounded above.
\end{theorem}
\begin{proof}
Let $A$ and $B$ both be bounded above.
Then by definition $A$ and $B$ both have an upper bound $U_A$ and $U_B$ respectively.
Suppose $U_A \le U_B$.
Then:
:$\forall a \in A: a \le U_B$
and also, by definition:
:$\forall b \in B: b \le U_B$
and so $U_B$ is an upper bound for $A$.
Otherwise, suppose $U_A > U_B$.
Then:
:$\forall b \in B: b \le U_A$
and also, by definition:
:$\forall a \in A: a \le U_A$
Let $x \in A \cup B$.
Then from the above, either $x \le U_A$ or $x \le U_B$.
So either $U_A$ or $U_B$ is an upper bound for $A \cup B$.
Hence, by definition, $A \cup B$ is bounded above by either $U_A$ or $U_B$.
{{qed}}
\end{proof}
|
22996
|
\section{Union of Bounded Below Real Subsets is Bounded Below}
Tags: Set Union, Boundedness
\begin{theorem}
Let $A$ and $B$ be sets of real numbers.
Let $A$ and $B$ be bounded below.
Then $A \cup B$ is also bounded below.
\end{theorem}
\begin{proof}
Let $A$ and $B$ both be bounded below.
Then by definition $A$ and $B$ both have a lower bound $L_A$ and $L_B$ respectively.
Suppose $L_A \ge L_B$.
Then:
:$\forall a \in A: a \ge L_B$
and also, by definition:
:$\forall b \in B: b \ge L_B$
and so $L_B$ is a lower bound for $A$.
Otherwise, suppose $L_A < L_B$.
Then:
:$\forall b \in B: b \ge L_A$
and also, by definition:
:$\forall a \in A: a \ge L_A$
Let $x \in A \cup B$.
Then from the above, either $x \ge L_A$ or $x \ge L_B$.
So either $L_A$ or $L_B$ is a lower bound for $A \cup B$.
Hence, by definition, $A \cup B$ is bounded below by either $L_A$ or $L_B$.
{{qed}}
\end{proof}
|
22997
|
\section{Union of Chain of Proper Ideals is Proper Ideal}
Tags: Ring Theory, Ideal Theory
\begin{theorem}
Let $R$ be a ring with unity.
Let $\struct {P, \subseteq}$ be the ordered set consisting of all ideals of $R$, ordered by inclusion.
Let $\sequence {I_\alpha}_{\alpha \mathop \in A}$ be a non-empty chain of proper ideals in $P$.
Let $\ds I = \bigcup_{\alpha \mathop \in A} I_\alpha$ be their union.
Then $I$ is a proper ideal of $R$.
\end{theorem}
\begin{proof}
By Union of Chain of Ideals is Ideal, $I$ is an ideal.
It remains to show that $I \subsetneq R$.
The ideals $I_\alpha$ are all proper, so none of them contain the unity.
Thus $I$ does not contain $1$, which means $I \subsetneq R$.
{{qed}}
\end{proof}
|
22998
|
\section{Union of Class is Subset implies Class is Transitive}
Tags: Class Union, Class is Transitive iff Union is Subset, Transitive Classes
\begin{theorem}
Let $A$ be a class.
Let $\ds \bigcup A$ denote the union of $A$.
Let:
:$\ds \bigcup A \subseteq A$
Then $A$ is transitive.
\end{theorem}
\begin{proof}
Let $\ds \bigcup A \subseteq A$.
Let $x \in \ds \bigcup A$.
Then by definition:
:$\exists y \in A: x \in y$
By definition of subclass:
:$x \in A$
Thus we have that:
:$x \in y \land y \in A \implies x \in A$
It follows by definition that $A$ is a transitive class.
{{qed}}
\end{proof}
|
22999
|
\section{Union of Class is Transitive if Every Element is Transitive}
Tags: Class Union, Transitive Classes
\begin{theorem}
Let $A$ be a class.
Let $\bigcup A$ denote the union of $A$.
Let $A$ be such that every element of $A$ is transitive.
Then $\bigcup A$ is also transitive.
\end{theorem}
\begin{proof}
Let $A$ be such that every $y \in A$ is transitive.
Let $x \in \bigcup A$.
Then $x$ is an element of some element $y$ of $A$.
By hypothesis, $y$ is transitive.
Hence, by definition of transitive, $x \subseteq y$.
Because $y \in A$, by definition of union of class, $y \subseteq \bigcup A$.
So $x \subseteq \bigcup A$.
As this is true for all $x \in A$, it follows by definition that $\bigcup A$ is transitive.
{{qed}}
\end{proof}
|
23000
|
\section{Union of Closed Intervals of Positive Reals is Set of Positive Reals}
Tags: Real Intervals
\begin{theorem}
Let $\R_{>0}$ be the set of strictly positive real numbers.
For all $x \in \R_{> 0}$, let $B_x$ be the closed real interval $\closedint 0 x$.
Then:
:$\ds \bigcup_{x \mathop \in \R_{>0} } B_x = \R_{\ge 0}$
\end{theorem}
\begin{proof}
Let $\ds B = \bigcap_{x \mathop \in \R_{>0} } B_x$.
Let $y \in B$.
Then by definition of union of family:
:$\exists x \in \R_{>0}: y \in B_x$
As $B_x \subseteq \R_{>0}$ it follows by definition of subset that:
:$y = 0$
or
:$y \in \R_{>0}$
In either case, $y \in \R_{\ge 0}$
So:
:$\ds \bigcap_{x \mathop \in \R_{>0} } B_x \subseteq \R_{\ge 0}$
{{qed|lemma}}
Let $y \in \R_{\ge 0}$.
If $y = 0$ then $y \in \R_{\ge 0}$ by definition.
Otherwise, by the Archimedean Property:
:$\exists z \in \N: z > y$
and so:
:$y \in B_z$
That is by definition of union of family:
:$\ds y \in \bigcap_{x \mathop \in \R_{\ge 0} } B_x$
So by definition of subset:
:$\ds \R_{\ge 0} \subseteq \bigcap_{x \mathop \in \R_{>0} } B_x$
{{qed|lemma}}
By definition of set equality:
:$\ds \bigcup_{x \mathop \in \R_{>0} } B_x = \R_{\ge 0}$
{{qed}}
\end{proof}
|
23001
|
\section{Union of Closure with Closure of Complement is Whole Space}
Tags: Set Closures, Relative Complements, Set Union, Union, Relative Complement
\begin{theorem}
Let $T = \left({S, \tau}\right)$ be a topological space.
Let $H \subseteq S$ be a subset of $S$.
Let $H^-$ denote the closure of $H$ in $T$.
Let $S \setminus H$ denote the complement of $H$ relative to $S$.
Then:
:$H^- \cup \left({S \setminus H}\right)^- = S$
\end{theorem}
\begin{proof}
We have that:
:$H^- \cup \left({S \setminus H}\right)^- \subseteq S$
by definition of $S$.
From Union with Relative Complement:
:$H \cup \left({S \setminus H}\right) = S$
From Set is Subset of its Topological Closure:
{{begin-eqn}}
{{eqn | l = H
| o = \subseteq
| r = H^-
}}
{{eqn | l = \left({S \setminus H}\right)
| o = \subseteq
| r = \left({S \setminus H}\right)^-
}}
{{end-eqn}}
From Set Union Preserves Subsets:
:$H \subseteq H^-, \left({S \setminus H}\right) \subseteq \left({S \setminus H}\right)^- \implies H \cup \left({S \setminus H}\right) \subseteq H^- \cup \left({S \setminus H}\right)^-$
which means:
:$S \subseteq H^- \cup \left({S \setminus H}\right)^-$
The result follows by definition of equality of sets.
{{qed}}
Category:Set Closures
Category:Set Union
Category:Relative Complement
\end{proof}
|
23002
|
\section{Union of Closures of Singleton Rationals is Rational Space}
Tags: Singletons, Rational Number Space
\begin{theorem}
Let $\struct {\Q, \tau_d}$ be the rational number space under the usual (Euclidean) topology $\tau_d$.
Let $B_\alpha$ denote the singleton containing the rational number $\alpha$.
Then the union of the closures in the set of real numbers $\R$ of all $B_\alpha$ is $\Q$:
:$\ds \bigcup_{\alpha \mathop \in \Q} \map \cl {B_\alpha} = \Q$
\end{theorem}
\begin{proof}
Let $\alpha \in \Q$.
By Real Number is Closed in Real Number Line, $B_\alpha = \set \alpha$ is closed in $\R$.
From Closed Set Equals its Closure, it follows that:
:$B_\alpha = \map \cl {B_\alpha}$
Hence the result.
{{qed}}
\end{proof}
|
23003
|
\section{Union of Conjugacy Classes is Normal}
Tags: Normal Subgroups
\begin{theorem}
Let $G$ be a group.
Let $H \le G$.
Then $H$ is normal in $G$ {{iff}} $H$ is a union of conjugacy classes of $G$.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = H
| o = \lhd
| r = G
| c = where $\lhd$ denotes that $H$ is normal in $G$
}}
{{eqn | ll= \leadstoandfrom
| q = \forall g \in G
| l = g H g^{-1}
| o = \subseteq
| r = H
| c = {{Defof|Normal Subgroup}}
}}
{{eqn | ll= \leadstoandfrom
| q = \forall x \in H: \forall g \in G
| l = g x g^{-1}
| o = \in
| r = H
| c =
}}
{{eqn | ll= \leadstoandfrom
| q = \forall x \in H
| l = \conjclass x
| o = \subseteq
| r = H
| c = where $\conjclass x$ is the conjugacy class of $x \in G$
}}
{{eqn | ll= \leadstoandfrom
| l = H
| r = \bigcup_{x \mathop \in H} \conjclass x
| c =
}}
{{end-eqn}}
Hence the result.
{{qed}}
\end{proof}
|
23004
|
\section{Union of Connected Sets with Common Point is Connected}
Tags: Connected Spaces, Union of Connected Sets with Common Point is Connected
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space.
Let $\family {B_\alpha}_{\alpha \mathop \in A}$ be a family of connected sets of $T$.
Let $\exists x \in \ds \bigcap \family {B_\alpha}_{\alpha \mathop \in A}$.
Then
:$\ds \bigcup \family {B_\alpha}_{\alpha \mathop \in A}$ is a connected set of $T$.
\end{theorem}
\begin{proof}
Let $U, V$ be disjoint open sets of the subspace of $\displaystyle \bigcup_{\alpha \in A} B_\alpha$ such that:
:$U \cup V = \displaystyle \bigcup_{\alpha \in A} B_\alpha$
{{WLOG}} assume $x \in U$.
From Set is Subset of Union,
:$\displaystyle \forall \alpha \in A : B_\alpha \subseteq \bigcup_{\alpha \in A} B_\alpha = U \cup V$
From Leigh.Samphier/Sandbox/Connected Subset of Union of Disjoint Open Sets,
:$\forall \alpha \in A : B_\alpha \subseteq U$
From Union is Smallest Superset,
:$\displaystyle \bigcup_{\alpha \in A} B_\alpha = U$
Hence $\displaystyle \bigcup_{\alpha \in A} B_\alpha$ admits no separation.
And so $\displaystyle \bigcup_{\alpha \in A} B_\alpha$ is a connected set of $T$ by definition.
{{qed}}
\end{proof}
|
23005
|
\section{Union of Connected Sets with Non-Empty Intersections is Connected}
Tags: Connected Spaces, Space with Connected Intersection has Connected Union, Union of Connected Sets with Non-Empty Intersections is Connected, Topology, Connectedness
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space.
Let $I$ be an indexing set.
Let $\AA = \family {A_\alpha}_{\alpha \mathop \in I}$ be an indexed family of subsets of $S$, all connected in $T$.
Let $\AA$ be such that no two of its elements are disjoint:
:$\forall B, C \in \AA: B \cap C \ne \O$
Then $\ds \bigcup \AA$ is itself connected.
\end{theorem}
\begin{proof}
Let $A := \ds \bigcup \AA$.
Let $D = \set {0, 1}$, with the discrete topology.
Let $f: A \to D$ be continuous.
To show that $A$ is connected, we need to show that $f$ is not a surjection.
Since each $C \in \AA$ is connected and the restriction $f \restriction_C$ is continuous:
:$\map f C = \set {\map \epsilon C}$
where $\map \epsilon C = 0$ or $1$.
But, for all $B, C \in \AA$:
:$B \cap C \ne \O$
Hence $\map \epsilon B = \map \epsilon C$.
Thus $f$ is constant on $A$ as required.
{{qed}}
\end{proof}
|
23006
|
\section{Union of Countable Sets of Sets}
Tags: Set Union, Union of Countable Sets of Sets, Countable Sets, Union
\begin{theorem}
Let $\AA$ and $\BB$ be countable sets of sets.
Then:
:$\set {A \cup B: A \in \AA, B \in \BB}$
is also countable.
\end{theorem}
\begin{proof}
Since $\mathcal A$ is countable, its contents can be arranged in a sequence:
:$\mathcal A = \left\{{A_1, A_2, \ldots}\right\}$
Let $B \in \mathcal B$.
Consider the sequence of sets:
:$\left \langle A_1 \cup B, A_2 \cup B, \ldots \right \rangle$
We may leave out any possible repetitions, and obtain a countable set:
:$\left({A \cup B: A \in \mathcal A}\right\}$
for every $B \in \mathcal B$.
Thus as $B$ varies over all the elements of $\mathcal B$, we obtain the countable family:
:$\left\{{A \cup B: A \in \mathcal A}\right\}_{\left({B \in \mathcal B}\right)}$
From Union of Countable Sets, their union is countable.
{{qed}}
{{AoC|Union of Countable Sets}}
\end{proof}
|
23007
|
\section{Union of Derivatives is Subset of Derivative of Union}
Tags: Set Derivatives, T1 Spaces
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space.
Let:
:$\FF \subseteq \powerset S$ be a set of subsets of $S$
where $\powerset S$ denotes the power set of $S$.
Then:
:$\ds \bigcup_{A \mathop \in \FF} A' \subseteq \paren {\bigcup_{A \mathop \in \FF} A}'$
where $A'$ denotes the derivative of $A$.
\end{theorem}
\begin{proof}
Let $\ds x \in \bigcup_{A \mathop \in \FF} A'$.
Then by definition of union there exists $A \in \FF$ such that:
:$(1): \quad x \in A'$
By Set is Subset of Union:
:$\ds A \subseteq \bigcup_{A \mathop \in \FF} A$
Then by Derivative of Subset is Subset of Derivative:
:$\ds A' \subseteq \paren {\bigcup_{A \mathop \in \FF} A}'$
Hence by $(1)$ the result:
:$\ds x \in \paren {\bigcup_{A \mathop \in \FF} A}'$
follows by definition of subset.
{{qed}}
\end{proof}
|
23008
|
\section{Union of Disjoint Singletons is Doubleton}
Tags: Doubletons, Set Union, Singletons, Union
\begin{theorem}
Let $\set a$ and $\set b$ be singletons such that $a \ne b$.
Then:
:$\set a \cup \set b = \set {a, b}$
\end{theorem}
\begin{proof}
Let $x \in \set a \cup \set b$.
Then by the Axiom of Unions:
:$x = a \lor x = b$
It follows from the Axiom of Pairing that:
:$x \in \set {a, b}$
Thus by definition of subset:
:$\set a \cup \set b \subseteq \set {a, b}$
{{qed|lemma}}
Let $x \in \set {a, b}$.
Then by the Axiom of Pairing:
:$x = a \lor x = b$
So by the Axiom of Unions:
:$x \in \set a \cup \set b$
Thus by definition of subset:
:$\set {a, b} \subseteq \set a \cup \set b$
{{qed|lemma}}
The result follows by definition of set equality.
{{qed}}
\end{proof}
|
23009
|
\section{Union of Elements of Power Set}
Tags: Set Union, Power Set, Union
\begin{theorem}
Let $S$ be a set.
Then:
:$\ds S = \bigcup_{X \mathop \in \powerset S} X$
where $\powerset S$ denotes the power set of $S$.
\end{theorem}
\begin{proof}
By Subset of Union:
:$\ds \forall X \in \powerset S: X \subseteq \bigcup_{X \mathop \in \powerset S} X$
From Set is Subset of Itself, $S \subseteq S$ and so $S \in \powerset S$.
So:
:$\ds S \subseteq \bigcup_{X \mathop \in \powerset S} X$
From Union is Smallest Superset:
:$\ds \paren {\forall X \in \mathbb S: X \subseteq T} \iff \bigcup \mathbb S \subseteq T$
where $\mathbb S \subseteq \powerset S$.
But as $\powerset S \subseteq \powerset S$ from Set is Subset of Itself:
:$\ds \paren {\forall X \in \powerset S: X \subseteq S} \iff \bigcup \powerset S \subseteq S$
The {{LHS}} is no more than the definition of the power set, making it a tautological statement, and so:
:$\ds \bigcup \powerset S \subseteq S$
The result follows by definition of set equality.
{{qed}}
\end{proof}
|
23010
|
\section{Union of Empty Set}
Tags: Set Union, Empty Set, Union
\begin{theorem}
Consider the set of sets $\mathbb S$ such that $\mathbb S$ is the empty set $\O$.
Then the union of $\mathbb S$ is $\O$:
:$\mathbb S = \O \implies \ds \bigcup \mathbb S = \O$
\end{theorem}
\begin{proof}
Let $\mathbb S = \O$.
Then from the definition:
:$\ds \bigcup \mathbb S = \set {x: \exists X \in \mathbb S: x \in X}$
from which it follows directly:
:$\ds \bigcup \mathbb S = \O$
as there are no sets in $\mathbb S$.
{{qed}}
\end{proof}
|
23011
|
\section{Union of Equivalence Classes is Whole Set}
Tags: Quotient Sets, Equivalence Relations, Fundamental Theorem on Equivalence Relations
\begin{theorem}
Let $\RR \subseteq S \times S$ be an equivalence on a set $S$.
Then the set of $\RR$-classes constitutes the whole of $S$.
\end{theorem}
\begin{proof}
We have that:
{{begin-eqn}}
{{eqn | q = \forall x \in S
| l = x
| o = \in
| r = \eqclass x \RR
| c = {{Defof|Equivalence Class}}
}}
{{eqn | n = 1
| ll= \leadsto
| l = x
| o = \in
| r = \set {y \in S: x \mathrel \RR y}
| c = {{Defof|Equivalence Class}}
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = \eqclass x \RR
| r = \set {y: x \mathrel \RR y}
| c = {{Defof|Equivalence Class}}
}}
{{eqn | n = 2
| o = \subseteq
| r = S
| c = {{Defof|Subset}}
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = S
| r = \bigcup_{x \mathop \in S} \set x
| c = {{Defof|Union of Set of Sets}}
}}
{{eqn | o = \subseteq
| r = \bigcup_{x \mathop \in S} \eqclass x \RR
| c = from $(1)$
}}
{{eqn | o = \subseteq
| r = \bigcup_{x \mathop \in S} S
| c = from $(2)$
}}
{{eqn | r = S
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
23012
|
\section{Union of Event with Complement is Certainty}
Tags: Set Union, Complementary Events
\begin{theorem}
Let the probability space of an experiment $\EE$ be $\struct {\Omega, \Sigma, \Pr}$.
Let $A \in \Sigma$ be an events of $\EE$, so that $A \subseteq \Omega$.
Then:
:$A \cup \overline A = \Omega$
where $\overline A$ is the complementary event to $A$.
That is, $A \cup \overline A$ is a certainty.
\end{theorem}
\begin{proof}
By definition:
:$A \subseteq \Omega$
and:
:$\overline A = \relcomp \Omega A$
From Union with Relative Complement:
:$A \cup \overline A = \Omega$
We then have from Kolmogorov axiom $(2)$ that:
:$\map \Pr \Omega = 1$
The result follows by definition of certainty.
{{qed}}
{{LEM|Union with Relative Complement}}
Category:Complementary Events
Category:Set Union
\end{proof}
|
23013
|
\section{Union of Filtered Sets is Filtered}
Tags: Preorder Theory
\begin{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 filtered
and
:$\forall X, Y \in A: \exists Z \in A: X \cup Y \subseteq Z$
Then:
:$\bigcup A$ is also filtered.
\end{theorem}
\begin{proof}
Let $x, y \in \bigcup A$.
By definition of union:
:$\exists X \in A: x \in X$
and
:$\exists Y \in A: y \in Y$
By assumption:
:$\exists Z \in A: X \cup Y \subseteq Z$
By definition of union:
:$x, y \in X \cup Y$
By definition of subset:
:$x, y \in Z$
By assumption:
:$Z$ is filtered.
By definition of filtered:
:$\exists z \in Z: z \preceq x \land z \preceq y$
Thus by definition of union:
:$z \in \bigcup A$
Hence
:$\bigcup A$ is filtered.
{{qed}}
\end{proof}
|
23014
|
\section{Union of Finite Sets is Finite}
Tags: Union of Finite Sets is Finite, Set Union, Union, Finite Sets, Cardinals
\begin{theorem}
Let $S$ and $T$ be finite sets.
Then $S \cup T$ is a finite set.
\end{theorem}
\begin{proof}
Note that $\left|{ S \cup T }\right| \le \left|{ S \times T }\right|$ by Cardinal of Union Less than Cardinal of Cartesian Product.
The theorem follows from the fact that $S \times T$ is finite by Product of Finite Sets is Finite.
{{qed}}
\end{proof}
|
23015
|
\section{Union of Functions Theorem}
Tags: Mapping Theory, Named Theorems
\begin{theorem}
Let $X$ be a set.
Let $\sequence {X_i: i \in \N}$ be an exhausting sequence of sets on $X$.
For each $i \in \N$, let $g_i: X_i \to Y$ be a mapping such that:
:$g_{i + 1} \restriction X_i = g_i$
where $g_{i + 1} \restriction X_i$ denotes the restriction of $g_{i + 1}$ to $g_i$.
Then:
:$\ds \bigcup \set {g_i: i \in \N}$
is a mapping from $X$ to $Y$.
\end{theorem}
\begin{proof}
By definition, $\ds g = \bigcup \set {g_i: i \in \N}$ is a relation whose domain is $X$.
{{AimForCont}} $g$ is not a mapping.
Then for some $x \in X$ and $i, h \in \N$:
:$(1): \quad x \in X_i, \map {g_i} x \ne \map {g_{i + h} } x$
Let $k \in \N$ be the smallest such that:
:$\map {g_i} x \ne g_{i + k}$
where $x$ and $i$ are the same as in $(1)$.
Then:
:$\map {g_{i + k - 1} } x = g_i$
But then:
:$g_{i + k} \restriction X_{i + k - 1} = g_{i + k - 1}$
From this contradiction it follows that our supposition that $g$ is not a mapping must be false.
That is:
:$\ds \bigcup \set {g_i: i \in \N}$
is a mapping from $X$ to $Y$.
{{qed}}
\end{proof}
|
23016
|
\section{Union of Functions Theorem/Corollary}
Tags: Inverse Mappings
\begin{theorem}
Let $X$ be a set.
Let $\sequence {X_i: i \in \N}$ be an exhausting sequence of sets on $X$.
For each $i \in \N$, let $g_i: X_i \to Y$ be a mapping such that:
:$g_{i + 1} \restriction X_i = g_i$
where $g_{i + 1} \restriction X_i$ denotes the restriction of $g_{i + 1}$ to $g_i$.
For each $i \in \N$, let $g_i : X_i \to Y$ be invertible.
Then $\ds \bigcup \set {g_i: i \in \N}$ is invertible and:
:$\ds \paren {\bigcup \set {g_i: i \in \N} }^{-1} = \bigcup \set {g_i^{-1}: i \in \N}$
\end{theorem}
\begin{proof}
{{ProofWanted}}
Category:Inverse Mappings
\end{proof}
|
23017
|
\section{Union of Horizontal Sections is Horizontal Section of Union}
Tags: Set Union, Horizontal Section of Sets
\begin{theorem}
Let $X$, $Y$ and $A$ be sets.
Let $\set {E_\alpha : \alpha \in A}$ be a set of subsets of $X \times Y$.
Let $y \in Y$.
Then:
:$\ds \paren {\bigcup_{\alpha \in A} E_\alpha}^y = \bigcup_{\alpha \in A} \paren {E_\alpha}^y$
where:
:$\ds \paren {\bigcup_{\alpha \in A} E_\alpha}^y$ is the $y$-horizontal section of $\ds \bigcup_{\alpha \in A} E_\alpha$
:$\paren {E_\alpha}^y$ is the $y$-horizontal section of $E_\alpha$.
\end{theorem}
\begin{proof}
Note that:
:$\ds x \in \bigcup_{\alpha \in A} \paren {E_\alpha}^y$
{{iff}}:
:$x \in \paren {E_\alpha}^y$ for some $\alpha \in A$.
From the definition of the $y$-horizontal 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 $y$-horizontal section, this is the case {{iff}}:
:$\ds x \in \paren {\bigcup_{\alpha \in A} E_\alpha}^y$
So:
:$\ds x \in \bigcup_{\alpha \in A} \paren {E_\alpha}^y$ {{iff}} $\ds x \in \paren {\bigcup_{\alpha \in A} E_\alpha}^y$
giving:
:$\ds \paren {\bigcup_{\alpha \in A} E_\alpha}^y = \bigcup_{\alpha \in A} \paren {E_\alpha}^y$
{{qed}}
Category:Set Union
Category:Horizontal Section of Sets
\end{proof}
|
23018
|
\section{Union of Indexed Family of Sets Equal to Union of Disjoint Sets}
Tags: Union of Indexed Family of Sets Equal to Union of Disjoint Sets, Indexed Families
\begin{theorem}
Let $\family {E_n}_{n \mathop \in \N}$ be a countable indexed family of sets where at least two $E_n$ are distinct.
Then there exists a countable indexed family of disjoint sets $\family {F_n}_{n \mathop \in \N}$ defined by:
:$\ds F_k = E_k \setminus \paren {\bigcup_{j \mathop = 0}^{k \mathop - 1} E_j}$
satisfying:
:$\ds \bigsqcup_{n \mathop \in \N} F_n = \bigcup_{n \mathop \in \N} E_n$
where $\bigsqcup$ denotes disjoint union.
\end{theorem}
\begin{proof}
Denote:
{{begin-eqn}}
{{eqn | l = E
| r = \bigcup_{k \mathop \in \N} E_k
}}
{{eqn | l = F
| r = \bigcup_{k \mathop \in \N} F_k
}}
{{end-eqn}}
where:
:$\ds F_k = E_k \setminus \paren {\bigcup_{j \mathop = 0}^{k \mathop - 1} E_j}$
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:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \bigcup_{k \mathop \in \N} F_k
}}
{{eqn | ll= \leadsto
| q = \exists k \in \N
| l = x
| o = \in
| r = F_k
}}
{{eqn | ll= \leadsto
| q = \exists k \in \N
| l = x
| o = \in
| r = E_k
}}
{{eqn | lo= \land
| l = x
| o = \notin
| r = \paren {E_0 \cup E_1 \cup E_2 \cup \cdots \cup E_{k - 1} }
}}
{{eqn | ll= \leadsto
| q = \exists k \in \N
| l = x
| o = \in
| r = E_k
| c = Rule of Simplification
}}
{{end-eqn}}
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 $x \in F_k$ for some $F_k$.
By the Well-Ordering Principle, there exists a smallest such $k$.
Then:
:$\forall j < k: x \notin F_j$
Choose any distinct $\ell, m \in \N$.
We have:
If $m > \ell$, then:
{{begin-eqn}}
{{eqn | l = x \in F_\ell
| o = \implies
| r = x \in E_\ell
}}
{{eqn | l = x \in F_m
| o = \implies
| r = x \notin E_m
}}
{{end-eqn}}
If $m < \ell$, then:
{{begin-eqn}}
{{eqn | l = x \in F_m
| o = \implies
| r = x \in E_m
}}
{{eqn | l = x \in F_\ell
| o = \implies
| r = x \notin E_\ell
}}
{{end-eqn}}
So the sets $F_\ell, F_m$ are disjoint.
Thus $F$ is the disjoint union of sets equal to $E$:
:$\ds \bigcup_{k \mathop \in \N} E_k = \bigsqcup_{k \mathop \in \N} F_k$
{{qed}}
\end{proof}
|
23019
|
\section{Union of Indexed Family of Sets Equal to Union of Disjoint Sets/General Result}
Tags: Union of Indexed Family of Sets Equal to Union of Disjoint Sets
\begin{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 $\alpha \ne \beta$.
\end{theorem}
\begin{proof}
Denote:
{{begin-eqn}}
{{eqn | l = E
| r = \bigcup_{\beta \mathop \in I} E_\beta
}}
{{eqn | l = F
| r = \bigcup_{\beta \mathop \in I} F_\beta
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \bigcup_{\beta \mathop \in I} F_\beta
}}
{{eqn | ll= \leadsto
| q = \exists \beta \in I
| l = x
| o = \in
| r = F_\beta
}}
{{eqn | ll= \leadsto
| q = \exists \beta \in I
| l = x
| o = \in
| r = E_\beta
}}
{{eqn | lo= \land
| l = x
| o = \notin
| r = \paren {\bigcup_{\gamma \mathop \prec \beta} E_\gamma}
}}
{{eqn | ll= \leadsto
| q = \exists \beta \in I
| l = x
| o = \in
| r = E_\beta
| c = Rule of Simplification
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = x \in F_\eta
| o = \implies
| r = x \in E_\eta
}}
{{eqn | l = x \in F_\zeta
| o = \implies
| r = x \notin E_\zeta
}}
{{end-eqn}}
If $\zeta < \eta$, then:
{{begin-eqn}}
{{eqn | l = x \in F_\zeta
| o = \implies
| r = x \in E_\zeta
}}
{{eqn | l = x \in F_\eta
| o = \implies
| r = x \notin E_\eta
}}
{{end-eqn}}
So the sets $F_\eta, F_\zeta$ are disjoint.
Thus $F$ is the disjoint union of sets equal to $E$:
:$\ds \bigcup_{\alpha \mathop \in I} E_\alpha = \bigsqcup_{\alpha \mathop \in I} F_\alpha$
{{qed}}
\end{proof}
|
23020
|
\section{Union of Initial Segments is Initial Segment or All of Woset}
Tags: Order Theory, Union of Initial Segments is Initial Segment or All of Woset
\begin{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 $X$.
\end{theorem}
\begin{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 $b$.
{{AimForCont}} 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 $b \preccurlyeq y$.
{{stub}}
\end{proof}
|
23021
|
\section{Union of Interiors is Subset of Interior of Union}
Tags: Set Union, Set Interiors
\begin{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 $
\end{theorem}
\begin{proof}
In the following, $H^-$ denotes the closure of the set $H$.
{{begin-eqn}}
{{eqn | l = \paren {\bigcup_{H \mathop \in \mathbb H} H}^\circ
| r = T \setminus \paren {T \setminus \bigcup_{H \mathop \in \mathbb H} H}^-
| c = Complement of Interior equals Closure of Complement
}}
{{eqn | r = T \setminus \paren {\paren {\bigcap_{H \mathop \in \mathbb H} \paren {T \setminus H} }^-}
| c = De Morgan's Laws: Difference with Union
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = T \setminus \paren {\bigcap_{H \mathop \in \mathbb H} \paren {T \setminus H}^-}
| r = T \setminus \paren {\bigcap_{H \mathop \in \mathbb H} T \setminus H^\circ}
| c = Complement of Interior equals Closure of Complement
}}
{{eqn | r = T \setminus \paren {T \setminus \paren {\bigcup_{H \mathop \in \mathbb H} H^\circ} }
| c = De Morgan's Laws: Difference with Union
}}
{{eqn | r = \bigcup_{H \mathop \in \mathbb H} H^\circ
| c = Relative Complement of Relative Complement
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
23022
|
\section{Union of Inverse of Relations is Inverse of their Union}
Tags: Set Union, Inverse Relations, Set Unions
\begin{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:
:${\RR_1}^{-1} \cup {\RR_2}^{-1} = \paren {\RR_1 \cup \RR_2}^{-1}$
\end{theorem}
\begin{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:
:$\tuple {t, s} \in \paren {\RR_1 \cup \RR_2}^{-1}$
{{qed}}
Category:Inverse Relations
Category:Set Union
\end{proof}
|
23023
|
\section{Union of Inverses of Mappings is Inverse of Union of Mappings}
Tags: Set Union, Inverse Relations, Indexed Families, Union
\begin{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:
:$\ds \paren {\bigcup \family {f_i: i \in I} }^{-1} = \bigcup \family {f_i^{-1}: i \in I}$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \tuple {y, x}
| o = \in
| r = \paren {\bigcup \family {f_i: i \in I} }^{-1}
}}
{{eqn | ll= \leadstoandfrom
| l = \tuple {x, y}
| o = \in
| r = \family {f_i: i \in I}
| c = {{Defof|Inverse Relation}}
}}
{{eqn | ll= \leadstoandfrom
| q = \exists i \in I
| l = \tuple {x, y}
| o = \in
| r = f_i
| c = {{Defof|Union of Family}}
}}
{{eqn | ll= \leadstoandfrom
| q = \exists i \in I
| l = \tuple {y, x}
| o = \in
| r = f_i^{-1}
| c = {{Defof|Inverse Relation}}
}}
{{eqn | ll= \leadstoandfrom
| l = \tuple {y, x}
| o = \in
| r = \bigcup \family {f_i^{-1}: i \in I}
| c = {{Defof|Union of Family}}
}}
{{end-eqn}}
{{qed}}
Category:Inverse Relations
Category:Set Union
Category:Indexed Families
\end{proof}
|
23024
|
\section{Union of Left-Total Relations is Left-Total}
Tags: Relation Theory
\begin{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.
\end{theorem}
\begin{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:
:$\tuple {s, t} \in \RR_i \subseteq \RR \implies \tuple {s, t} \in \RR$
{{qed}}
\end{proof}
|
23025
|
\section{Union of Local Bases is Basis}
Tags: Topology, Topological Bases
\begin{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 $\tau$.
\end{theorem}
\begin{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 $\tau$.
{{qed}}
\end{proof}
|
23026
|
\section{Union of Many-to-One Relations with Disjoint Domains is Many-to-One}
Tags: Relation Theory
\begin{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 $\paren {S_1 \cup S_2} \times \paren {T_1 \cup T_2}$.
\end{theorem}
\begin{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.
{{qed}}
Category:Relation Theory
\end{proof}
|
23027
|
\section{Union of Mappings which Agree is Mapping}
Tags: Set Union, Mapping Theory
\begin{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.
\end{theorem}
\begin{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 $x \in A \cup B$.
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.
{{qed}}
\end{proof}
|
23028
|
\section{Union of Mappings with Disjoint Domains is Mapping}
Tags: Set Union, Mapping Theory, Mappings, Union
\begin{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 $S_1 \cup S_2$.
\end{theorem}
\begin{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.
{{qed}}
\end{proof}
|
23029
|
\section{Union of Matroid Base with Element of Complement is Dependent}
Tags: Matroid Theory
\begin{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 $B$
\end{theorem}
\begin{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:
:$B \cup \set x \notin \mathscr I$
{{qed}}
Category:Matroid Theory
\end{proof}
|
23030
|
\section{Union of Meager Sets is Meager Set}
Tags: Set Union, Meager Spaces
\begin{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 $T$.
\end{theorem}
\begin{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 $\ds A = \bigcup_{n \in \N} U_n$.
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 $\ds B = \bigcup_{m \in \N} V_m$.
Then:
:$\ds A \cup B = \paren {\bigcup_{n \in \N} U_n} \cup \paren {\bigcup_{m \in \N} V_m}$
The {{RHS}} is a countable union of sets nowhere dense in $T$, so:
:$A \cup B$ is meager in $T$.
{{qed}}
Category:Meager Spaces
Category:Set Union
\end{proof}
|
23031
|
\section{Union of Nest of Orderings is Ordering}
Tags: Set Theory, Order Theory
\begin{theorem}
Let $S$ be a set.
Let $C$ be a nonempty nest of orderings on $S$.
Then $\bigcup C$ is an ordering on $S$.
\end{theorem}
\begin{proof}
Let $\preceq$ be an arbitrary element of $C$.
Let $\sim \, = \, \bigcup C$.
Checking in turn each of the criteria for an ordering:
Let $a, b \in S$.
Let $a \sim b$ and $b \sim a$.
Since $a \sim b$ and $b \sim a$, there exist $\preceq_1, \preceq_2 \, \in \, C$ such that $a \preceq_1 b$ and $b \preceq_2 a$.
Since $C$ is a chain:
:$\preceq_1 \, \subset \, \preceq_2$
or:
:$\preceq_2 \, \subset \, \preceq_1$
{{WLOG}}, suppose $\preceq_1 \, \subset \, \preceq_2$.
Then it must hold that $a \preceq_2 b$.
Since:
:$a \preceq_2 b$
:$b \preceq_2 a$
:$\preceq_2 \, \in \, T$ is an ordering
it follows that:
:$a = b$
and so $\sim$ is antisymmetric.
{{qed|lemma}}
\end{proof}
|
23032
|
\section{Union of Non-Disjoint Bounded Subsets of Metric Space is Bounded}
Tags: Bounded Metric Spaces
\begin{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:
:$\map \diam {B \cup C} \le \map \diam B + \map \diam C$
\end{theorem}
\begin{proof}
{{AimForCont}} 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 $\map d {x, y} \le \map \diam C$.
{{WLOG}}, let $x \in B$ and $y \in C$.
Let $z \in B \cap C$.
This is possible because $B \cap C \ne \O$ {{hypothesis}}.
Then:
{{begin-eqn}}
{{eqn | l = \map d {x, z}
| o = \le
| r = \map \diam B
| c = {{Defof|Diameter of Subset of Metric Space|Diameter of $B$}}
}}
{{eqn | l = \map d {z, y}
| o = \le
| r = \map \diam C
| c = {{Defof|Diameter of Subset of Metric Space|Diameter of $C$}}
}}
{{eqn | ll= \leadsto
| l = \map d {x, z} + \map d {z, y}
| o = \le
| r = \map \diam B + \map \diam C
| c =
}}
{{eqn | ll= \leadsto
| l = \map d {x, y}
| o = \le
| r = \map \diam B + \map \diam C
| c = {{Metric-space-axiom|2}}: $\map d {x, y} \le \map d {x, z} + \map d {z, y}$
}}
{{end-eqn}}
This contradicts our supposition that $\map d {x, y} > \map \diam B + \map \diam C$.
Hence:
{{begin-eqn}}
{{eqn | q = \forall x, y \in B \cup C
| l = \map d {x, y}
| o = \le
| r = \map \diam B + \map \diam C
| c =
}}
{{eqn | ll= \leadsto
| l = \sup \set {\map d {x, y}: x, y \in B \cup C}
| o = \le
| r = \map \diam B + \map \diam C
| c = {{Defof|Supremum}}
}}
{{eqn | ll= \leadsto
| l = \map \diam {B \cup C}
| o = \le
| r = \map \diam B + \map \diam C
| c = {{Defof|Diameter of Subset of Metric Space}}
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
23033
|
\section{Union of Non-Disjoint Convex Sets is Convex Set}
Tags: Set Union, Convex Sets (Order Theory), Convex Sets, Union
\begin{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.
\end{theorem}
\begin{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.
{{qed}}
\end{proof}
|
23034
|
\section{Union of One-to-Many Relations with Disjoint Images is One-to-Many}
Tags: Relation Theory
\begin{theorem}
Let $S_1, S_2, T_1, T_2$ be sets or classes.
Let $\RR_1$ be a one-to-many relation on $S_1 \times T_1$.
Let $\RR_2$ be a one-to-many relation on $S_2 \times T_2$.
Suppose that the images of $\RR_1$ and $\RR_2$ are disjoint.
Then $\RR_1 \cup \RR_2$ is a one-to-many relation on $\paren {S_1 \cup S_2} \times \paren {T_1 \cup T_2}$.
\end{theorem}
\begin{proof}
Let $\QQ = \RR_1 \cup \RR_2$.
Then $\QQ \subseteq \paren {S_1 \times T_1} \cup \paren {S_2 \times T_2} \subseteq \paren {S_1 \cup S_2} \times \paren {T_1 \cup T_2}$.
Thus $\QQ$ is a relation on $\paren {S_1 \cup S_2} \times \paren {T_1 \cup T_2}$.
Let $T'_1$ and $T'_2$ be the images of $\RR_1$ and $\RR_2$, respectively.
Let $\tuple {x_1, y}, \tuple {x_2, y} \in \QQ$.
Then $y \in T'_1$ or $y \in T'_2$.
If $y \in T'_1$ then $y \notin T'_2$, so neither $\tuple {x_1, y}$ nor $\tuple {x_2, y}$ is in $\RR_2$, so these pairs are both in $\RR_1$.
As $\RR_1$ is one-to-many, $x_1 = x_2$.
A similar argument leads to the same result for $y \in T'_2$.
As this holds for all such $x_1, x_2, y$: $\QQ$ is a one-to-many relation.
{{qed}}
Category:Relation Theory
\end{proof}
|
23035
|
\section{Union of Open Intervals of Positive Reals is Set of Strictly Positive Reals}
Tags: Real Intervals
\begin{theorem}
Let $\R_{>0}$ be the set of strictly positive real numbers.
For all $x \in \R_{>0}$, let $A_x$ be the open real interval $\openint 0 x$.
Then:
:$\ds \bigcup_{x \mathop \in \R_{>0} } A_x = \R_{>0}$
\end{theorem}
\begin{proof}
Let $\ds A = \bigcap_{x \mathop \in \R_{>0} } A_x$.
Let $y \in A$.
Then by definition of union of family:
:$\exists x \in \R_{>0}: y \in A_x$
As $A_x \subseteq \R_{>0}$ it follows by definition of subset that:
:$y \in \R_{>0}$
So:
:$\ds \bigcap_{x \mathop \in \R_{>0} } A_x \subseteq \R_{>0}$
{{qed|lemma}}
Let $y \in \R_{>0}$.
By the Archimedean Property:
:$\exists z \in \N: z > y$
and so:
:$y \in A_z$
That is by definition of union of family:
:$y \in \ds \bigcap_{x \mathop \in \R_{>0} } A_x$
So by definition of subset:
:$\ds \R_{>0} \subseteq \bigcap_{x \mathop \in \R_{>0} } A_x$
{{qed|lemma}}
By definition of set equality:
:$\ds \bigcup_{x \mathop \in \R_{>0} } A_x = \R_{>0}$
{{qed}}
\end{proof}
|
23036
|
\section{Union of Open Irreducible Non-Disjoint Subspaces is Irreducible}
Tags: Irreducible Spaces
\begin{theorem}
Let $T = \left({S, \tau}\right)$ be an irreducible toplogical space.
Let $U$ and $V$ be open irreducible subspaces.
Let their intersection $U \cap V$ be non-empty.
Then their union $U \cup V$ is irreducible.
\end{theorem}
\begin{proof}
{{ProofWanted}}
Category:Irreducible Spaces
\end{proof}
|
23037
|
\section{Union of Open Sets of Metric Space is Open}
Tags: Set Union, Open Sets, Metric Spaces, Union
\begin{theorem}
Let $M = \struct {A, d}$ be a metric space.
The union of a set of open sets of $M$ is open in $M$.
\end{theorem}
\begin{proof}
Let $I$ be any indexing set.
Let $U_i$ be open in $M$ for all $i \in I$.
Let $\ds x \in \bigcup_{i \mathop \in I} U_i$.
Then by definition of set union, $x \in U_k$ for some $k \in I$.
Since $U_k$ is open in $M$:
:$\ds \exists \epsilon > 0: \map {B_\epsilon} x \subseteq U_k$
where $\map {B_\epsilon} x$ is the open $\epsilon$-ball of $x$ in $M$.
By Set is Subset of Union:
:$\ds U_k \subseteq \bigcup_{i \mathop \in I} U_i$
Hence by Subset Relation is Transitive:
:$\ds \map {B_\epsilon} x \subseteq \bigcup_{i \mathop \in I} U_i$
and the result follows by definition of open set in metric space.
{{qed}}
\end{proof}
|
23038
|
\section{Union of Open Sets of Neighborhood Space is Open}
Tags: Neighborhood Spaces
\begin{theorem}
Let $S$ be a neighborhood space.
Let $I$ be an indexing set.
Let $\family {U_\alpha}_{\alpha \mathop \in I}$ be a family of open sets of $\struct {S, \NN}$ indexed by $I$.
Then their union $\ds \bigcup_{\alpha \mathop \in I} U_i$ is an open set of $\struct {S, \NN}$.
\end{theorem}
\begin{proof}
Let $\ds x \in \bigcup_{\alpha \mathop \in I} U_\alpha$.
Then $x \in U_\beta$ for some $\beta \in I$.
By definition of open set, $U_\beta$ is a neighborhood of $x$.
But from Set is Subset of Union:
:$\ds U_\beta \subseteq x \in \bigcup_{\alpha \mathop \in I} U_\alpha$
By neighborhood space axiom $N 3$ it follows that $\ds \bigcup_{\alpha \mathop \in I} U_\alpha$ is a neighborhood of $x$.
As $x$ is arbitrary, it follows that the above is true for all $\ds x \in \bigcup_{\alpha \mathop \in I} U_\alpha$.
It follows by definition that $\ds \bigcup_{\alpha \mathop \in I} U_\alpha$ is an open set of $\struct {S, \NN}$.
{{qed}}
\end{proof}
|
23039
|
\section{Union of Open Sets of Normed Vector Space is Open}
Tags: Set Union, Open Sets
\begin{theorem}
Let $M = \struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
The union of a set of open sets of $M$ is open in $M$.
\end{theorem}
\begin{proof}
Let $I$ be any indexing set.
Let $U_i$ be open in $M$ for all $i \in I$.
Let $\ds x \in \bigcup_{i \mathop \in I} U_i$.
Then $x \in U_k$ for some $k \in I$.
Since $U_k$ is open in $M$:
:$\ds \exists \epsilon > 0: \map {B_\epsilon} x \subseteq U_k \subseteq \bigcup_{i \mathop \in I} U_i$
where $\map {B_\epsilon} x$ is the open $\epsilon$-ball of $x$ in $M$.
The result follows.
{{qed}}
\end{proof}
|
23040
|
\section{Union of Ordinals is Least Upper Bound}
Tags: Ordinals
\begin{theorem}
Let $A \subset \On$.
That is, let $A$ be a class of ordinals (every member of $A$ is an ordinal).
Then $\bigcup A$, the union of $A$, is the least upper bound of $A$:
:$\ds \forall x \in A: x \le A$
:$\ds \forall y \in A: y \le x \implies \bigcup A \le x$
\end{theorem}
\begin{proof}
First, we must show that $\ds \bigcup A$ is an upper bound.
Take any member $a \in A$.
Then by Subset of Union:
:$\ds a \subseteq \bigcup A$
By Ordering on Ordinal is Subset Relation:
:$a \le A$
By generalizing for all $a \in A$:
:$\ds \forall x \in A: x \le \bigcup A$
Similarly, suppose now that $x$ is an upper bound of $A$.
We shall denote $<$ for ordering on the ordinal numbers.
By Ordering on Ordinal is Subset Relation and Transitive Set is Proper Subset of Ordinal iff Element of Ordinal, $<$ is the same as both $\in$ and $\subsetneq$.
Then:
{{begin-eqn}}
{{eqn | l = z \in \bigcup A
| o = \leadsto
| r = \exists y: \paren {z \in y \land y \in A}
| c = {{Defof|Union of Set of Sets}}
}}
{{eqn | o = \leadsto
| r = \exists y: \paren {z \in y \land y < x}
| c = {{hypothesis}} (as $y \in A$, $y < x$)
}}
{{eqn | o = \leadsto
| r = z \in x
| c = transitivity of $\in$: see Alternative Definition of Ordinal
}}
{{end-eqn}}
Thus, by definition of subset:
:$\ds \bigcup A \subseteq x$
Therefore:
:$\ds \forall y \in A: y \le x \implies \bigcup A \le x$
{{qed}}
\end{proof}
|
23041
|
\section{Union of Ordinals is Ordinal}
Tags: Ordinals
\begin{theorem}
Let $y$ be a set.
Let $\map F z$ be a mapping such that:
:$\ds \forall z \in y: \map F z \in \On$
Then:
:$\ds \bigcup_{z \mathop \in y} \map F z \in \On$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \bigcup_{z \mathop \in y} \map F z
| c =
}}
{{eqn | ll= \leadsto
| q = \exists z \in y
| l = x
| o = \in
| r = \map F z
| c = {{Defof|Set Union}}
}}
{{eqn | ll= \leadsto
| l = x
| o = \in
| r = \On
| c = {{hypothesis}}
}}
{{end-eqn}}
So:
:$\ds \bigcup_{z \mathop \in y} \map F z \subseteq \On$
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \bigcup_{z \mathop \in y} \map F z
| c =
}}
{{eqn | ll= \leadsto
| q = \exists z \in y
| l = x
| o = \in
| r = \map F z
| c = {{Defof|Set Union}}
}}
{{eqn | ll= \leadsto
| q = \exists z \in y
| l = x
| o = \subseteq
| r = \map F z
| c = Ordinals are Transitive
}}
{{eqn | ll= \leadsto
| l = x
| o = \subseteq
| r = \bigcup_{z \mathop \in y} \map F z
| c = Union Preserved Under Subset Relation
}}
{{end-eqn}}
So $\ds \bigcup_{z \mathop \in y} \map F z$ is a transitive set.
Therefore $\ds \bigcup_{z \mathop \in y} \map F z$ is an ordinal.
:$\ds \bigcup_{z \mathop \in y} \map F z = \bigcup \Img y$
Let $U$ denote the universal class.
{{begin-eqn}}
{{eqn | l = y
| o = \in
| r = V
| c =
}}
{{eqn | ll= \leadsto
| l = \bigcup \Img y
| o = \in
| r = V
| c = Axiom of Replacement Equivalents
}}
{{eqn | ll= \leadsto
| l = \bigcup \bigcup \Img y
| o = \in
| r = V
| c = Axiom of Unions Equivalents
}}
{{eqn | ll= \leadsto
| l = \bigcup_{z \mathop \in y} \map F z
| o = \in
| r = V
| c = Equality given above
}}
{{end-eqn}}
Therefore $\ds \bigcup_{z \mathop \in y} \map F z$ is a set, so it is a member of the ordinal class.
{{qed}}
Category:Ordinals
\end{proof}
|
23042
|
\section{Union of Overlapping Convex Sets in Toset is Convex}
Tags: Total Orderings
\begin{theorem}
Let $\struct {S, \preceq}$ be a totally ordered set.
Let $U$ and $V$ be convex sets in $S$.
Let $U \cap V \ne \O$.
Then $U \cup V$ is also convex.
\end{theorem}
\begin{proof}
Let $a,b,c \in S$
Let $a,c \in U \cup V$.
Let $a \prec b \prec c$.
If $a, c \in U$ then $b \in U$ because $U$ is convex.
Thus $b \in U \cup V$ by the definition of union.
Similarly, if $a,c \in V$ then $b \in U \cup V$.
Otherwise, {{WLOG}}, suppose that $a \in U$ and $c \in V$.
Since $U \cap V$ is nonempty by the premise, it has an element $p$.
Since $\preceq$ is a total ordering:
:$b \preceq p$ or $p \preceq b$.
If $b \preceq p$, then since $a \prec b$, $a,p \in U$, and $U$ is convex, we can conclude that
:$b \in U$
so $b \in U \cup V$.
A similar argument shows that it $p \preceq b$ then $b \in V$, so $b \in U \cup V$.
Thus in all cases we can conclude that $b \in U \cup V$, so $U \cup V$ is convex.
{{qed}}
Category:Total Orderings
\end{proof}
|
23043
|
\section{Union of Overlapping Convex Sets in Toset is Convex/Infinite Union}
Tags: Total Orderings
\begin{theorem}
Let $\struct {S, \preceq}$ be a totally ordered set.
Let $\AA$ be a set of convex subsets of $S$.
For any $P, Q \in \AA$, let there be elements $C_0, \dotsc, C_n \in \AA$ such that:
:$C_0 = P$
:$C_n = Q$
:For $k = 0, \dotsc, n - 1: C_k \cap C_{k + 1} \ne \O$
Then $\bigcup \AA$ is convex in $S$.
\end{theorem}
\begin{proof}
Let $a, c \in \bigcup \AA$.
Let $b \in S$.
Let $a \prec b \prec c$.
Since $a, c \in \bigcup \AA$, there are $P, Q \in \AA$ such that $a \in P$ and $c \in Q$.
By the premise, there are elements $C_0, \dots, C_n \in \AA$ such that:
:$C_0 = P$
:$C_n = Q$
:For $k = 0, \dotsc, n - 1: C_k \cap C_{k + 1} \ne \O$
{{explain|details of induction. Consider putting the finite chain case into a separate lemma.}}
Applying Union of Overlapping Intervals is Interval inductively:
:$\ds \bigcup_{k \mathop = 0}^n C_k$ is convex.
Since $\ds a, c \in \bigcup_{k \mathop = 0}^n C_k$, by the definition of convexity:
:$\ds b \in \bigcup_{k \mathop = 0}^n C_k$
Thus:
:$\ds b \in \bigcup \AA$
Since this holds for all such triples $a, b, c$, it follows that $\AA$ is convex.
{{qed}}
Category:Total Orderings
\end{proof}
|
23044
|
\section{Union of Path-Connected Sets with Common Point is Path-Connected}
Tags: Connected Spaces
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space.
Let $\family {B_\alpha}_{\alpha \mathop \in A}$ be a family of path-connected sets of $T$.
Let $\exists x \in \ds \bigcap \family {B_\alpha}_{\alpha \mathop \in A}$.
Then
:$\ds \bigcup \family {B_\alpha}_{\alpha \mathop \in A}$ is a path-connected set of $T$.
\end{theorem}
\begin{proof}
Let $B = \ds \bigcup \family {B_\alpha}_{\alpha \mathop \in A}$.
Let $a, b \in B$.
Then
:$\exists \alpha, \beta \in A: a \in B_\alpha \land b \in B_\beta$.
As $B_\alpha$ is a path-connected set in $T$ then $a$ and $x$ are path-connected points.
Similarly, $x$ and $b$ are path-connected points.
From Joining Paths makes Another Path, $a$ and $b$ are path-connected points.
Since $a$ and $b$ were arbitrary points then $B$ is a path-connected set of $T$.
{{qed}}
\end{proof}
|
23045
|
\section{Union of Power Sets not always Equal to Powerset of Union}
Tags: Set Union, Power Set, Subsets
\begin{theorem}
The union of the power sets of two sets $S$ and $T$ is not necessarily equal to the power set of their union.
\end{theorem}
\begin{proof}
Proof by Counterexample:
Let $S = \set {1, 2, 3}, T = \set {2, 3, 4}, X = \set {1, 2, 3, 4}$.
{{begin-eqn}}
{{eqn | l = S \cup T
| r = \set {1, 2, 3, 4}
| c =
}}
{{eqn | ll= \leadsto
| l = X
| o = \subseteq
| r = S \cup T
| c =
}}
{{eqn | ll= \leadsto
| l = X
| o = \in
| r = \powerset {S \cup T}
| c =
}}
{{end-eqn}}
But note that $X \nsubseteq S \land X \nsubseteq T$.
Thus:
{{begin-eqn}}
{{eqn | l = X
| o = \nsubseteq
| r = S \land X \nsubseteq T
| c =
}}
{{eqn | ll= \leadsto
| l = X
| o = \notin
| r = \powerset S \land X \notin \powerset T
| c =
}}
{{eqn | ll= \leadsto
| l = \neg (X
| o = \in
| r = \powerset S \lor X \in \powerset T)
| c =
}}
{{eqn | ll= \leadsto
| l = X
| o = \notin
| r = \powerset S \cup \powerset T
| c =
}}
{{eqn | ll= \leadsto
| l = \powerset {S \cup T}
| o = \nsubseteq
| r = \powerset S \cup \powerset T
| c =
}}
{{end-eqn}}
So:
:$\powerset {S \cup T} \ne \powerset S \cup \powerset T$
{{qed}}
\end{proof}
|
23046
|
\section{Union of Primitive Recursive Sets}
Tags: Set Union, Primitive Recursive Functions, Union
\begin{theorem}
Let $A, B \subseteq \N$ be subsets of the set of natural numbers $\N$.
Let $A$ and $B$ both be primitive recursive.
Then $A \cup B$, the union of $A$ and $B$, is primitive recursive.
\end{theorem}
\begin{proof}
$A$ and $B$ are primitive recursive, therefore so are their characteristic functions $\chi_A$ and $\chi_B$.
Let $n \in \N$ be a natural number.
Then $n \in A \cup B \iff \chi_A \left({n}\right) + \chi_B \left({n}\right) > 0$.
So:
{{begin-eqn}}
{{eqn | l = \chi_{A \cup B} \left({n}\right)
| r = \operatorname{sgn} \left({\chi_A \left({n}\right) + \chi_B \left({n}\right)}\right)
| c = Signum Function is Primitive Recursive
}}
{{eqn | r = \operatorname{sgn} \left({\operatorname{add} \left({\chi_A \left({n}\right), \chi_B \left({n}\right)}\right)}\right)
| c = Addition is Primitive Recursive
}}
{{end-eqn}}
Thus $A \cup B$ is defined by substitution from the primitive recursive functions $\operatorname{sgn}$, $\operatorname{add}$, $\chi_A$ and $\chi_B$.
Hence the result.
{{qed}}
Category:Primitive Recursive Functions
Category:Set Union
\end{proof}
|
23047
|
\section{Union of Real Intervals is not necessarily Real Interval}
Tags: Real Intervals
\begin{theorem}
Let $I_1$ and $I_2$ be real intervals.
Then $I_1 \cup I_2$ is not necessarily a real interval.
\end{theorem}
\begin{proof}
Proof by Counterexample:
Consider the real intervals:
:$I_1 = \left({0 \,.\,.\, 2}\right)$
:$I_2 = \left({4 \,.\,.\, 6}\right)$
Then we have that:
:$1 < 3 < 5$
where:
:$1 \in I_1 \cup I_2$
:$5 \in I_1 \cup I_2$
but:
:$3 \notin I_1 \cup I_2$
Thus $I_1 \cup I_2$ is not a real interval.
{{qed}}
\end{proof}
|
23048
|
\section{Union of Reflexive Relations is Reflexive}
Tags: Reflexive Relations, Intersection, Set Union, Union, Relations
\begin{theorem}
The union of two reflexive relations is also a reflexive relation.
\end{theorem}
\begin{proof}
Let $\RR_1$ and $\RR_2$ be reflexive relations on a set $S$.
From Relation Contains Diagonal Relation iff Reflexive, we have that:
: $\Delta_S \subseteq \RR_1$
: $\Delta_S \subseteq \RR_2$
Hence from Subset Relation is Transitive:
: $\Delta_S \subseteq \RR_1 \cup \RR_2$
Hence the result, by definition of reflexive relation.
{{qed}}
Category:Reflexive Relations
Category:Set Union
\end{proof}
|
23049
|
\section{Union of Regular Open Sets is not necessarily Regular Open}
Tags: Open Sets, Regular Open Sets, Set Union
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space.
Let $U$ and $V$ be regular open sets of $T$.
Then $U \cup V$ is not also necessarily a regular open set of $T$.
\end{theorem}
\begin{proof}
Proof by Counterexample:
By Open Real Interval is Regular Open, the open real intervals:
:$\openint 0 {\dfrac 1 2}, \openint {\dfrac 1 2} 1$
are both regular open sets of $\R$.
Consider $A$, the union of the adjacent open intervals:
:$A := \openint 0 {\dfrac 1 2} \cup \openint {\dfrac 1 2} 1$
From Interior of Closure of Interior of Union of Adjacent Open Intervals:
:$A^{- \circ} = \openint 0 1$
Thus $A$ is not a regular open set of $\R$.
{{qed}}
\end{proof}
|
23050
|
\section{Union of Relations Compatible with Operation is Compatible}
Tags: Compatible Relations
\begin{theorem}
Let $\struct {S, \circ}$ be a closed algebraic structure.
Let $\FF$ be a family of relations on $S$.
Let each element of $\FF$ be compatible with $\circ$.
Let $\QQ = \bigcup \FF$.
Then $Q$ is a relation compatible with $\circ$.
\end{theorem}
\begin{proof}
Let $x, y, z \in S$.
Let $x \mathrel \QQ y$.
Then for some $\RR \in \FF$:
:$x \mathrel \RR y$.
Since $\RR$ is a relation compatible with $\circ$:
:$\tuple {x \circ z} \mathrel \RR \tuple {y \circ z}$
Since $\RR \in \FF$:
:$\RR \subseteq \bigcup \FF = \QQ$
Thus
:$\tuple {x \circ z} \mathrel \QQ \tuple {y \circ z}$
We have shown that:
:$\forall x, y, z \in S: x \mathrel \QQ y \implies \tuple {x \circ z} \mathrel \QQ \tuple {y \circ z}$
A similar argument shows that:
:$\forall x, y, z \in S: x \mathrel \QQ y \implies \tuple {z \circ x} \mathrel \QQ \tuple {z \circ y}$
So $Q$ is a relation compatible with $\circ$.
{{qed}}
Category:Compatible Relations
\end{proof}
|
23051
|
\section{Union of Relations is Relation}
Tags: Relation Theory
\begin{theorem}
Let $S$ and $T$ be sets.
Let $\FF$ be a family of relations from $S$ to $T$.
Let $\ds \RR = \bigcup \FF$, the union of all the elements of $\FF$.
Then $\RR$ is a relation from $S$ to $T$.
{{expand|Binary case}}
\end{theorem}
\begin{proof}
By the definition of a relation from $S$ to $T$, each element of $\FF$ is a subset of $S \times T$.
By Union of Subsets is Subset: Set of Sets:
:$\RR \subseteq S \times T$
Therefore, by the definition of a relation from $S$ to $T$, $\RR$ is a relation from $S$ to $T$.
{{qed}}
Category:Relation Theory
\end{proof}
|
23052
|
\section{Union of Relative Complements of Nested Subsets}
Tags: Set Union, Relative Complement, Union
\begin{theorem}
Let $R \subseteq S \subseteq T$ be sets with the indicated inclusions.
Then:
:$\complement_T \left({S}\right) \cup \complement_S \left({R}\right) = \complement_T \left({R}\right)$
where $\complement$ denotes relative complement.
Phrased via Set Difference as Intersection with Relative Complement:
:$\left({T \setminus S}\right) \cup \left({S \setminus R}\right) = T \setminus R$
where $\setminus$ denotes set difference.
\end{theorem}
\begin{proof}
From Union with Set Difference:
:$T = T \setminus S \cup S$
and therefore by Set Difference is Right Distributive over Union:
:$T \setminus R = \left({\left({T \setminus S}\right) \setminus R}\right) \cup \left({S \setminus R}\right)$
Now, by Set Difference with Union and Union with Superset is Superset:
:$\left({T \setminus S}\right) \setminus R = T \setminus \left({S \cup R}\right) = T \setminus S$
Combining the above yields:
:$T \setminus R = \left({T \setminus S}\right) \cup \left({S \setminus R}\right)$
{{qed}}
\end{proof}
|
23053
|
\section{Union of Right-Total Relations is Right-Total}
Tags: Relation Theory
\begin{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 right-total relations.
Then $\RR_1 \cup \RR_2$ is right-total.
\end{theorem}
\begin{proof}
Define the predicates $L$ and $R$ by:
:$\map L X \iff \text {$X$ is left-total}$
:$\map R X \iff \text {$X$ is right-total}$
{{begin-eqn}}
{{eqn | l = \map R {\RR_1} \land \map R {\RR_2}
| o = \leadsto
| r = \map L {\RR_1^{-1} } \land \map L {\RR_2^{-1} }
| c = Inverse of Right-Total Relation is Left-Total
}}
{{eqn | o = \leadsto
| r = \map L {\RR_1^{-1} \cup \RR_2^{-1} }
| c = Union of Left-Total Relations is Left-Total
}}
{{eqn | o = \leadsto
| r = \map L {\paren {\RR_1 \cup \RR_2}^{-1} }
| c = Union of Inverse of Relations is Inverse of their Union
}}
{{eqn | o = \leadsto
| r = \map R {\RR_1 \cup \RR_2}
| c = Inverse of Right-Total Relation is Left-Total
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
23054
|
\section{Union of Set of Dense-in-itself Sets is Dense-in-itself}
Tags: Set Derivatives, Denseness
\begin{theorem}
Let $T$ be a topological space.
Let $\FF \subseteq \powerset T$ such that
:every element of $\FF$ is dense-in-itself.
Then the union $\bigcup \FF$ is also dense-in-itself.
\end{theorem}
\begin{proof}
By Dense-in-itself iff Subset of Derivative:
:$\forall A \in \FF: A \subseteq A'$
where $A'$ denotes the derivative of $A$.
Then by Set Union Preserves Subsets:
:$\ds \bigcup \FF \subseteq \bigcup_{A \mathop \in \FF} A'$
By Union of Derivatives is Subset of Derivative of Union:
:$\ds \bigcup_{A \mathop \in \FF} A' \subseteq \paren {\bigcup \FF}'$
Then by Subset Relation is Transitive:
:$\ds \bigcup \FF \subseteq \paren {\bigcup \FF}'$
The result follows by Dense-in-itself iff Subset of Derivative.
{{qed}}
\end{proof}
|
23055
|
\section{Union of Set of Sets when a Set Intersects All}
Tags: Set Union, Union
\begin{theorem}
Let $F$ be a set of sets.
Let $S$ be a set or class.
Suppose that:
:$\forall A \in F: A \cap S \ne \O$
Then:
:$\ds F = \bigcup_{x \mathop \in S} \set {A \in F: x \in A}$
\end{theorem}
\begin{proof}
Suppose that $B \in F$.
Then $B \cap S$ has an element $x_B$.
Thus $B \in \set {A \in F: x_B \in A}$.
By the definition of union:
:$\ds B \in \bigcup_{x \mathop \in S} \set {A \in F: x \in A}$
Suppose instead that $\ds B \in \bigcup_{x \mathop \in S} \set {A \in F: x \in A}$.
Then by the definition of union, there exists an $x_B \in S$ such that $B \in \set {A \in F: x_B \in A} \subseteq F$.
Thus $B \in F$.
We have shown that $\ds \forall B: \paren {B \in F \iff B \in \bigcup_{x \mathop \in S} \set {A \in F: x \in A} }$.
Therefore $\ds F = \bigcup_{x \mathop \in S} \set {A \in F: x \in A}$ by the Axiom of Extension.
{{qed}}
Category:Set Union
\end{proof}
|
23056
|
\section{Union of Singleton}
Tags: Set Union, Singletons, Union
\begin{theorem}
Consider the set of sets $\mathbb S$ such that $\mathbb S$ consists of just one set $S$.
Then the union of $\mathbb S$ is $S$:
:$\ds \mathbb S = \set S \implies \bigcup \mathbb S = S$
\end{theorem}
\begin{proof}
Let $\mathbb S = \set S$.
Then from the definition of set union:
:$\ds \bigcup \mathbb S = \set {x: \exists X \in \mathbb S: x \in X}$
from which it follows directly that:
:$\ds \bigcup \mathbb S = \set {x: x \in S}$
as $S$ is the only set in $\mathbb S$.
That is:
:$\ds \bigcup \mathbb S = S$
{{qed}}
\end{proof}
|
23057
|
\section{Union of Small Classes is Small}
Tags: Set Union, Zermelo-Fraenkel Class Theory, Union
\begin{theorem}
Let $x$ and $y$ be small classes.
Then $x \cup y$ is also small.
\end{theorem}
\begin{proof}
Let $\map {\mathscr M} A$ denote that $A$ is small.
{{begin-eqn}}
{{eqn | l = \map {\mathscr M} x \land \map {\mathscr M} y
| o = \leadsto
| r = \map {\mathscr M} {\set {x, y} }
| c = Axiom of Pairing
}}
{{eqn | o = \leadsto
| r = \map {\mathscr M} {\bigcup \set {x, y} }
| c = Axiom of Unions
}}
{{eqn | o = \leadsto
| r = \map {\mathscr M} {x \cup y}
| c = Union of Doubleton
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
23058
|
\section{Union of Subclass is Subset of Union of Class}
Tags: Class Union
\begin{theorem}
Let $A$ and $B$ be classes.
Let $\ds \bigcup A$ and $\ds \bigcup B$ denote the union of $A$ and union of $B$ respectively.
Let $A$ be a subclass of $B$:
:$A \subseteq B$
Then $\ds \bigcup A$ is a subset of $\ds \bigcup B$:
:$\ds \bigcup A \subseteq \ds \bigcup B$
\end{theorem}
\begin{proof}
By the axiom of unions, both $\ds \bigcup A$ and $\ds \bigcup B$ are sets.
Let $x \in \ds \bigcup A$.
Then:
:$\exists y \in A: x \in y$
But as $A \subseteq B$ it follows that $y \in B$.
That is:
:$\exists y \in B: x \in y$
That is:
:$x \in \ds \bigcup B$
Hence the result by definition of subset.
{{Qed}}
\end{proof}
|
23059
|
\section{Union of Subgroups}
Tags: Subgroups, Group Theory, Set Union, Union of Subgroups, Union
\begin{theorem}
Let $\struct {G, \circ}$ be a group.
Let $H, K \le G$ be subgroups of $G$.
Let neither $H \subseteq K$ nor $K \subseteq H$.
Then $H \cup K$ is ''not'' a subgroup of $G$.
\end{theorem}
\begin{proof}
As neither $H \subseteq K$ nor $K \subseteq H$, it follows from Set Difference with Superset is Empty Set that neither $H \setminus K = \O$ nor $K \setminus H = \O$.
So, let $h \in H \setminus K, k \in K \setminus H$.
Thus, $h \notin K, k \notin H$.
If $\struct {H \cup K, \circ}$ is a group, then it must be closed.
If $\struct {H \cup K, \circ}$ is closed, then $h \circ k \in H \cup K \implies h \circ k \in H \lor h \circ k \in K$.
If $h \circ k \in H$ then $h^{-1} \circ h \circ k \in H \implies k \in H$.
If $h \circ k \in K$ then $h \circ k \circ k^{-1} \in K \implies h \in K$.
So $h \circ k$ can be in neither $H$ nor $K$.
Therefore $\struct {H \cup K, \circ}$ is not closed.
Therefore $H \cup K$ is not a subgroup of $G$.
{{qed}}
\end{proof}
|
23060
|
\section{Union of Subgroups/Corollary 1}
Tags: Union of Subgroups
\begin{theorem}
Let $\struct {G, \circ}$ be a group.
Let $H, K \le G$.
Let $H \cup K$ be a subgroup of $G$.
Then either $H \subseteq K$ or $K \subseteq H$.
\end{theorem}
\begin{proof}
{{AimForCont}} neither $H \subseteq K$ nor $K \subseteq H$.
Then from Union of Subgroups it follows that $H \cup K$ is not a subgroup of $G$.
The result follows by Proof by Contradiction.
{{qed}}
\end{proof}
|
23061
|
\section{Union of Subgroups/Corollary 2}
Tags: Set Union, Union of Subgroups, Subgroups, Union
\begin{theorem}
Let $\struct {G, \circ}$ be a group.
Let $H, K \le G$.
Let $H \vee K$ be the join of $H$ and $K$.
Then $H \vee K = H \cup K$ {{iff}} $H \subseteq K$ or $K \subseteq H$.
\end{theorem}
\begin{proof}
From the definition of join, $H \vee K$ is the smallest subgroup of $G$ containing $H \cup K$.
The result follows from Union of Subgroups.
{{qed}}
\end{proof}
|
23062
|
\section{Union of Subset of Ordinals is Ordinal}
Tags: Ordinals
\begin{theorem}
{{explain|If $A$ is a proper class, then its union may be the class of all ordinals, which we do not consider an ordinal.}}
Let $A$ be a class of ordinals.
That is, $A \subseteq \On$, where $\On$ denotes the ordinal class.
Then $\bigcup A$ is an ordinal.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = x \in \bigcup A
| o = \leadsto
| r = \exists y \in A: x \in y
| c =
}}
{{eqn | o = \leadsto
| r = \exists y \subseteq \bigcup A: x \subseteq y
| c =
}}
{{eqn | o = \leadsto
| r = x \subseteq \bigcup A
| c =
}}
{{end-eqn}}
From this, we conclude that $\ds \bigcup A$ is a transitive class.
From Class is Transitive iff Union is Subset, it follows that:
:$\ds \bigcup A \subseteq A \subseteq \On$
By Subset of Well-Ordered Set is Well-Ordered, $A$ is also well-ordered by $\Epsilon$.
Thus by Alternative Definition of Ordinal, $\bigcup A$ is an ordinal.
\end{proof}
|
23063
|
\section{Union of Subset of Ordinals is Ordinal/Corollary}
Tags: Ordinals
\begin{theorem}
Let $y$ be a set.
Let $\On$ be the class of all ordinals.
Let $F: y \to \On$ be a mapping.
Then:
:$\ds \bigcup \map F y \in \On$
where $\map F y$ is the image of $y$ under $F$.
\end{theorem}
\begin{proof}
By the Axiom of Replacement, $\map F y$ is a set.
Thus by the Axiom of Unions, $\ds \bigcup \map F y$ is a set.
By Union of Subset of Ordinals is Ordinal, $\ds \bigcup \map F y$ is transitive.
By the epsilon relation $\ds \bigcup \map F y$ is well-ordered.
Thus $\ds \bigcup \map F y$ is a member of $\On$, the ordinal class.
{{qed}}
Category:Ordinals
\end{proof}
|
23064
|
\section{Union of Subsets is Subset}
Tags: Subsets, Set Union, Union, Union of Subsets is Subset, Subset
\begin{theorem}
Let $S_1$, $S_2$, and $T$ be sets.
Let $S_1$ and $S_2$ both be subsets of $T$.
Then:
:$S_1 \cup S_2 \subseteq T$
That is:
:$\paren {S_1 \subseteq T} \land \paren {S_2 \subseteq T} \implies \paren {S_1 \cup S_2} \subseteq T$
\end{theorem}
\begin{proof}
{{improve|The version for set of sets doesn't rely on the more sophisticated Set Union Preserves Subsets. Change this to match.}}
Let $\left({S_1 \subseteq T}\right) \land \left({S_2 \subseteq T}\right)$.
Then:
{{begin-eqn}}
{{eqn
| l=S_1 \cup S_2
| o=\subseteq
| r=T \cup T
| c=Set Union Preserves Subsets
}}
{{eqn | ll=\implies
| l=S_1 \cup S_2
| o=\subseteq
| r=T
| c=as $T = T \cup T$ from Union is Idempotent Operation
}}
{{end-eqn}}
So:
:$\left({S_1 \subseteq T}\right) \land \left({S_2 \subseteq T}\right) \implies \left({S_1 \cup S_2}\right) \subseteq T$
{{qed}}
\end{proof}
|
23065
|
\section{Union of Subsets is Subset/Family of Sets}
Tags: Subsets, Set Union, Union, Union of Subsets is Subset, Subset
\begin{theorem}
Let $\family {S_i}_{i \mathop \in I}$ be a family of sets indexed by $I$.
Then for all sets $X$:
:$\ds \paren {\forall i \in I: S_i \subseteq X} \implies \bigcup_{i \mathop \in I} S_i \subseteq X$
where $\ds \bigcup_{i \mathop \in I} S_i$ is the union of $\family {S_i}$.
\end{theorem}
\begin{proof}
Suppose that $\forall i \in I: S_i \subseteq X$.
Consider any $\ds x \in \bigcup_{i \mathop \in I} S_i$.
By definition of set union:
:$\exists i \in I: x \in S_i$
But as $S_i \subseteq X$ it follows that $x \in X$.
Thus it follows that:
:$\ds \bigcup_{i \mathop \in I} S_i \subseteq X$
So:
:$\ds \paren {\forall i \in I: S_i \subseteq X} \implies \bigcup_{i \mathop \in I} S_i \subseteq X$
{{qed}}
\end{proof}
|
23066
|
\section{Union of Subsets is Subset/Set of Sets}
Tags: Subsets, Set Union, Union, Union of Subsets is Subset, Subset
\begin{theorem}
Let $T$ be a set.
Let $\mathbb S$ be a set of sets.
Suppose that for each $S \in \mathbb S$, $S \subseteq T$.
Then:
:$\ds \bigcup \mathbb S \subseteq T$
\end{theorem}
\begin{proof}
Let $x \in \ds \bigcup \mathbb S$.
By the definition of union, there exists an $S \in \mathbb S$ such that $x \in S$.
By premise, $S \subseteq T$.
By the definition of subset, $x \in T$.
Since this result holds for each $x \in \ds \bigcup \mathbb S$:
:$\ds \bigcup \mathbb S \subseteq T$
{{qed}}
Category:Union of Subsets is Subset
\end{proof}
|
23067
|
\section{Union of Subsets is Subset/Subset of Power Set}
Tags: Set Union, Subsets, Subset, Union
\begin{theorem}
Let $S$ and $T$ be sets.
Let $\powerset S$ be the power set of $S$.
Let $\mathbb S$ be a subset of $\powerset S$.
Then:
:$\ds \paren {\forall X \in \mathbb S: X \subseteq T} \implies \bigcup \mathbb S \subseteq T$
\end{theorem}
\begin{proof}
Let $\mathbb S \subseteq \powerset S$.
Suppose that $\forall X \in \mathbb S: X \subseteq T$.
Consider any $\ds x \in \bigcup \mathbb S$.
By definition of set union, it follows that:
:$\exists X \in \mathbb S: x \in X$
But as $X \subseteq T$ it follows that $x \in T$.
Thus it follows that:
:$\ds \bigcup \mathbb S \subseteq T$
So:
:$\ds \paren {\forall X \in \mathbb S: X \subseteq T} \implies \bigcup \mathbb S \subseteq T$
{{qed}}
Category:Set Union
Category:Subsets
\end{proof}
|
23068
|
\section{Union of Successor Ordinal}
Tags: Ordinals
\begin{theorem}
Let $x$ be an ordinal.
Let $x^+$ denote the successor of $x$.
Then:
:$\ds \map \bigcup {x^+} = x$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \map \bigcup {x^+}
| r = \map \bigcup {x \cup \set x}
| c = {{Defof|Successor Set}}
}}
{{eqn | r = \paren {\bigcup x \cup \bigcup \set x}
| c = Union Distributes over Union/Sets of Sets
}}
{{eqn | r = \paren {\bigcup x \cup x}
| c = Union of Singleton
}}
{{eqn | r = x
| c = Class is Transitive iff Union is Subset
}}
{{end-eqn}}
{{qed}}
Category:Ordinals
\end{proof}
|
23069
|
\section{Union of Symmetric Differences}
Tags: Set Union, Symmetric Difference, Union
\begin{theorem}
Let $R, S, T$ be sets.
Then:
:$\paren {R \symdif S} \cup \paren {S \symdif T} = \paren {R \cup S \cup T} \setminus \paren {R \cap S \cap T}$
where $R \symdif S$ denotes the symmetric difference between $R$ and $S$.
\end{theorem}
\begin{proof}
From the definition of symmetric difference, we have:
:$R \symdif S = \paren {R \setminus S} \cup \paren {S \setminus R}$
Thus, expanding:
{{begin-eqn}}
{{eqn | l = \paren {R \symdif S} \cup \paren {S \symdif T}
| r = \paren {R \setminus S} \cup \paren {S \setminus R} \cup \paren {S \setminus T} \cup \paren {T \setminus S}
}}
{{eqn | r = \paren {\paren {R \setminus S} \cup \paren {T \setminus S} } \cup \paren {\paren {S \setminus R} \cup \paren {S \setminus T} }
}}
{{eqn | r = \paren {\paren {R \cup T} \setminus S} \cup \paren {\paren {S \setminus R} \cup \paren {S \setminus T} }
| c = Set Difference is Right Distributive over Union
}}
{{eqn | r = \paren {\paren {R \cup T} \setminus S} \cup \paren {S \setminus \paren {R \cap T} }
| c = De Morgan's Laws: Difference with Intersection
}}
{{eqn | r = \paren {\paren {R \cup S \cup T} \setminus S} \cup \paren {S \setminus \paren {R \cap T} }
| c = Set Difference with Union is Set Difference
}}
{{eqn | r = \paren {R \cup S \cup T} \setminus \paren {R \cap S \cap T}
| c = De Morgan's Laws for Difference with Intersection: Corollary
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
23070
|
\section{Union of Symmetric Relations is Symmetric}
Tags: Set Union, Relations, Symmetric Relations, Union
\begin{theorem}
The union of two symmetric relations is also a symmetric relation.
\end{theorem}
\begin{proof}
Let $\RR_1$ and $\RR_2$ be symmetric relations on a set $S$.
Let $\RR_3 = \RR_1 \cup \RR_2$.
Then:
{{begin-eqn}}
{{eqn | l = \tuple {x, y}
| o = \in
| r = \RR_3
| c =
}}
{{eqn | ll= \leadsto
| l = \tuple {x, y}
| o = \in
| r = \RR_1
| c = {{Defof|Set Union}}
}}
{{eqn | lo= \lor
| l = \tuple {x, y}
| o = \in
| r = \RR_2
| c =
}}
{{eqn | ll= \leadsto
| l = \tuple {y, x}
| o = \in
| r = \RR_1
| c = {{Defof|Symmetric Relation}}
}}
{{eqn | lo= \lor
| l = \tuple {y, x}
| o = \in
| r = \RR_2
| c =
}}
{{eqn | ll= \leadsto
| l = \tuple {y, x}
| o = \in
| r = \RR_3
| c = {{Defof|Set Union}}
}}
{{end-eqn}}
{{qed}}
Category:Symmetric Relations
Category:Set Union
\end{proof}
|
23071
|
\section{Union of Topologies is not necessarily Topology}
Tags: Topology
\begin{theorem}
Let $\tau_1$ and $\tau_2$ be topologies on a set $S$.
Then $\tau_1 \cup \tau_2$ is not necessarily also a topology on $S$.
\end{theorem}
\begin{proof}
Let $S := \set {0, 1, 2}$ be a set.
Let:
:$\tau_1 := \set {\O, \set 0, \set 1, \set {0, 1}, S}$
:$\tau_2 := \set {\O, \set 0, \set 2, \set {0, 2}, S}$
be topologies on $S$.
Then:
:$\tau := \tau_1 \cup \tau_2 = \set {\O, \set 0, \set 1, \set 2, \set {0, 1} \set {0, 2}, S}$
For $\tau$ to be a topology the union of any number of elements of $\tau$ should also be in $\tau$.
But:
:$\set 1 \cup \set 2 = \set {1, 2} \not \in \tau$
Therefore $\tau$ is not atopology on $S$.
Hence the result.
{{qed}}
\end{proof}
|
23072
|
\section{Union of Topologies on Singleton or Doubleton is Topology}
Tags: Doubletons, Singletons, Topology, Set Union
\begin{theorem}
Let $S$ be a set which is either a singleton or a doubleton.
Let $\family {\tau_i}_{i \mathop \in I}$ be an arbitrary non-empty indexed set of topologies on $S$.
Then $\tau := \ds \bigcup_{i \mathop \in I} {\tau_i}$ is also a topology on $S$.
\end{theorem}
\begin{proof}
Let $S$ be a singleton.
Let $S = \set a$.
From Topology on Singleton is Indiscrete Topology, the only possible topology on $S$ is the indiscrete topology.
From Union is Idempotent, the union of any number of indiscrete topologies on $S$ is the indiscrete topology.
Thus the union of any number of topologies on a singleton is a topology on that singleton.
{{qed|lemma}}
Let $S$ be a doubleton.
Let $S = \set {a, b}$.
By Topologies on Doubleton, there are $4$ possible different topologies on $S$:
{{begin-eqn}}
{{eqn | l = \tau_1
| r = \set {\O, \set {a, b} }
| c = Indiscrete Topology
}}
{{eqn | l = \tau_2
| r = \set {\O, \set a, \set {a, b} }
| c = Sierpiński Topology
}}
{{eqn | l = \tau_3
| r = \set {\O, \set b, \set {a, b} }
| c = Sierpiński Topology
}}
{{eqn | l = \tau_4
| r = \set {\O, \set a, \set b, \set {a, b} }
| c = Discrete Topology
}}
{{end-eqn}}
It remains to show that every union of elements in $\set {\tau_1, \tau_2, \tau_3, \tau_4}$ is a topology on $S$.
By inspection, the following statements are seen to hold:
:$\tau_1 \cup \tau_2 = \tau_2$
:$\tau_1 \cup \tau_3 = \tau_3$
:$\tau_1 \cup \tau_4 = \tau_4$
:$\tau_2 \cup \tau_3 = \tau_4$
:$\tau_2 \cup \tau_4 = \tau_4$
:$\tau_3 \cup \tau_4 = \tau_4$
Now let $\family {\tau_i}_{i \mathop \in I}$ be an arbitrary non-empty indexed set of topologies for a set $S$.
By the above, if $\tau_4$ is one of the $\tau_i$:
:$\ds \tau = \bigcup_{i \mathop \in I} \tau_i = \tau_4$
So assume that $\tau_4$ is not one of the $\tau_i$.
If both $\tau_2$ and $\tau_3$ are in $\family {\tau_i}_{i \mathop \in I}$, then by the above:
:$\ds \tau = \bigcup_{i \mathop \in I} \tau_i = \tau_4$
Now suppose that $\tau_3$ is in $\family {\tau_i}_{i \mathop \in I}$ and $\tau_2$ is not.
Then by the above it follows that:
:$\tau = \tau_3$
Now suppose that $\tau_2$ is in $\family {\tau_i}_{i \mathop \in I}$ and $\tau_3$ is not.
Then by the above it follows that:
:$\tau = \tau_2$
Finally, assume that $\tau_2$ and $\tau_3$ are not in $\family {\tau_i}_{i \mathop \in I}$.
Then the only topology in $\family {\tau_i}_{i \mathop \in I}$ is $\tau_1$.
In this, we find:
:$\tau = \tau_1$
Hence the result.
{{qed}}
\end{proof}
|
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