nyuuzyou/wikis huggingface dataset:
Set Union Preserves Subsets/Proof 1

== Theorem ==
== Proof ==
Let $A \subseteq B$ and $S \subseteq T$.
Then:
Now we invoke the Constructive Dilemma of propositional logic:
:$p \implies q, \ r \implies s \vdash p \lor r \implies q \lor s$
applying it as:
:$\paren {x \in A \implies x \in B, \ x \in S \implies x \in T} \implies \paren {x \in A \lor x \in S \implies x \in B \lor x \in T}$
The result follows directly from the definition of set union:
:$\paren {x \in A \implies x \in B, \ x \in S \implies x \in T} \implies \paren {x \in A \cup S \implies x \in B \cup T}$
and from the definition of subset:
:$A \subseteq B, \ S \subseteq T \implies A \cup S \subseteq B \cup T$
Category:Set Union Preserves Subsets
Set Union Preserves Subsets/Proof 2

== Theorem ==
== Proof ==
By Subset Relation is Transitive, $\subseteq$ is a transitive relation.
By the corollary to Set Union Preserves Subsets (Proof 2), $\subseteq$ is compatible with $\cup$.
Thus the theorem holds by Operating on Transitive Relationships Compatible with Operation.
Category:Set Union Preserves Subsets
Set Union can be Derived using Axiom of Abstraction

== Theorem ==
Let $a$ be a set of sets.
By application of the , the union $\bigcup a$ can be formed.
Hence the union $\bigcup a$ can be derived as a valid object in Frege set theory.
== Proof ==
Let $P$ be the property defined as:
:$\forall x: \map P x := \paren {\exists y: y \in a \land x \in y}$
where $\land$ is the conjunction operator.
That is, $\map P x$ :
:$x$ is an element of some set $y$, where $y$ is one of the sets that comprise the elements of $a$.
Hence, using the , we form the set:
:$\bigcup a := \set {x: \exists y: y \in a \land x \in y}$
== Sources ==
* : Chapter $1$: General Background: $\S 7$ Frege set theory
Category:Set Union
Category:Axiom of Abstraction
Set Union expressed as Intersection Complement

== Theorem ==
Let $A$ and $B$ be subsets of a universal set $\Bbb U$.
Let $\uparrow$ denote the operation on $A$ and $B$ defined as:
:$\paren {A \uparrow B} \iff \paren {\relcomp {\Bbb U} {A \cap B} }$
where $\relcomp {\Bbb U} A$ denotes the complement of $A$ in $\Bbb U$.
Then:
:$A \cup B = \paren {A \uparrow A} \uparrow \paren {B \uparrow B}$
== Proof ==
== Sources ==
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: Exercise $16$
Category:Set Union
Category:Set Intersection
Category:Set Complement
Set Union is Idempotent

== Theorem ==
Set union is idempotent:
:$S \cup S = S$
== Proof ==
== Also see ==
* Set Intersection is Idempotent
== Sources ==
* : $\S 4$: Unions and Intersections
* : Exercise $1.1: \ 8 \ \text{(a)}$
* : Introduction to Set Theory: $1$. Elementary Operations on Sets
* : Chapter $1$: Set Theory: $1.3$: Set operations
* : Chapter $\text{I}$: Sets and Functions: Exercise $\text{B i}$
* : $\text I$: Random Events: $1$. Relations among Random Events: Problem $1$
* : Chapter $1$: The Notation and Terminology of Set Theory: $\S 6 \ \text{(c)}$
* : $\S 1.6$: Set Identities and Other Set Relations: Exercise $2 \ \text{(d)}$
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: $\text{(a)}$
* : Chapter $1$: Sets and mappings: $\S 1.2$: Sets: Exercise $1 \ \text{(ii)}$
* : $\S 1$: Naive Set Theory: $\S 1.2$: Operations on Sets: Exercise $1.2.1 \ \text{(i)}$
* : $\S 2$
* : $\text {(v)}$
* : $\text {(i)}$
Category:Set Union
Category:Examples of Idempotence‎
Set Union is Self-Distributive

== Theorem ==
Set union is self-distributive:
:$\forall A, B, C: \paren {A \cup B} \cup \paren {A \cup C} = A \cup B \cup C = \paren {A \cup C} \cup \paren {B \cup C}$
where $A, B, C$ are sets.
=== Sets of Sets ===
=== Families of Sets ===
=== General Result ===
== Proof ==
We have:
* Union is Associative
* Union is Commutative
* Set Union is Idempotent
The result follows from Associative Commutative Idempotent Operation is Self-Distributive.
Category:Set Union
Category:Examples of Self-Distributive Operations
Category:Set Union is Self-Distributive
Set Union is Self-Distributive/Families of Sets

== 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 $\family {A_\alpha}_{\alpha \mathop \in I}$.
== Proof ==
Thus by definition of subset:
:$\ds \map {\bigcup_{\alpha \mathop \in I} } {A_\alpha \cup B_\alpha} \subseteq \paren {\bigcup_{\alpha \mathop \in I} A_\alpha} \cup \paren {\bigcup_{\alpha \mathop \in I} B_\alpha}$
Thus by definition of subset:
:$\ds \paren {\bigcup_{\alpha \mathop \in I} A_\alpha} \cup \paren {\bigcup_{\alpha \mathop \in I} B_\alpha} \subseteq \map {\bigcup_{\alpha \mathop \in I} } {A_\alpha \cup B_\alpha}$
By definition of set equality:
:$\ds \map {\bigcup_{\alpha \mathop \in I} } {A_\alpha \cup B_\alpha} = \paren {\bigcup_{\alpha \mathop \in I} A_\alpha} \cup \paren {\bigcup_{\alpha \mathop \in I} B_\alpha}$
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 4$: Indexed Families of Sets: Exercise $1 \ \text{(c)}$
* : $\S 1$: Naive Set Theory: $\S 1.4$: Sets of Sets: Exercise $1.4.4 \ \text{(iii)}$
Category:Set Union is Self-Distributive
Category:Indexed Families
Set Union is Self-Distributive/General Result

== Theorem ==
Let $\family {\mathbb S_i}_{i \mathop \in I}$ be an $I$-indexed family of sets of sets.
Then:
:$\ds \bigcup_{i \mathop \in I} \bigcup \mathbb S_i = \bigcup \bigcup_{i \mathop \in I} \mathbb S_i$
== 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 of Family, 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.
Category:Set Union is Self-Distributive
Set Union is Self-Distributive/Sets of Sets

== 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 $A$.
== 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:
:$s \in \set {x: \exists X \in B: x \in X}$
, 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:
:$\ds \bigcup \paren {A \cup B} = \paren {\bigcup A} \cup \paren {\bigcup B}$
Category:Set Union is Self-Distributive
Set Union is not Cancellable

== Theorem ==
Set union is not a cancellable operation.
That is, for a given $A, B, C \subseteq S$ for some $S$, it is not always the case that:
:$A \cup B = A \cup C \implies B = C$
== Proof ==
Proof by Counterexample:
Let $S = \set {a, b}$.
Let:
:$A = \set {a, b}$
:$B = \set a$
:$C = \set b$
Then:
but:
== Sources ==
* : $\S 2.1$: Sets: Exercise $\text{B} \ 2$
Category:Set Union
Set between Connected Set and Closure is Connected

== Theorem ==
Let $T$ be a topological space.
Let $H$ be a connected set of $T$.
Let $H \subseteq K \subseteq H^-$, where $H^-$ denotes the closure of $H$. 
Then $K$ is connected.
=== Corollary 1 ===
=== Corollary 2 ===
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness
Category:Set Closures
Category:Connected Sets
Category:Set between Connected Set and Closure is Connected
Set between Connected Set and Closure is Connected/Proof 1

== Theorem ==
== Proof ==
Let $D$ be the discrete space $\set {0, 1}$.
Let $f: K \to D$ be an arbitrary continuous mapping.
From Continuity of Composite with Inclusion, the restriction $f \restriction_H$ is continuous.
We have that:
:$H$ is connected
:$f \restriction_H$ is continuous
Thus by definition of connected set:
:$f \sqbrk H = \set 0$ or $f \sqbrk H = \set 1$ 
, let $\map f H = \set 0$.
 $\exists k \in K: \map f k = 1$.
By definition of discrete space, $\set 1$ is open in $D$.
Hence by definition of continuous mapping:
:$f^{-1} \sqbrk {\set 1}$ is open in $K$.
Let $K$ be given the subspace topology.
Then for some $U$ open in $T$:
:$f^{-1} \sqbrk {\set 1} = K \cap U$
We have that:
:$k \in f^{-1} \sqbrk {\set 1} \subseteq U$
and:
:$k \in H^-$
By definition of topology:
:$\exists x \in H \cap U$
As $x \in H$, we have that:
:$\map f x = 0$
But because $x \in H \cap U \subseteq K \cap U = f^{-1} \sqbrk {\set 1}$:
:$\map f x = 1$
This contradicts the definition of mapping.
Thus by Proof by Contradiction, $f: K \to D$ can not be a surjection.
Thus $K$ is connected.
== Sources ==
* : $6.2$: Connectedness: Proposition $6.2.18$
Category:Set between Connected Set and Closure is Connected
Set between Connected Set and Closure is Connected/Proof 2

== Theorem ==
== Proof ==
Let $T_K = \struct {K, \tau_K}$ be the topological subspace of $T$ whose underlying set is $K$.
Let $\map {\cl_K} H$ denote the closure of $H$ in $K$.
From Closure of Subset in Subspace:
:$\map {\cl_K} H = K \cap H^-$
By hypothesis:
:$K \subseteq H^-$
and so by Intersection with Subset is Subset‎:
:$\map {\cl_K} H = K$
Let $D$ be the discrete space $\set {0, 1}$.
Let $f: K \to D$ be any continuous mapping.
From Continuity of Composite with Inclusion, the restriction $f \restriction_H$ is continuous.
SWe have that:
:$H$ is connected
:$f \restriction_H$ is continuous
Thus by definition of connected set:
:$f \sqbrk H = \set 0$ or $f \sqbrk H = \set 1$ 
, let $f \sqbrk H = \set 0$.
From Continuity Defined by Closure:
:$f \sqbrk {\map {\cl_K} H} \subseteq \map {\cl_K} {f \sqbrk H} = \set 0^-$
where $\set 0^-$ denotes the closure of $\set 0$ in $D$.
As $D$ is the discrete space, it follows from Set in Discrete Topology is Clopen that $\set 0$ is closed in $D$.
Thus by Set is Closed iff Equals Topological Closure:
:$\set 0^- = \set 0$
That is, $f \sqbrk K = \set 0$.
Thus $K$ is connected by definition.
== Sources ==
* : $6.2$: Connectedness: Proposition $6.2.18$
Category:Set between Connected Set and Closure is Connected
Set equals Union of Power Set

== Theorem ==
Let $x$ be a set of sets.
Let $\powerset x$ denote the power set of $x$.
Let $\map \bigcup {\powerset x}$ denote the union of $\powerset x$.
Then:
:$x = \map \bigcup {\powerset x}$
== Proof ==
From Set is Element of its Power Set:
:$x \in \powerset x$
From Element of Class is Subset of Union of Class it follows that:
:$x \subseteq \map \bigcup {\powerset x}$
Let $z \in \map \bigcup {\powerset x}$
Then by definition of union|union:
:$\exists y \in \powerset x: z \in y$
By definition of $\powerset x$
:$\exists y \subseteq x: z \in y$
But by definition of subset, that means:
:$z \in x$
Thus, again by definition of subset:
:$\map \bigcup {\powerset x} \subseteq x$
We have that:
:$x \subseteq \map \bigcup {\powerset x}$
and:
:$\map \bigcup {\powerset x} \subseteq x$
Hence by definition of set equality:
:$x = \map \bigcup {\powerset x}$
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 6$ The power axiom: Exercise $6.1. \ \text {(b)}$
Category:Power Set
Category:Set Union
Set has Rank/Proof 1

== Theorem ==
== Proof ==
The proof shall proceed by Epsilon Induction on $S$.
Suppose that all the elements $a \in S$ have a rank.
That is, $a \in \map V x$ for some $x$.
Let:
:$\ds \map F a = \inf \set {x \in \On : a \in \map V x}$
be the rank of $a$.
Let:
:$\ds y = \sup \set {\map F a : a \in S}$
be the least level of the Von Neumann Hierarchy containing all elements of $S$.
Then, for any $a \in S$:
Therefore $S \in \map V z$ for some ordinal $z = y + 1$.
Thus by Epsilon Induction every set has a rank.
== Sources ==
* : $\S 9.13$
Category:Set has Rank
Set has Rank/Proof 2

== Theorem ==
== Proof ==
Let $G$ be the smallest transitive set containing $S$ as a subset.
By Set Contained in Smallest Transitive Set, $G$ must exist.
By Transitive Set Contained in Von Neumann Hierarchy Level, $G \subseteq V_i$ for some ordinal $i$.
Therefore:
:$G \in V_{i + 1}$
We have that Ordinals are Well-Ordered
From Proper Well-Ordering determines Smallest Elements, there exists a smallest ordinal $k$ such that $G \in V_{k + 1}$.
Hence, by definition, $G$ has rank $k$.
Category:Set has Rank
Set in Discrete Topology is Clopen

== Theorem ==
Let $T = \struct {S, \tau}$ be a discrete topological space.
:$\forall U \subseteq S: U$ is both closed and open in $\struct {S, \tau}$. 
== Proof ==
Let $U \subseteq S$.
By definition of discrete topological space, $U \in \tau$.
By definition of closed set, $\relcomp S U$ is closed in $T$, where $\relcomp S U$ is the relative complement of $U$ in $S$. 
But from Set Difference is Subset:
:$\relcomp S U = S \setminus U \subseteq S$
and so:
:$\relcomp S U \in \tau$
That is, $\relcomp S U$ is both closed and open in $T$. 
Then by Relative Complement of Relative Complement:
:$\relcomp S {\relcomp S U} = U$
which is seen to be both closed and open in $T$.
== Sources ==
* : Chapter $\text {I}$: Topological Spaces: $1$. Open Sets and Closed Sets
* : Chapter $3$: Topological Spaces: $\S 2$: Topological Spaces: Exercise $6$
* : Part $\text {II}$: Counterexamples: $1 \text { - } 3$: Discrete Topology
Category:Discrete Topology
Category:Clopen Sets
Set in Standard Discrete Metric Space has no Limit Points

== Theorem ==
Let $M = \struct {S, d}$ be the standard discrete metric space on a set $A$.
Let $H \subseteq S$ be a subset of $S$.
Then $H$ has no limit points.
== Proof ==
By definition of the standard discrete metric:
:$\map d {x, y} = \begin {cases} 0 & : x = y \\ 1 & : x \ne y \end {cases}$
Let $\alpha \in S$.
By definition, $\alpha$ is a limit point of $H$ every deleted $\epsilon$-neighborhood $\map {B_\epsilon} \alpha \setminus \set \alpha$ of $\alpha$ contains a point in $A$:
:$\forall \epsilon \in \R_{>0}: \paren {\map {B_\epsilon} \alpha \setminus \set \alpha} \cap H \ne \O$
Let $\map {B_1} \alpha$ be the open $1$-ball of $\alpha$ in $M$.
Thus:
So there exists a deleted $\epsilon$-neighborhood $\map {B_\epsilon} \alpha \setminus \set \alpha$ of $\alpha$ containing no points of $H$.
Hence $\alpha$ is not a limit point of $H$
As $\alpha$ is arbitrary, the result follows.
Category:Standard Discrete Metric
Category:Examples of Limit Points
Set is Clopen iff Boundary is Empty

== Theorem ==
Let $T$ be a topological space, and let $H \subseteq T$.
Then $H$ is both closed and open in $T$ :
:$\partial H = \O$
where $\partial H$ is the boundary of $H$.
== Proof ==
From Set is Open iff Disjoint from Boundary we have that:
:$H$ is open in $T$ $\partial H \cap H = \O$
From Set is Closed iff it Contains its Boundary we have that:
:$H$ is closed in $T$ $\partial H \subseteq H$
From Intersection with Subset is Subset‎:
:$\partial H \subseteq H \iff \partial H \cap H = \partial H $
From the above it follows that $\partial H = \O$ $H$ is closed and $H$ is open.
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 30 \ \text {(ii)}$
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Closures and Interiors
Category:Clopen Sets
Category:Set Boundaries
Set is Closed iff Equals Topological Closure

== Theorem ==
Let $T$ be a topological space.
Let $H \subseteq T$.
Then $H$ is closed in $T$ :
:$H = \map \cl H$
That is, a closed set equals its closure (which makes semantic sense).
== Proof 1 ==
== Proof 2 ==
Category:Closed Sets
Category:Set Closures
Category:Set is Closed iff Equals Topological Closure
Set is Closed iff Equals Topological Closure/Proof 1

== Theorem ==
== Proof ==
Let $H'$ denote the derived set of $H$.
By Closed Set iff Contains all its Limit Points, $H$ is closed in $T$ $H' \subseteq H$.
By Union with Superset is Superset, $H' \subseteq H$ $H = H \cup H'$.
The result follows from the definition of closure.
== Sources ==
* : $2.27 b$
Category:Set is Closed iff Equals Topological Closure
Set is Closed iff Equals Topological Closure/Proof 2

== Theorem ==
== Proof ==
Let $H^\complement$ denote the relative complement of $H$ in $T$.
By definition, we have that $H$ is closed in $T$ $H^\complement$ is open in $T$.
By Set is Open iff Neighborhood of all its Points, this is equivalent to:
:$\forall x \in H^\complement: \exists U \in \tau: x \in U \subseteq H^\complement$
By Empty Intersection iff Subset of Complement, we have that:
:$U \subseteq H^\complement \iff U \cap H = \O$
By Condition for Point being in Closure, it follows that $H^{\complement}$ is open in $T$ :
:$\forall x \in H^\complement: x \notin \map \cl H$
By the Rule of Transposition, this is equivalent to $\map \cl H \subseteq H$.
From Set is Subset of its Topological Closure, we have that $H \subseteq \map \cl H$.
By definition of set equality, this is equivalent to:
:$H = \map \cl H$
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Proposition $3.7.15 \ \text{(a)}$
Category:Set is Closed iff Equals Topological Closure
Set is Closed iff it Contains its Boundary

== Theorem ==
Let $T$ be a topological space, and let $H \subseteq T$.
Then $H$ is closed in $T$ :
:$\partial H \subseteq H$
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:
:$\partial H \subseteq H^-$
Then from Closed Set Equals its Closure, $H$ is closed in $T$ $H = H^-$.
Hence the result.
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 30 \ \text {(i)}$
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Closures and Interiors
Category:Closed Sets
Category:Set Boundaries
Set is Closed in Metric Space iff Closed in Induced Topological Space

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $\tau$ be the topology induced by the metric $d$.
Let $F$ be a subset of $M$.
Then:
:$F$ is closed in $M$ $F$ is closed in $\struct {A, \tau}$
== Proof ==
By definition of a closed set in $M$:
:$F$ is closed set in $M$ $A \setminus F$ is open in $M$
By definition of the topology $\tau$ induced by the metric $d$:
:$A \setminus F$ is open in $M$ $A \setminus F$ is open in $\struct {A, \tau}$
By definition of a closed set in $\struct{A, \tau}$:
:$A \setminus F$ is open in $\struct {A, \tau}$ $F$ is closed set in $\struct {A, \tau}$
The result follows.
Category:Closed Sets
Category:Metric Spaces
Set is Coarser than Image of Mapping of Infima

== Theorem ==
Let $\struct {S, \wedge, \preceq}$ be a meet semilattice.
Let $f, g:\N \to S$ be mappings such that:
:$\forall n \in \N: \map g n = \inf \set {\map f m: m \in \N \land m \le n}$
Then $f \sqbrk \N$ is coarser than $g \sqbrk \N$
where $f \sqbrk \N$ denotes the image of mapping $f$.
== Proof ==
Let $x \in f \sqbrk \N$.
By definition of image of mapping:
:$\exists n \in \N: x = \map f n$
By definition of $g$:
:$\map g n = \inf \set {\map f m: m \in \N \land m \le n}$
By definition of reflexivity:
:$n \le n$
Then
:$\map f n \in \set {\map f m: m \in \N \land m \le n}$
By definitions of infimum and lower bound:
:$\map g n \preceq x$
By definition of image of mapping:
:$\map g n \in g \sqbrk \N$
Thus
:$\exists y \in g \sqbrk \N: y \preceq x$
== Sources ==
* 
Category:Join and Meet Semilattices
Set is Countable if Cardinality equals Cardinality of Countable Set

== Theorem ==
Let $X, Y$ be sets.
Let:
:$\card X = \card Y$
where $\card X$ denotes the cardinality of $X$.
If $X$ is countable then $Y$ is countable.
== Proof ==
Assume that $X$ is countable.
By definition of countable set there exists an injection:L
:$f: X \to \N$
By definition of cardinality the sets $Y$ and $X$ are equivalent:
:$Y \sim X$
Then by definition of set equivalence there exists a bijection:
:$g: Y \to X$
By definition of bijection:
:$g$ is an injection.
Hence by Composite of Injections is Injection:
:$f \circ g: Y \to \N$ is an injection.
Thus by definition:
:$Y$ is countable.
Category:Countable Sets
Set is Countable iff Cardinality not greater Aleph Zero

== Theorem ==
Let $X$ be a set.
Then:
:$X$ is countable $\size X \le \aleph_0$
where
:$\size X$ denotes the cardinality of $X$,
:$\aleph$ denotes the aleph mapping.
== Proof ==
:$X$ is countable
:there exists an injection $X \to \N$ by definition of countable set
:$\size X \le \size \N$ by Injection iff Cardinal Inequality
:$\size X \le \aleph_0$ by Aleph Zero equals Cardinality of Naturals.
Category:Countable Sets
Set is Element of Successor

== Theorem ==
Let $x$ be a set.
Let $x^+$ be the successor of $x$.
Then $x \in x^+$.
== Proof ==
By the definition of successor set:
: $x^+ = x \cup \{x\}$.
By the definition of singleton, $x \in \{x\}$.
Thus by the definition of union, $x \in x^+$.
Category:Ordinals
Set is Element of its Power Set

== Theorem ==
A set is an element of its power set:
:$S \in \powerset S$
== Proof ==
== Also see ==
* Empty Set is Element of Power Set
== Sources ==
* : Introduction to Set Theory: $1$. Elementary Operations on Sets
* : $\S 1.8$. Sets of sets: Example $25$
* : Chapter $1$: Theory of Sets: $\S 2$: Sets and Subsets: Exercise $1 \ \text{(a)}$
Category:Power Set
Set is Equivalent to Itself

== Theorem ==
Let $S$ be a set.
Then:
:$S \sim S$
where $\sim$ denotes set equivalence.
== Proof ==
From Identity Mapping is Bijection, the identity mapping $I_S: S \to S$ is a bijection from $S$ to $S$.
Thus there exists a bijection from $S$ to itself
Hence by definition $S$ is therefore equivalent to itself.
== Sources ==
* : Introduction to Set Theory: $2$. Set Theoretical Equivalence and Denumerability
* : $\S 3.7$. Similar sets
* : Chapter $\text {III}$: The Natural Numbers: $\S 17$: Finite Sets: Theorem $17.1$
* : Part $1$: Not Entirely Naive Set Theory: Chapter $3$: Cardinality: Exercise $1 \ \text{(a)}$
* : $2$. Definition of Equivalence. The Concept of Cardinality. The Axiom of Choice: Theorem $2.1$
Category:Set Equivalence behaves like Equivalence Relation