nyuuzyou/wikis huggingface dataset:
Power Set can be Derived using Axiom of Abstraction

== Theorem ==
Let $a$ be a set.
By application of the , the power set $\powerset a$ can be formed.
Hence the power set $\powerset a$ can be derived as a valid object in Frege set theory.
== Proof ==
Let $P$ be the property defined as:
:$\forall x: \map P x := \paren {x \subseteq a}$
where $\lor$ is the disjunction operator.
Hence, using the , we form the set:
:$\powerset a := \set {x: x \subseteq a}$
== Sources ==
* : Chapter $1$: General Background: $\S 7$ Frege set theory
Category:Power Set
Category:Axiom of Abstraction
Power Set is Algebra of Sets

== Theorem ==
Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Then $\powerset S$ is an algebra of sets where $S$ is the unit.
== Proof ==
From Power Set is Closed under Intersection and Power Set is Closed under Symmetric Difference, we have that:
:$(1): \quad \forall A, B \in \powerset S: A \cap B \in \powerset S$
:$(2): \quad \forall A, B \in \powerset S: A * B \in \powerset S$
From the definition of power set:
:$\forall A \in \powerset S: A \subseteq S$
and so $S$ is the unit of $\powerset S$.
Thus we see that $\powerset S$ is a ring of sets with a unit.
Hence the result, by definition of an algebra of sets.
Category:Power Set
Category:Algebras of Sets
Power Set is Boolean Ring

== Theorem ==
Let $S$ be a set, and let $\powerset S$ be its power set.
Denote with $\symdif$ and $\cap$ symmetric difference and intersection, respectively.
Then $\struct {S, \symdif, \cap}$ is a Boolean ring.
== Proof ==
From Symmetric Difference with Intersection forms Ring, $\struct {S, \symdif, \cap}$ is a ring with unity.
By Set Intersection is Idempotent, $\cap$ is idempotent.
It follows that $\struct {S, \symdif, \cap}$ is a Boolean ring.
== Sources ==
* : $\S 3$
Category:Power Set
Category:Boolean Rings
Power Set is Closed under Complement

== Theorem ==
Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Then:
:$\forall A \in \powerset S: \relcomp S A \in \powerset S$
where $\relcomp S A$ denotes the complement of $A$ relative to $S$.
== Proof ==
== Also see ==
* Power Set is Closed under Union
* Power Set is Closed under Intersection
* Power Set is Closed under Set Difference
Category:Power Set
Category:Relative Complement
Power Set is Closed under Countable Unions

== Theorem ==
Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Then:
:$\forall A_n \in \powerset S: n = 1, 2, \ldots: \ds \bigcup_{n \mathop = 1}^\infty A_n \in \powerset S$
== Proof ==
Let $\sequence {A_i}$ be a countably infinite sequence of sets in $\powerset S$.
Consider an element of the union of all the sets in this $\sequence {A_i}$:
:$\ds x \in \bigcup_{i \mathop \in \N} A_i$
By definition of union:
:$\exists i \in \N: x \in A_i$
But $A_i \in \powerset S$ and so by definition $A_i \subseteq S$.
By definition of subset, it follows that $x \in S$.
Hence, again by definition of subset:
:$\ds \bigcup_{i \mathop \in \N} A_i \subseteq S$
Finally, by definition of power set:
:$\ds \bigcup_{i \mathop \in \N} A_i \in \powerset S$
== Also see ==
* Power Set is Closed under Union
* Power Set is Closed under Intersection
* Power Set is Closed under Set Difference
Category:Power Set
Category:Set Union
Category:Countable Sets
Power Set is Closed under Intersection

== Theorem ==
Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Then:
:$\forall A, B \in \powerset S: A \cap B \in \powerset S$
== Proof ==
Let $A, B \in \powerset S$.
Then by the definition of power set, $A \subseteq S$ and $B \subseteq S$.
From Intersection is Subset we have that $A \cap B \subseteq A$.
It follows from Subset Relation is Transitive that $A \cap B \subseteq S$.
Thus $A \cap B \in \powerset S$ and closure is proved.
== Also see ==
* Power Set is Closed under Union
* Power Set is Closed under Set Difference
== Sources ==
* : $\S 4.1$. Binary operations on a set: Example $60$
* 
Category:Power Set
Category:Set Intersection
Power Set is Closed under Set Complement

== Theorem ==
Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Then:
:$\forall A \in \powerset S: \relcomp S A \in \powerset S$
== Proof ==
Let $A \in \powerset S$.
Then by the definition of power set, $A \subseteq S$.
By definition of relative complement:
:$\relcomp S A = \set {x \in S: x \notin A}$
Hence $\relcomp S A$ is a subset of $S$.
That is:
:$\relcomp S A \in \powerset S$
and closure is proved.
== Also see ==
* Power Set is Closed under Union
* Power Set is Closed under Intersection
* Power Set is Closed under Set Difference
== Sources ==
* 
Category:Power Set
Category:Set Complement
Power Set is Closed under Set Difference

== Theorem ==
Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Then:
:$\forall A, B \in \powerset S: A \setminus B \in \powerset S$
where $A \setminus B$ denotes the set difference of $A$ and $B$.
== Proof ==
Let $A, B \in \powerset S$.
Then by the definition of power set, $A \subseteq S$ and $B \subseteq S$.
We also have $A \setminus B \subseteq A$ from Set Difference is Subset.
Thus by Subset Relation is Transitive, $A \setminus B \subseteq S$.
Thus $A \setminus B \in \powerset S$, and closure is proved.
== Also see ==
* Power Set is Closed under Intersection
* Power Set is Closed under Union
== Sources ==
* : $\S 4.1$. Binary operations on a set: Example $60$
Category:Power Set
Category:Set Difference
Power Set is Closed under Symmetric Difference

== Theorem ==
Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Then:
:$\forall A, B \in \powerset S: A \symdif B \in \powerset S$
where $A \symdif B$ is the symmetric difference between $A$ and $B$.
== Proof ==
Let $A, B \in \powerset S$.
Then by definition of power set:
:$A, B \subseteq S$
Then:
== Sources ==
* : Introduction to Set Theory: $1$. Elementary Operations on Sets
Category:Power Set
Category:Symmetric Difference
Power Set is Closed under Union

== Theorem ==
Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Then:
:$\forall A, B \in \powerset S: A \cup B \in \powerset S$
== Proof ==
Let $A, B \in \powerset S$.
Then by the definition of power set, $A \subseteq S$ and $B \subseteq S$.
We also have $A \cup B \subseteq S \iff A \subseteq S \land B \subseteq S$ from Union is Smallest Superset.
Thus $A \cup B \in \powerset S$, and closure is proved.
== Also see ==
* Power Set is Closed under Intersection
* Power Set is Closed under Set Difference
== Sources ==
* : $\S 4.1$. Binary operations on a set: Example $60$
* 
Category:Power Set
Category:Set Union
Category:Closed Algebraic Structures
Power Set is Complete Lattice

== Theorem ==
Let $S$ be a set.
Let $\struct {\powerset S, \subseteq}$ be the relational structure defined on the power set $\powerset S$ of $S$ by the relation $\subseteq$.
Then:
:$\struct {\powerset S, \subseteq}$ is a complete lattice
where for every subset $\mathbb S$ of $\powerset S$:
:the infimum of $\mathbb S$ necessarily admitted by $\mathbb S$ is $\bigcap \mathbb S$.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Power Set is Lattice
Category:Examples of Complete Lattices
Category:Power Set
Category:Power Set is Complete Lattice
Power Set is Complete Lattice/Proof 1

== Theorem ==
== Proof ==
From Subset Relation on Power Set is Partial Ordering, we have that $\subseteq$ is a partial ordering.
We note in passing that for any set $S$:
:From Supremum of Power Set, $\powerset S$ has a supremum, that is, $S$ itself
:From Infimum of Power Set, $\powerset S$ has an infimum, that is, $\O$.
These are also the maximal and minimal elements of $\powerset S$.
Let $\mathbb S$ be a subset of $\powerset S$.
Then from Union is Smallest Superset:
:$\paren {\forall X \in \mathbb S: X \subseteq T} \iff \bigcup \mathbb S \subseteq T$
and from Intersection is Largest Subset:
:$\paren {\forall X \in \mathbb S: T \subseteq X} \iff T \subseteq \bigcap \mathbb S$
So $\bigcap \mathbb S$ is the infimum and $\bigcup \mathbb S$ is the supremum of $\struct {\mathbb S, \subseteq}$.
Hence by definition $\powerset S$ is a complete lattice.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Example $14.3$
Category:Power Set is Complete Lattice
Power Set is Complete Lattice/Proof 2

== Theorem ==
== Proof ==
From Set is Subset of Itself:
:$S \in \powerset S$
Let $\mathbb S$ be a non-empty subset of $\powerset S$.
From Intersection is Subset:
:$\bigcap \mathbb S \in \powerset S$
Hence, from Set of Subsets which contains Set and Intersection of Subsets is Complete Lattice:
:$\struct {\powerset S, \subseteq}$ is a complete lattice
where $\bigcap \mathbb S$ is the infimum of $\mathbb S$.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.11 \ \text{(c) (3)}$
Category:Power Set is Complete Lattice
Power Set is Filter in Lattice of Power Set

== Theorem ==
Let $X$ be a set.
Let $L = \struct {\powerset X, \cup, \cap, \subseteq}$ be a inclusion lattice of power set of $X$.
Then $\powerset X$ is a filter on $L$.
== Proof ==
=== Filtered ===
By Set is Element of its Power Set:
:$X \in \powerset X$
Thus by definition:
:$\powerset X$ is a non-empty set.
Let $x, y \in \powerset X$.
By Intersection is Subset:
:$x \cap y \subseteq x$ and $x \cap y \subseteq y$
By Subset Relation is Transitive:
:$x \cap y \in \powerset X$
Thus
:$\exists z \in \powerset X: z \subseteq x \land z \subseteq y$
=== Upper ===
Thus we have:
:$\forall x, y \in \powerset X: x \subseteq y \implies y \in \powerset X$
Thus by definition of filter in ordered set:
:$\powerset X$ is a filter on $L$.
== Sources ==
* 
Category:Power Set
Power Set is Lattice

== Theorem ==
Let $S$ be a set.
Let $\struct {\powerset S, \subseteq}$ be the relational structure defined on $\powerset S$ by the subset relation $\subseteq$.
Then $\struct {\powerset S, \subseteq}$ is a lattice.
== Proof ==
From Subset Relation on Power Set is Partial Ordering, we have that $\subseteq$ is a partial ordering.
Let $X, Y \in \powerset S$.
Then from Union is Smallest Superset:
:$X \subseteq T, Y \subseteq T \iff X \cup Y \subseteq T$
and from Intersection is Largest Subset:
:$X \subseteq T, Y \subseteq T \iff T \subseteq X \cap Y$
So $X \cap Y$ is the infimum and $X \cup Y$ is the supremum of $\set {X, Y}$.
Hence by definition $\powerset S$ is a lattice.
== Also see ==
* Power Set is Complete Lattice
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Example $14.3$
* : $\S 7$: Example $7.7$
Category:Lattice Theory
Category:Power Set
Power Set is Magma of Sets

== Theorem ==
Let $X$ be a set.
Let $\family {\phi_i}: i \in I$ be an indexed family of mappings.
Then $\powerset X$, the power set of $X$, is a magma of sets for $\family {\phi_i}: i \in I$ on $X$.
== Proof ==
For every $i \in I$, let $J_i$ be an index set, and let:
:$\phi_i: \powerset X^{J_i} \to \powerset X$
be a partial mapping.
For each $i \in I$, for each $\family {S_{j_i} }_{j_i \mathop \in J_i} \in \powerset X^{J_i} \cap \DD_i$, that:
:$\map {\phi_i} {\family {S_{j_i} }_{j_i \mathop \in J_i} } \in \powerset X$
follows directly from the fact that $\powerset X$ is the codomain of $\phi_i$,
for every indexed family $\family {S_j}_{j \mathop \in J_i} \in \SS^{J_i}$ in the domain of $\phi_i$.
Hence $\powerset X$ is a magma of sets for $\family {\phi_i}: i \in I$ on $X$.
Category:Magmas of Sets
Power Set is Sigma-Algebra

== Theorem ==
The power set of a set is a sigma-algebra.
== Proof ==
Let $S$ be a set, and let $\powerset S$ be its power set.
We have that a power set is an algebra of sets, and so:
:$(1): \quad \forall A, B \in \powerset S: A \cup B \in \powerset S$
:$(2): \quad \relcomp S A \in \powerset S$
Let $\sequence {A_i}$ be a countably infinite sequence of sets in $\powerset S$.
Then from Power Set is Closed under Countable Unions:
:$\ds \bigcup_{i \mathop \in \N} A_i \in \powerset S$
So, by definition, $\powerset S$ is a sigma-algebra.
== Sources ==
* : $3.3 \ \text{(i)}$
* : $1.1$: Algebras and Sigma-Algebras
Category:Sigma-Algebras
Category:Examples of Sigma-Algebras
Power Set less Empty Set has no Smallest Element iff not Singleton

== Theorem ==
Let $S$ be a set which is non-empty.
Let $\CC = \powerset S \setminus \O$, that is, the power set of $S$ without the empty set.
Then the ordered structure $\struct {\CC, \subseteq}$ has no smallest element $S$ is not a singleton.
== Proof ==
=== Necessary Condition ===
Let $S$ not be a singleton.
Then $\exists x, y \in S: x \ne y$.
Let $Z \in \CC$ be the smallest element of $\CC$.
Then:
:$\forall T \in \CC: Z \subseteq T$
But by Singleton of Power Set less Empty Set is Minimal Subset, both $\set x$ and $\set y$ are minimal elements of $\struct {\CC, \subseteq}$.
Therefore it cannot be the case that $Z \subseteq \set x$ and $Z \subseteq \set y$.
Therefore $\struct {\CC, \subseteq}$ has no smallest element.
=== Sufficient Condition ===
Let the ordered structure $\struct {\CC, \subseteq}$ has no smallest element.
 $S$ is a singleton.
Let $S = \set x$.
Then:
:$\CC = \set {\set x}$
Then:
:$\forall y \in \CC: y \subseteq \set x$
trivially.
Thus $\struct {\CC, \subseteq}$ has a smallest element which is $\set x$.
By Proof by Contradiction it follows that $S$ is not a singleton.
== Sources ==
* : $\S 14$: Order
Category:Power Set
Power Set of Doubleton

== Theorem ==
Let $x, y$ be distinct objects.
Then the power set of the doubleton $\set {x, y}$ is:
:$\powerset {\set {x, y} } = \set {\O, \set x, \set y, \set {x, y} }$
== Proof ==
By definition of a subset:
:$\set x , \set y, \set {x, y} \subseteq \set {x, y}$
Let $A \subseteq \set {x, y}$:
:$A \ne \set x, \set y, \set {x, y}$
From set equality:
:$\set {x, y} \nsubseteq A$
From Doubleton of Elements is Subset:
:either $x \notin A$ or $y \notin A$.
 assume that $x \notin A$.
From Intersection With Singleton is Disjoint if Not Element:
:$A \cap \set x = \O$
From Subset of Set Difference iff Disjoint Set:
:$A \subseteq \set {x, y} \setminus \set x$
From Set Difference of Doubleton and Singleton is Singleton:
:$A \subseteq \set y$
From set equality:
:$\set y \nsubseteq A$
From Singleton of Element is Subset:
:$y \notin A$.
From Intersection With Singleton is Disjoint if Not Element:
:$A \cap \set y = \O$
From Subset of Set Difference iff Disjoint Set:
:$A \subseteq \set y \setminus \set y$
From Set Difference with Self is Empty Set:
:$A \subseteq \O$
From Empty Set is Subset of All Sets:
:$\O \subseteq A$
From set equality:
:$A = \O$
It follows that:
:$\powerset {\set {x, y} } = \set {\O, \set x, \set y, \set {x, y} }$
Category:Power Set
Category:Doubletons
Power Set of Empty Set

== Theorem ==
The power set of the empty set $\O$ is the set $\set \O$.
== Proof ==
From Empty Set is Element of Power Set and Set is Element of its Power Set:
:$\O \in \powerset \O$
From Empty Set is Subset of All Sets:
:$S \subseteq \O \implies S = \O$
That is:
:$S \in \powerset \O \implies S = \O$
Hence the only element of $\powerset \O$ is $\O$.
== Sources ==
* : Introduction to Set Theory: $1$. Elementary Operations on Sets
* : Chapter $\text I$: Algebraic Structures: $\S 1$: The Language of Set Theory: Exercise $1.8$
Category:Power Set
Category:Empty Set
Power Set of Finite Set is Finite

== Theorem ==
Let $S$ be a finite set.
Then the power set of $S$ is likewise finite.
== Proof ==
Let $S$ be a finite set.
Then by definition:
:$\exists n \in \N: \card S = n$
where $\card S$ denotes the cardinality of $S$.
From Cardinality of Power Set of Finite Set:
:$\card {\powerset S} = 2^n$
where $\powerset S$ denotes the power set of $S$.
As $n \in \N$ it follows that $2^n \in \N$ and so $\powerset S$ is also by definition a finite set.
Category:Power Set
Power Set of Natural Numbers has Cardinality of Continuum

== Theorem ==
Let $\N$ denote the set of natural numbers.
Let $\powerset \N$ denote the power set of $\N$.
Let $\card {\powerset \N}$ denote the cardinality of $\powerset \N$.
Let $\mathfrak c = \card \R$ denote the cardinality of the continuum.
Then:
:$\mathfrak c = \card {\powerset \N}$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $1$: General Background: $\S 5$ The continuum problem
Category:Power Set of Natural Numbers has Cardinality of Continuum
Category:Cardinality of Continuum
Category:Power Set
Category:Natural Numbers
Category:Infinite Sets
Power Set of Natural Numbers has Cardinality of Continuum/Proof 1

== Theorem ==
== Proof ==
=== Outline ===
$\powerset \N$ is demonstrated to have the same cardinality as the set of real numbers.
This is done by identifying a real number with its basis expansion in binary notation.
Such a basis expansion is a sequence of $0$s and $1$s.
Each $1$ of a real number $x$ expressed in such a way can be identified with the integer which identifies the power of $2$ representing the $1$ in that place.
Hence $x$ corresponds to the subset of $\N$ whose elements are the positions in $x$ which hold a $1$.
Thus $\R$ and $\powerset \N$ are equivalent.
From the Cantor-Dedekind Hypothesis, the real numbers are in one-to-one correspondence with the points on a line segment.
The points on a line segment form a continuum.
Category:Power Set of Natural Numbers has Cardinality of Continuum
Power Set of Natural Numbers has Cardinality of Continuum/Proof 2

== Theorem ==
== Proof ==
By Reals are Isomorphic to Dedekind Cuts there exists bijection:
:$f: \R \to \mathscr D$
where:
:$\mathscr D$ denotes the set of all Dedekind cuts of $\struct {\Q, \le}$.
Dedekind's cuts are subsets of $\Q$.
Therefore by definition of power set:
:$\mathscr D \subseteq \powerset \Q$
By Subset implies Cardinal Inequality:
:$\card {\mathscr D} \le \card {\powerset \Q}$
By Rational Numbers are Countably Infinite:
:$\Q$ is countably infinite.
Then by definition of countably infinite there exists a bijection:
:$g: \Q \to \N$
By definition of set equivalence:
:$\Q \sim \N$
Hence by definition of cardinality:
:$\card \Q = \card \N$
Then by Cardinality of Power Set is Invariant:
:$\card {\powerset \Q} = \card {\powerset \N}$
By definition of set equivalence:
:$\R \sim \mathscr D$
Hence by definition of cardinality:
:$\card \R = \card {\mathscr D}$
Thus:
:$\mathfrak c \le \card {\powerset \N}$
Define a mapping $h: \map {\operatorname {Fin} } \N \times \powerset \N \to \R^+$:
:$\ds \forall F \in \map {\operatorname {Fin} } \N, A \in \powerset \N: \map h {F, A} = \sum_{i \mathop \in F} 2^i + \sum_{i \mathop \in A} \paren {\frac 1 2}^i$
where $\map {\operatorname {Fin} } \N$ denotes the set of all finite subsets of $\N$.
A pair $\tuple {F, A}$ corresponds to binary denotation of a real number $\map h {F, A}$.
It means that $h$ is a surjection.
By Surjection iff Cardinal Inequality:
:$\card {\map {\operatorname {Fin} } \N \times \powerset \N} \le \card {\R^+}$
By definition of subset:
:$\map {\operatorname {Fin} } \N \subseteq \powerset \N$
Then by Subset implies Cardinal Inequality:
:$\card {\map {\operatorname {Fin} } \N} \le \card {\powerset \N}$
Because $\R^+ \subseteq \R$, we have by Subset implies Cardinal Inequality:
:$\card {\R^+} \le \card \R$
Thus:
:$\card {\powerset \N} \le \mathfrak c$
Hence the result:
:$\mathfrak c = \card {\powerset \N}$
== Sources ==
* 
Category:Power Set of Natural Numbers has Cardinality of Continuum
Power Set of Natural Numbers is Uncountable

== Theorem ==
The power set $\powerset \N$ of the natural numbers $\N$ is uncountable.
== Proof ==
There is no bijection from a set to its power set.
From Injection from Set to Power Set, we have that there exists an injection $f: \N \to \powerset \N$.
From the Cantor-Bernstein-Schröder Theorem, there can be no injection $g: \powerset \N \to \N$.
So, by definition, $\powerset \N$ is uncountable.
== Sources ==
* : Chapter $1$: Mappings: $\S 15$
Category:Power Set
Category:Natural Numbers
Category:Uncountable Sets
Power Set of Sample Space is Event Space

== Theorem ==
Let $\EE$ be an experiment whose sample space is $\Omega$.
Let $\powerset \Omega$ be the power set of $\Omega$.
Then $\powerset \Omega$ is an event space of $\EE$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $1$: Events and probabilities: $1.2$: Outcomes and events: Example $4$
Category:Event Spaces
Category:Power Set of Sample Space is Event Space
Power Set of Sample Space is Event Space/Proof 1

== Theorem ==
== Proof ==
Let $\powerset \Omega := \Sigma$.
;Event Space Axiom $(\text {ES} 1)$:
From Empty Set is Subset of All Sets we have that $\O \subseteq \Omega$.
By the definition of power set:
:$\O \in \Sigma$
thus fulfilling axiom $(\text {ES} 1)$.
;Event Space Axiom $(\text {ES} 2)$:
Let $A \in \Sigma$.
Then by the definition of power set:
:$A \subseteq \Omega$
From Set with Relative Complement forms Partition:
:$\Omega \setminus A \subseteq \Omega$
and so by the definition of power set:
:$\Omega \setminus A \in \Sigma$
thus fulfilling axiom $(\text {ES} 2)$.
;Event Space Axiom $(\text {ES} 3)$:
Let $\sequence {A_i}$ be a countably infinite sequence of sets in $\Sigma$.
Then from Power Set is Closed under Countable Unions:
:$\ds \bigcup_{i \mathop \in \N} A_i \in \Sigma$
thus fulfilling axiom $(\text {ES} 3)$.
All the event space axioms are seen to be fulfilled by $\powerset \Omega$.
Hence the result.
Category:Power Set of Sample Space is Event Space
Power Set of Sample Space is Event Space/Proof 2

== Theorem ==
== Proof ==
For $\powerset \Omega$ to be an event space of $\EE$, it needs to fulfil the following properties:
:$(1): \quad \powerset \Omega \ne \O$, that is, an event space can not be empty.
:$(2): \quad$ If $A \in \powerset \Omega$, then $\relcomp \Omega A \in \powerset \Omega$, that is, the complement of $A$ relative to $\Omega$, is also in $\powerset \Omega$.
:$(3): \quad$ If $A_1, A_2, \ldots \in \powerset \Omega$, then $\ds \bigcup_{i \mathop = 1}^\infty A_i \in \powerset \Omega$, that is, the union of any countable collection of elements of $\powerset \Omega$ is also in $\powerset \Omega$.
These all follow directly from Power Set is Sigma-Algebra.
Category:Power Set of Sample Space is Event Space
Power Set of Set of Finite Strings is Uncountable

== Theorem ==
Let $\Sigma$ be a finite alphabet.
Let $\Sigma^*$ be the set of finite strings of $\Sigma$.
Let $\powerset {\Sigma^*}$ be the power set of $\Sigma^*$
Then $\powerset {\Sigma^*}$ is an uncountable set.
== Proof ==
From Set of Finite Strings is Countably Infinite, $\Sigma^*$ is a countably infinite set.
The result follows from Power Set of Countably Infinite Set is Uncountable.
== Sources ==
* : Chapter $1$: Preliminaries: $1.4$ Set Notation: Infinite sets
Category:Sets of Finite Strings
Category:Uncountable Sets
Power Set of Singleton

== Theorem ==
Let $x$ be an object.
Then the power set of the singleton $\set x$ is:
:$\powerset {\set x} = \set {\O, \set x}$
== Proof ==
From Empty Set is Subset of All Sets:
:$\O \in \powerset {\set x}$
Let $A \in \powerset {\set x}$ such that $A \ne \O$
That is:
So a subset of $\set x$ is either $\O$ or $\set x$.
Category:Power Set
Category:Singletons
Power Set of Subset

== Theorem ==
Let $S \subseteq T$ where $S$ and $T$ are both sets.
Then:
:$\powerset S \subseteq \powerset T$
where $\powerset S$ denotes the power set of $S$.
== Proof ==
== Sources ==
* : $\S 5$: Complements and Powers
* : $\S 2$. Sets of sets
* : Chapter $1$: Theory of Sets: $\S 2$: Sets and Subsets: Exercise $1 \ \text{(g)}$
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 6$ The power axiom: Exercise $6.1. \ \text {(c)}$
Category:Power Set
Category:Subsets
Power Set of Transitive Set is Transitive

== Theorem ==
Let $x$ be a transitive set.
Then its power set $\powerset x$ is also a transitive set.
== Proof ==
Let $x$ be transitive.
By Set is Transitive iff Subset of Power Set:
:$x \subseteq \powerset x$
Then by Power Set of Subset:
:$\powerset x \subseteq \powerset {\powerset x}$
Thus by Set is Transitive iff Subset of Power Set:
:$\powerset x$ is a transitive set.
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 10$ Some useful facts about transitivity: Theorem $10.5$
Category:Power Set
Category:Transitive Classes