nyuuzyou/wikis huggingface dataset:
Set is Equivalent to Proper Subset of Power Set

== Theorem ==
Every set is equivalent to a proper subset of its power set:
:$\forall S: \exists T \subset \powerset S: S \sim T$
== Proof ==
To show equivalence between two sets, we need to demonstrate that a bijection exists between them.
We will now define such a bijection.
Let $T = \set {\set x: x \in S}$.
Now we define the mapping $\phi: S \to T$:
:$\phi: S \to T: \forall x \in S: \map \phi x = \set x$
$\phi$ is an injection:
:$\forall x, y \in S: \set x = \set Y \implies x = y$ by definition of set equality
$\phi$ is a surjection:
:$\forall \set x \in T: \exists x \in S: \map \phi x = \set x$
So $\phi$, being both an injection and a surjection, is a bijection by definition.
To show that $T \subset \powerset S$, that is, is a proper subset of $\powerset S$, we merely note that $\O \in \powerset S$ by Empty Set is Element of Power Set, but $\O \notin T$.
Thus $T \subseteq \powerset S$ but $\powerset S \nsubseteq T$.
Hence the result.
Category:Power Set
Category:Proper Subsets
Set is G-Set iff Element of G-Ordered Set

== Theorem ==
Let $g$ be a progressing mapping.
Let $x$ be a set.
Then:
:$x$ is a $g$-set
:
:$x$ is an element of a $g$-ordered set.
== Proof ==
Let $M$ be the class of all $g$-sets.
Then $M$ is a $g$-tower.
=== Sufficient Condition ===
Let $x$ be a $g$-set.
Hence by $g$-tower is $g$-ordered, $M$ is $g$-ordered.
Hence for $x \in M$, the lower section $x^\subseteq$ of $x$ is also $g$-ordered.
As $x \in x^\subseteq$ it follows that $x$ is an element of a $g$-ordered set.
=== Necessary Condition ===
Let $y$ be a $g$-ordered set.
Then by Transfinite Induction on the $g$-ordering of $y$, every element of $y$ must be an element of $M$.
Indeed:
:$\O \in M$
Let $x \in y$ be an element of $M$ other than the greatest element of $y$
Then the immediate successor of $x$ is $\map g x$.
But $\map g x$ is an element of $M$.
Let $z$ be a limit element $y$.
Let each element of $z^\subset$ be an element of $M$.
Then:
:because $z = \bigcup x^\subset$
:$M$ is closed under chain unions.
$z$ is an element of $M$.
Thus every element of $y$ is an element of $M$.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {III}$ -- The existence of minimally superinductive classes: $\S 8$ Another characterization of $g$-sets
Category:G-Sets
Category:G-Ordered Classes
Set is Infinite iff exist Subsets of all Finite Cardinalities

== Theorem ==
A set $S$ is infinite for all $n \in \N$, there exists a subset of $S$ whose cardinality is $n$.
== Proof ==
Let the cardinality of a set $S$ be denoted $\card S$.
=== Necessary Condition ===
Suppose $S$ is infinite.
We use mathematical induction on $n$.
For all $n \in \N$, let $\map P n$ be the proposition:
:There exists a subset of $S$ whose cardinality is $n$.
==== Basis for the Induction ====
From Empty Set is Subset of All Sets:
:$\O \subseteq S$
From Cardinality of Empty Set, its cardinality is $0$.
This is our basis for the induction.
==== Induction Hypothesis ====
Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:There exists a subset of $S$ whose cardinality is $k$.
Then we need to show:
:There exists a subset of $S$ whose cardinality is $k + 1$.
==== Induction Step ====
This is our induction step:
By the induction hypothesis, there exists $T \subseteq S$ such that $\card T = k$.
First, note that $S \ne T$, otherwise by definition of infinite $S$ would be finite.
So $T$ is therefore a proper subset of $S$.
So:
As $x \notin T$ it follows that:
:$\set x \cap T = \O$
Thus by Cardinality of Set Union:
:$\card {T \cup \set x} = k + 1$
That is, $T \cup \set x$ is a subset of $S$ whose cardinality is $k + 1$.
This is the set whose existence was to be be proved.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:For all $n \in \N$, there exists a subset of $S$ whose cardinality is $n$.
=== Sufficient Condition ===
Suppose that for all $n \in \N$, there exists a subset of $S$ whose cardinality is $n$.
 that $S$ is finite.
Let $N = \card S$.
As $N \in \N$ it follows that $N + 1 \in \N$.
By hypothesis, there exists a subset $T \subseteq S$ whose cardinality is $N + 1$.
From Cardinality of Subset of Finite Set, $\card S \ge \card T$.
But then $\card S = N \ge N + 1 = \card T$, which contradicts the fact that $N < N + 1$.
From this contradiction it follows that $S$ can not be finite.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 18$: Induced $N$-ary Operations: Exercise $18.18$
Category:Infinite Sets
Set is Neighborhood of Subset iff Neighborhood of all Points of Subset

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $N \subseteq S$ be a subset of $T$.
Let $A \subseteq N$ be a subset of $T$.
Then:
:$N$ is a neighborhood of $A$ in $T$
:
:$N$ is a neighborhood of all points in $A$
== Proof ==
=== Necessary Condition ===
Let $N$ be a neighborhood of $A$ in $T$.
By definition of neighborhood of set:
:$\exists U \in \tau : A \subseteq U \subseteq N$
Let $z \in A$.
By definition of subset:
:$z \in U$
From Set is Open iff Neighborhood of all its Points:
:$U$ is a neighborhood of $z$
From Superset of Neighborhood in Topological Space is Neighborhood:
:$N$ is a neighborhood of $z$
Since $z$ was arbitrary:
:$N$ is a neighborhood of all points in $A$
=== Sufficient Condition ===
Suppose that for all points of $z \in A$, $N$ is a neighborhood of $z$.
That is, for all $z \in A$ there exists an open set $U_z \subseteq N$ of $T$ such that $z \in U_z$.
Now by Union is Smallest Superset: Family of Sets:
:$\ds \bigcup_{z \mathop \in A} U_z \subseteq N$
as $\forall z \in A: U_z \subseteq N$.
If $z \in A$, then $z \in U_z$ by definition of $U_z$.
So:
:$\ds z \in \bigcup_{z \mathop \in A} U_z$
Thus we also have:
:$\ds A \subseteq \bigcup_{z \mathop \in A} U_z$
Let $U = \ds \bigcup_{z \mathop \in A} U_z$.
By :
:$U$ is open in $T$
Since $A \subseteq U \subseteq N$, then $N$ is neighborhood of $A$ in $T$ by definition.
Category:Neighborhoods
Set is Not Element of Itself

== Theorem ==
There cannot exist a set which is an element of itself.
That is:
:$\neg \exists a: a \in a$
== Proof ==
 $a$ is such a set.
Then $a \in a$ and $a \in \set a$.
$a \ne \O$ because the empty set has no elements by definition.
It is also seen that:
By the Axiom of Foundation:
:$a \cap \set a = \O$
Thus $a = \O$.
But it was previously established that $a \ne \O$.
From this contradiction it follows that there cannot exist such a set.
Hence the result.
Category:Axiom of Foundation
Set is Open iff Disjoint from Boundary

== Theorem ==
Let $T$ be a topological space.
Let $H \subseteq T$.
Then $H$ is open in $T$ :
:$\partial H \cap H = \O$
where $\partial H$ is the boundary of $H$.
== Proof ==
From Boundary is Intersection of Closure with Closure of Complement:
:$\partial H = H^- \cap \paren {T \setminus H}^-$
where $H^-$ is the closure of $H$.
Hence from Intersection is Subset we have that:
:$\partial H \subseteq \paren {T \setminus H}^-$
But from Closed Set Equals its Closure, $\paren {T \setminus H}^- = T \setminus H$ $T \setminus H$ is closed in $T$.
That is, $H$ is open in $T$.
So $\partial H \subseteq T \setminus H$ $H$ is open in $T$.
From Intersection with Complement is Empty iff Subset it follows that $\partial H \cap H = \O$ $H$ is open in $T$.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Closures and Interiors
Category:Open Sets
Category:Set Boundaries
Set is Open iff Neighborhood of all its Points

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $V \subseteq S$ be a subset of $T$.
Then:
:$V$ is an open set of $T$
:
:$V$ is a neighborhood of all the points in $V$.
== Proof ==
=== Necessary Condition ===
Let $V$ be open in $T$.
Let $z \in V$.
By definition, a neighborhood of $z$ is any subset of $S$ containing an open set which itself contains $z$.
But $V$ is itself an open set which itself contains $z$.
Hence by Set is Subset of Itself, $V$ is a subset of $S$ which contains an open set which itself contains $z$.
So for all points of $z \in V$, $V$ is a neighborhood of $z$.
=== Sufficient Condition ===
Suppose that for all points of $z \in V$, $V$ is a neighborhood of $z$.
That is, for all $z \in V$ there exists an open set $T_z \subseteq V$ of $T$ such that $z \in T_z$.
Now by Union is Smallest Superset: Family of Sets:
:$\ds \bigcup_{z \mathop \in V} T_z \subseteq V$
as $\forall z \in V: T_z \subseteq V$.
If $z \in V$, then $z \in T_z$ by definition of $T_z$.
So:
:$\ds z \in \bigcup_{z \mathop \in V} T_z$
Thus we also have:
:$\ds V \subseteq \bigcup_{z \mathop \in V} T_z$
Hence by definition of set equality:
:$\ds V = \bigcup_{z \mathop \in V} T_z$
Thus $V$ can be expressed as a union of open sets.
Hence $V$ is open in $T$, by of a topological space.
== Also see ==
* Space is Neighborhood of all its Points
== Sources ==
* : Chapter $\text {I}$: Topological Spaces: $1$. Open Sets and Closed Sets: Lemma $1$
* : Chapter $3$: Topological Spaces: $\S 2$: Topological Spaces: Corollary $2.3$
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction
* 
* 
Category:Open Sets
Category:Neighborhoods
Set is Open iff Union of Open Balls

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $U \subseteq A$.
Then $U$ is open in $M$ it is a union of open balls.
== Proof 1 ==
=== Necessary Condition ===
Let $U$ be open in $M$.
Let $a \in U$.
Then by definition of open set:
:$\exists \delta_a \in \R_{>0}: \map {B_{\delta_a} } {a, d} \subseteq U$
where $\map {B_{\delta_a} } {a, d}$ is the open $\delta_a$-ball of $a$ in $M$.
Therefore:
:$\ds U = \bigcup_{a \mathop \in U} \map {B_{\delta_a} } {a, d}$
and so $U$ is a union of open balls.
=== Sufficient Condition ===
Let $U$ be a union of open balls of $M$.
Let the centers of these open balls be the elements of an indexing set $I$.
Then $U$ can be written as:
:$\ds U = \bigcup_{a \mathop \in I} \map {B_{\delta_a} } {a, d}$
where $\delta_a \in \R_{>0}$ is the radius of the open ball of $a$.
Let $x \in U$.
Then by definition of union:
:$\exists a \in I: x \in \map {B_{\delta_a} } {a, d}$
From Open Ball is Neighborhood of all Points Inside, $\map {B_{\delta_a} } {a, d}$ is a neighborhood of $x$.
By Set is Subset of Union of Family:
:$\map {B_{\delta_a} } {a, d} \subseteq U$
From Superset of Neighborhood in Metric Space is Neighborhood, it follows that $U$ is a neighborhood of $x$.
Since $x$ is arbitrary, it follows that $U$ is a neighborhood of each of its points.
Hence by definition, $U$ is open in $M$.
== Proof 2 ==
=== Necessary Condition ===
See Proof 1.
=== Sufficient Condition ===
 $U$ is a union of open balls but not open.
As $U$ is not open, there is a $y \in U$ such that for any $\epsilon \in \R_{>0}$, $\map {B_\epsilon} y \setminus U \ne \O$.
As $U$ is a union of open balls, there is an $x \in U$ and $r \in \R_{>0}$ such that $y \in \map {B_r} x$.
By Open Ball of Point Inside Open Ball, there exists a $\delta \in \R_{>0}$ such that $\map {B_\delta} y \subseteq \map {B_r} x$.
This is contradicted by the assumption.
Thus, $U$ has to be open.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 6$: Open Sets and Closed Sets: Theorem $6.2$
* : $2$: Continuity generalized: metric spaces: Exercise $2.6: 10$
Category:Open Sets (Metric Spaces)
Category:Open Balls
Category:Metric Spaces
Set is Ordinal iff Every Transitive Proper Subset is Element of it

== Theorem ==
Let $x$ be a set.
Then:
:$x$ is an ordinal
:every transitive proper subset of $x$ is an element of $x$.
== Proof ==
=== Necessary Condition ===
Let $\alpha$ be an arbitrary ordinal.
Let $y$ be a proper subset of $\alpha$ such that $y$ is transitive.
From Transitive Set of Ordinals is Ordinal it follows that $y$ is an ordinal.
By definition of usual ordering of ordinals:
:$y < \alpha$
Hence from Strict Ordering of Ordinals is Equivalent to Membership Relation:
:$y \in \alpha$
As $\alpha$ is arbitrary, it follows that every transitive proper subset of $\alpha$ is an element of $\alpha$.
=== Sufficient Condition ===
Let $x$ be a set such that every transitive proper subset of $x$ is an element of $x$.
Let $\alpha$ be the smallest ordinal which is not in $x$.
This is always possible by Well-Ordering of Class of All Ordinals under Subset Relation.
Thus every ordinal which is smaller than $\alpha$ is an element of $x$.
Hence:
:$\alpha \subseteq x$
Suppose $\alpha$ is a proper subset of $x$.
From Ordinal is Transitive, $\alpha$ is then a transitive proper subset of $x$.
Hence $\alpha \in x$
But we have that $\alpha \notin x$.
Hence it must be the case that $\alpha = x$.
Thus $x$ is an ordinal.
The result follows.
== Sources ==
* : Chapter $5$: Ordinal Numbers: $\S 2$ Ordinals and transitivity: Theorem $2.3$
Category:Ordinals
Category:Transitive Classes
Set is Recursively Enumerable iff Domain of Recursive Function

== Theorem ==
Let $S \subseteq \N$ be a set of natural numbers.
Then:
:$S$ is recursively enumerable
:
:There exists a recursive function $f : \N \to \N$ such that $S = \Dom f$
where $\Dom f$ is the domain of $f$.
== Proof ==
=== Necessary Condition ===
Suppose $S$ is the empty set.
Define $\map f x = \map {\mu z} {z \ne z}$.
Then, $f$ is undefined for every $x \in \N$, and thus is defined on the empty set.
Additionally, $f$ is recursive by:
* Equality Relation is Primitive Recursive
* Set Operations on Primitive Recursive Relations
* Primitive Recursive Function is Total Recursive Function
Otherwise, let $S$ be non-empty.
Then, by Recursively Enumerable Set is Image of Primitive Recursive Function/Corollary, there exists a primitive recursive function $g : \N \to \N$ such that:
:$S = \Img g$
Define $\map f x = \map {\mu z} {\map g z = x}$.
It remains to be shown that $\Dom f = \Img f$.
Let $y \in \Dom f$.
Then, there exists $z \in \N$ such that $\map g z = y$.
But then, $y \in \Img g$.
Now, let $y \in \Img g$.
Then, there exists $z \in \N$ such that $\map g z = y$.
Let $z'$ be the smallest such $z$.
Then, $\map f y = z'$ be definition.
Thus, $y \in \Dom f$.
=== Sufficient Condition ===
Define $\map g x = \map {\pr_1^2} {x, \map f x}$.
As projection is a basic primitive recursive function, and $g$ is obtained by substitution, it is recursive by definition.
It remains to be shown that $\Img g = \Dom f$.
Let $y \in \Img g$.
Then, by definition of substitution, $\map f y$ is defined.
Therefore, $y \in \Dom f$.
Now, let $y \in \Dom f$.
Then, as $\map f y$ is defined:
:$\map g y = y$
Thus, $y \in \Img g$.
Category:Recursion Theory
Set is Subset of Finite Infima Set

== Theorem ==
Let $\left({S, \preceq}\right)$ be an ordered set.
Let $X$ be a subset of $S$.
Then $X \subseteq \operatorname{fininfs}\left({X}\right)$
where $\operatorname{fininfs}\left({X}\right)$ denotes the finite infima set of $X$.
== Proof ==
Let $x \in X$.
By Infimum of Singleton:
:$\left\{ {x}\right\}$ admits an infimum and $\inf \left\{ {x}\right\} = x$
By definitions of subset and singleton:
:$\left\{ {x}\right\} \subseteq X$
By Singleton is Finite:
:$\left\{ {x}\right\}$ is a finite set.
Thus by definition of finite infima set:
:$x \in \operatorname{fininfs}\left({X}\right)$
== Sources ==
* 
Category:Order Theory
Set is Subset of Finite Suprema Set

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $X$ be a subset of $S$.
Then $X \subseteq \map {\mathrm {finsups} } X$
where $\map {\mathrm {finsups} } X$ denotes finite suprema set of $X$.
== Proof ==
Let $x \in X$.
By Supremum of Singleton:
:$\set x$ admits a supremum and $\sup \set x = x$
By definitions of subset and singleton:
:$\set x \subseteq X$
By Singleton is Finite:
:$\set x$ is a finite set.
Thus by definition of finite suprema set:
:$x \in \map {\mathrm {finsups} } X$
== Sources ==
* 
Category:Suprema
Set is Subset of Intersection of Supersets

== Theorem ==
Let $S$, $T_1$ and $T_2$ be sets.
Let $S$ be a subset of both $T_1$ and $T_2$.
Then:
:$S \subseteq T_1 \cap T_2$
That is:
:$\paren {S \subseteq T_1} \land \paren {S \subseteq T_2} \implies S \subseteq \paren {T_1 \cap T_2}$
=== Set of Sets ===
=== General Result ===
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Union of Subsets is Subset
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Exercise $\text{B viii}$
* : $\S 1$: Naive Set Theory: $\S 1.2$: Operations on Sets: Exercise $1.2.1 \ \text{(iv)}$
* : $\S 1.6$: Set Identities and Other Set Relations: Exercise $4$
Category:Set Intersection
Category:Subsets
Category:Set is Subset of Intersection of Supersets
Set is Subset of Intersection of Supersets/General Result

== Theorem ==
Let $\family {S_i}_{i \mathop \in I}$ be a family of sets indexed by $I$.
Let $X$ be a set such that:
:$\forall i \in I: X \subseteq S_i$
Then:
:$\ds X \subseteq \bigcap_{i \mathop \in I} S_i$
where $\ds \bigcap_{i \mathop \in I} S_i$ is the intersection of $\family {S_i}$.
== Proof ==
Let $X \subseteq S_i$ for all $i \in I$.
Then:
== Sources ==
* : $\S 1$: Naive Set Theory: $\S 1.4$: Sets of Sets: Exercise $1.4.4 \ \text{(ii)}$
Category:Set is Subset of Intersection of Supersets
Set is Subset of Intersection of Supersets/Proof 1

== Theorem ==
== Proof ==
Let $S \subseteq T_1 \land S \subseteq T_2$.
Then:
== Sources ==
* : $\S 1.3$. Intersection: Example $13$
* : $\S 7.2 \ \text{(i)}$: Unions and Intersections
Category:Set is Subset of Intersection of Supersets
Set is Subset of Intersection of Supersets/Set of Sets

== Theorem ==
Let $T$ be a set.
Let $\mathbb S$ be a set of sets.
Suppose that for each $S \in \mathbb S$, $T \subseteq S$.
Then:
:$T \subseteq \ds \bigcap \mathbb S$
== Proof ==
Let $x \in T$.
We are given that:
:$\forall S \in \mathbb S: T \subseteq S$ 
Thus by definition of subset:
:$\forall S \in \mathbb S: x \in S$
Hence by definition of intersection:
:$x \in \ds \bigcap \mathbb S$
Thus by definition of subset:
:$T \subseteq \ds \bigcap \mathbb S$
Category:Set is Subset of Intersection of Supersets
Set is Subset of Itself

== Theorem ==
Every set is a subset of itself:
:$\forall S: S \subseteq S$
Thus, by definition, the relation is a subset of is reflexive.
== Proof ==
== Sources ==
* : $\S 1$: The Axiom of Extension
* : $\S 1$: Sets and Set Inclusion
* : $\S 1.1$
* : $\S 2.1$: Sets
* : Chapter $1$: Set Theory: $1.2$: Sets and subsets
* : $\S 1.2$. Subsets
* : Chapter $\text I$: Algebraic Structures: $\S 1$: The Language of Set Theory
* : $\S 0.2$. Sets
* : Chapter $\text{I}$: Sets and Functions: Exercise $\text{B i}$
* : $\S 1.3$: Subsets
* : Chapter $1$: Sets: Subsets
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: Theorem $1.1 \ \text{(a)}$
* : Chapter $1$: Theory of Sets: $\S 2$: Sets and Subsets
* : $\S 6.1$: Subsets
* : Chapter $1$: Sets and mappings: $\S 1.2$: Sets
* : $2$. Graphs: Sets: Definition $3$
* 
Category:Subsets
Set is Subset of Lower Closure

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $X$ be a subset of $S$.
Then $X \subseteq X^\preceq$
where $X^\preceq$ denotes the lower closure of $X$.
== Proof ==
Let $x \in X$.
By definition of reflexivity:
:$x \preceq x$
Thus by definition of lower closure:
:$x \in X^\preceq$
== Sources ==
* 
Category:Order Theory
Set is Subset of Power Set of Union

== Theorem ==
Let $x$ be a set of sets.
Let $\bigcup x$ denote the union of $x$.
Let $\powerset {\bigcup x}$ denote the power set of $\bigcup x$.
Then:
:$x \subseteq \powerset {\bigcup x}$
== Proof ==
Let $z \in x$.
By Element of Class is Subset of Union of Class:
:$z \subseteq \bigcup x$
By definition of power set:
:$z \in \powerset {\bigcup x}$
The result follows by definition of subset.
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 6$ The power axiom: Exercise $6.1. \ \text {(a)}$
Category:Power Set
Category:Set Union
Set is Subset of Union

== Theorem ==
The union of two sets is a superset of each:
:$S \subseteq S \cup T$
:$T \subseteq S \cup T$
=== General Result ===
=== Set of Sets ===
=== Indexed Family of Sets ===
In the context of a family of sets, the result can be presented as follows:
== Proof ==
Similarly for $T$.
== Sources ==
* : Chapter $0$: Subsets and Complements; Union and Intersection
* : $\S 1.1$: Theorem $1.4$
* : Exercise $1.1: \ 8 \ \text{(f)}$
* : $\S 1.4$. Union: Example $15$
* : Chapter $1$: The Notation and Terminology of Set Theory: $\S 6$
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems
* : $\S 7$: Unions and Intersections
* : $\S 1$: Naive Set Theory: $\S 1.2$: Operations on Sets: Exercise $1.2.1 \ \text{(ii)}$
Category:Set is Subset of Union
Category:Set Union
Category:Subsets
Set is Subset of Union/Family of Sets

== Theorem ==
Let $\family {S_\alpha}_{\alpha \mathop \in I}$ be a family of sets indexed by $I$.
Then:
:$\ds \forall \beta \in I: S_\beta \subseteq \bigcup_{\alpha \mathop \in I} S_\alpha$
where $\ds \bigcup_{\alpha \mathop \in I} S_\alpha$ is the union of $\family {S_\alpha}$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 4$: Indexed Families of Sets: Exercise $1 \ \text{(a)}$
Category:Set is Subset of Union
Category:Set Union
Category:Subsets
Category:Indexed Families
Set is Subset of Union/Family of Sets/Proof 1

== Theorem ==
== Proof ==
Let $x \in S_\beta$ for some $\beta \in I$.
Then:
As $\beta$ was arbitrary, it follows that:
:$\ds \forall \beta \in I: S_\beta \subseteq \bigcup_{\alpha \mathop \in I} S_\alpha$
Category:Set is Subset of Union
Set is Subset of Union/General Result

== Theorem ==
Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Let $\mathbb S \subseteq \powerset S$.
Then:
:$\ds \forall T \in \mathbb S: T \subseteq \bigcup \mathbb S$
== Proof ==
Let $x \in T$ for some $T \in \mathbb S$.
Then: 
As $T$ was arbitrary, it follows that:
:$\ds \forall T \in \mathbb S: T \subseteq \bigcup \mathbb S$
Category:Set Union
Category:Subsets
Set is Subset of Union/Set of Sets

== Theorem ==
Let $\mathbb S$ be a set of sets.
Then:
:$\ds \forall T \in \mathbb S: T \subseteq \bigcup \mathbb S$
== Proof ==
Let $T$ be any element of $\mathbb S$.
We wish to show that $T \subseteq S$.
Let $x \in T$.
Then: 
Since this holds for each $x \in T$:
As $T$ was arbitrary, it follows that:
:$\forall T \in \mathbb S: T \subseteq \bigcup \mathbb S$
Category:Set Union
Category:Subsets
Set is Subset of Upper Closure

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $X$ be a subset of $S$.
Then $X \subseteq X^\succeq$
where $X^\succeq$ denotes the upper closure of $X$.
== Proof ==
Let $x \in X$.
By definition of reflexivity:
:$x \preceq x$
Thus by definition of upper closure:
:$x \in X^\succeq$
== Sources ==
* 
Category:Order Theory
Set is Subset of its Topological Closure

== Theorem ==
Let $T$ be a topological space.
Let $H \subseteq T$.
Let $H^-$ be the closure of $H$ in $T$.
Then:
:$H \subseteq H^-$
== Proof ==
From the definition of closure, we have:
:$H^-$ is the union of $H$ and its limit points.
From Subset of Union it follows directly that:
:$H \subseteq H^-$
Category:Set Closures
Set is Transitive iff Subset of Power Set

== Theorem ==
A set $a$ is transitive :
:$a \subseteq \powerset a$
where $\powerset a$ denotes the power set of $a$.
== Proof ==
=== Necessary Condition ===
Let $a$ be transitive.
Let $x \in a$.
By definition of transitive set:
:$x \subseteq a$
Then by definition of power set:
:$x \in \powerset a$
Hence, by definition of subset:
:$a \subseteq \powerset a$
=== Sufficient Condition ===
Let $a \subseteq \powerset a$.
Let $x \in a$.
Then by definition of subset:
:$x \in \powerset a$
By definition of power set:
:$x \subseteq a$
As this is true for all $x \in a$, it follows by definition that $a$ is transitive.
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 10$ Some useful facts about transitivity: Theorem $10.4$
Category:Power Set
Category:Transitive Classes