nyuuzyou/wikis huggingface dataset:
Subset Products of Normal Subgroup with Normal Subgroup of Subgroup

== Theorem ==
Let $G$ be a group.
Let:
:$(1): \quad H$ be a subgroup of $G$
:$(2): \quad K$ be a normal subgroup of $H$
:$(3): \quad N$ be a normal subgroup of $G$
Then:
:$N K \lhd N H$
where:
: $N K$ and $N H$ denote subset product
: $\lhd$ denotes the relation of being a normal subgroup.
== Proof ==
Consider arbitrary $x_n \in N, x_h \in H$.
Thus:
:$x_n x_h \in N H$
We aim to show that:
:$x_n x_h N K \paren {x_n x_h}^{-1} \subseteq N K$
thus demonstrating $N K \lhd N H$ by the Normal Subgroup Test.
We have:
== Sources ==
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 46 \eta$
Category:Normal Subgroups
Category:Subset Products
Subset Relation is Antisymmetric

== Theorem ==
The subset relation is antisymmetric:
:$\paren {x \subseteq y} \land \paren {y \subseteq x} \iff x = y$
where $x$ and $y$ are sets.
== Proof ==
This is a direct statement of the definition of set equality:
:$x = y := \paren {x \subseteq y} \land \paren {y \subseteq x}$
== Sources ==
* : Introduction to Set Theory: $1$. Elementary Operations on Sets
* : $\S 1$: The Axiom of Extension
* 
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 1$ Introduction to well ordering
* 
Category:Subset Relation
Category:Examples of Antisymmetric Relations
Subset Relation is Compatible with Subset Product

== Theorem ==
Let $\struct {S, \circ}$ be a magma.
Let $\struct {\powerset S, \circ_\PP}$ be the power structure of $\struct {S, \circ}$.
Then the subset relation on $S$ is compatible with $\circ_\PP$.
That is:
=== Corollary 1 ===
=== Corollary 2 ===
== Proof ==
Let $x \in X, z \in Z$.
Then:
:$x \circ z \in X \circ Z$ and $z \circ x \in Z \circ X$
Now:
:$Y \circ Z = \set {y \circ z: y \in Y, z \in Z}$
:$Z \circ Y = \set {z \circ y: y \in Y, z \in Z}$
But by the definition of a subset:
:$x \in X \implies x \in Y$
Thus:
:$x \circ z \in Y \circ Z$ and $z \circ x \in Z \circ Y$
and the result follows.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 9$: Compositions Induced on the Set of All Subsets
* : Chapter $\text{II}$: Groups: Exercise $\text{G}$
Category:Power Structures
Category:Subset Products
Category:Compatible Relations
Subset Relation is Compatible with Subset Product/Corollary 1

== Theorem ==
Let $\struct {S, \circ}$ be a magma.
Let $\struct {\powerset S, \circ_\PP}$ be the power structure of $\struct {S, \circ}$.
Let $A, B, C, D \in \powerset S$.
Let $A \subseteq B$ and $C \subseteq D$.
Then:
:$A \circ_\PP C \subseteq B \circ_\PP D$
== Proof ==
By Subset Relation is Compatible with Subset Product, $\subseteq$ is compatible with $\circ_\PP$.
By Subset Relation is Transitive, $\subseteq$ is transitive.
Thus the theorem holds by Operating on Transitive Relationships Compatible with Operation.
Category:Subset Products
Category:Compatible Relations
Subset Relation is Compatible with Subset Product/Corollary 2

== Theorem ==
Let $\struct {S, \circ}$ be a magma.
Let $A, B \subseteq S$.
Let $A \subseteq B$.
Then:
== Proof ==
This follows from Subset Relation is Compatible with Subset Product and the definition of the subset product with a singleton.
Category:Subset Products
Subset Relation is Ordering

== Theorem ==
Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Let $\mathbb S \subseteq \powerset S$ be any subset of $\powerset S$, that is, an arbitrary set of subsets of $S$.
Then $\subseteq$ is an ordering on $\mathbb S$.
In other words, let $\struct {\mathbb S, \subseteq}$ be the relational structure defined on $\mathbb S$ by the subset relation $\subseteq$.
Then $\struct {\mathbb S, \subseteq}$ is an ordered set.
=== General Result ===
=== Class Theoretical Result ===
== Proof ==
To establish that $\subseteq$ is an ordering, we need to show that it is reflexive, antisymmetric and transitive.
So, checking in turn each of the criteria for an ordering:
=== Reflexivity ===
From Subset Relation is Reflexive:
:$\forall T \in \mathbb S: T \subseteq T$
So $\subseteq$ is reflexive.
=== Antisymmetry ===
From Subset Relation is Antisymmetric:
:$\forall S_1, S_2 \in \mathbb S: S_1 \subseteq S_2 \land S_2 \subseteq S_1 \iff S_1 = S_2$
So $\subseteq$ is antisymmetric.
=== Transitivity ===
From Subset Relation is Transitive:
:$\forall S_1, S_2, S_3 \in \mathbb S: S_1 \subseteq S_2 \land S_2 \subseteq S_3 \implies S_1 \subseteq S_3$
That is, $\subseteq$ is transitive.
So we have shown that $\subseteq$ is an ordering on $\mathbb S$.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $2 \ \text {(b)}$
Category:Subset Relation
Category:Orderings
Category:Subset Relation is Ordering
Subset Relation is Ordering/Class Theory

== Theorem ==
Let $C$ be a class.
Then the subset relation $\subseteq$ is an ordering on $C$.
== Proof ==
To establish that $\subseteq$ is an ordering, we need to show that it is reflexive, antisymmetric and transitive.
So, checking in turn each of the criteria for an ordering:
=== Reflexivity ===
From Subset Relation is Reflexive:
:$\forall x \in C: x \subseteq x$
So $\subseteq$ is reflexive.
=== Antisymmetry ===
From Subset Relation is Antisymmetric:
:$\forall x, y \in C: x \subseteq y \land y \subseteq x \iff x = y$
So $\subseteq$ is antisymmetric.
=== Transitivity ===
From Subset Relation is Transitive:
:$\forall x, y, z \in C: x \subseteq y \land y \subseteq z \implies x \subseteq z$
That is, $\subseteq$ is transitive.
So we have shown that $\subseteq$ is an ordering on $C$.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 1$ Introduction to well ordering
Category:Subset Relation is Ordering
Subset Relation is Ordering/General Result

== Theorem ==
Let $\mathbb S$ be a set of sets.
Then $\subseteq$ is an ordering on $\mathbb S$.
In other words, let $\struct {\mathbb S, \subseteq}$ be the relational structure defined on $\mathbb S$ by the subset relation $\subseteq$.
Then $\struct {\mathbb S, \subseteq}$ is an ordered set.
== Proof ==
To establish that $\subseteq$ is an ordering, we need to show that it is reflexive, antisymmetric and transitive.
So, checking in turn each of the criteria for an ordering:
=== Reflexivity ===
From Subset Relation is Reflexive:
:$\forall T \in \mathbb S: T \subseteq T$
So $\subseteq$ is reflexive.
=== Antisymmetry ===
From Subset Relation is Antisymmetric:
:$\forall S_1, S_2 \in \mathbb S: S_1 \subseteq S_2 \land S_2 \subseteq S_1 \iff S_1 = S_2$
So $\subseteq$ is antisymmetric.
=== Transitivity ===
From Subset Relation is Transitive:
:$\forall S_1, S_2, S_3 \in \mathbb S: S_1 \subseteq S_2 \land S_2 \subseteq S_3 \implies S_1 \subseteq S_3$
That is, $\subseteq$ is transitive.
So we have shown that $\subseteq$ is an ordering on $\mathbb S$.
== Sources ==
* : $\S 1$: Sets and Set Inclusion
* : $\S \text I.1$
* : $\S 1.5$: Relations
Category:Subset Relation is Ordering
Subset Relation is Reflexive

== Theorem ==
Let $C$ be a class.
The subset relation $\subseteq$ on $C$ is a reflexive relation on $C$.
== Proof ==
So $\subseteq$ is reflexive.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 1$ Introduction to well ordering
Category:Subset Relation
Category:Examples of Reflexive Relations
Subset Relation is Transitive

== Theorem ==
The subset relation is transitive:
:$\paren {R \subseteq S} \land \paren {S \subseteq T} \implies R \subseteq T$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : First Article: $\S 1$: The Conception of Power or Cardinal Number
* : Chapter $0$: Subsets and Complements; Union and Intersection
* : $\S 1$: The Axiom of Extension
* : $\S 1$: Sets and Set Inclusion
* : Introduction to Set Theory: $1$. Elementary Operations on Sets
* : Chapter $\text I$: Algebraic Structures: $\S 1$: The Language of Set Theory
* : Chapter $\text{I}$: Sets and Functions: Exercise $\text{B iii}$
* : Chapter $1$: Theory of Sets: $\S 2$: Sets and Subsets: Exercise $2$
* : Appendix $\text{A}.1$: Sets and Subsets: Problem Set $\text{A}.1$: $5$
* : $\S 6.2$: Subsets
* : Appendix $\text{A}.2$: Theorem $\text{A}.6$
* 
Category:Subset Relation
Category:Examples of Transitive Relations
Category:Subset Relation is Transitive
Subset Relation is Transitive/Proof 1

== Theorem ==
== Proof ==
== Sources ==
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: Theorem $1.1 \ \text{(b)}$
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 1$ Introduction to well ordering
Category:Subset Relation is Transitive
Subset Relation is Transitive/Proof 2

== Theorem ==
== Proof ==
Let $V$ be a basic universe.
By definition of basic universe, $R$, $S$ and $T$ are all elements of $V$.
By the , $R$, $S$ and $T$ are all classes.
We are given that $R \subseteq S$ and $S \subseteq T$.
Hence by Subclass of Subclass is Subclass, $R$ is a subclass of $T$.
By Subclass of Set is Set, it follows that $R$ is a subset of $T$.
Hence the result.
Category:Subset Relation is Transitive
Subset Relation on Power Set is Partial Ordering

== Theorem ==
Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Let $\struct {\powerset S, \subseteq}$ be the relational structure defined on $\powerset S$ by the subset relation $\subseteq$.
Then $\struct {\powerset S, \subseteq}$ is an ordered set.
The ordering $\subseteq$ is partial $S$ is neither empty nor a singleton; otherwise it is total.
== Proof ==
From Subset Relation is Ordering, we have that $\subseteq$ is an ordering on any set of subsets of a given set.
Suppose $S$ is neither a singleton nor the empty set.
Then $\exists a, b \in S$ such that $a \ne b$.
Then $\set a \in \powerset S$ and $\set b \in \powerset S$.
However, $\set a \nsubseteq \set b$ and $\set b \nsubseteq \set a$.
So by definition, $\subseteq$ is a partial ordering.
Now suppose $S = \O$.
Then $\powerset S = \set \O$ and, by Empty Set is Subset of All Sets, $\O \subseteq \O$.
Hence, trivially, $\subseteq$ is a total ordering on $\powerset S$.
Now suppose $S$ is a singleton: let $S = \set a$.
Then $\powerset S = \set {\O, \set a}$.
So there are only two elements of $\powerset S$, and we see that $\O \subseteq \set a$ from Empty Set is Subset of All Sets.
So, trivially again, $\subseteq$ is a total ordering on $\powerset S$.
== Sources ==
* : $\S 14$: Order
* : $\S 1.2$.Subsets: Example $11$
* : $\S 3.1$: Partially ordered sets: Example $3$
* : $\S 7$: Example $7.1$
* : Chapter $1$: Sets and mappings: $\S 1.5$: Ordered Sets: $\text{(i)}$
* : $\S 1$: Naive Set Theory: $\S 1.5$: Relations
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $2 \ \text {(a)}$
Category:Partial Orderings
Category:Power Set
Category:Subset Relation
Subset and Image Admit Infima and Mapping is Increasing implies Infimum of Image Succeeds Mapping at Infimum

== Theorem ==
Let $\struct {S, \preceq}$ and $\struct {T, \precsim}$ be ordered sets.
Let $f: S \to T$ be a increasing mapping.
Let $D \subseteq S$ such that
:$D$ admits a infimum in $S$ and $f \sqbrk D$ admits a infimum in $T$.
Then $\map f {\inf D} \precsim \map \inf {f \sqbrk D}$
== Proof ==
By definition of infimum:
:$\inf D$ is lower bound for $D$.
By Increasing Mapping Preserves Lower Bounds:
:$\map f {\inf D}$ is a lower bound for $f \sqbrk D$.
Thus by definition of infimum:
:$\map f {\inf D} \precsim \map \inf {f \sqbrk D}$
== Sources ==
* 
Category:Order Theory
Subset and Image Admit Suprema and Mapping is Increasing implies Supremum of Image Precedes Mapping at Supremum

== Theorem ==
Let $\struct {S, \preceq}$, $\struct {T, \precsim}$ be ordered sets.
Let $f: S \to T$ be a increasing mapping.
Let $D \subseteq S$ such that
:$D$ admits a supremum in $S$ and $f \sqbrk D$ admits a supremum in $T$.
Then:
:$\map \sup {f \sqbrk D} \precsim \map f {\sup D}$
== Proof ==
By definition of supremum:
:$\sup D$ is upper bound for $D$.
By Increasing Mapping Preserves Upper Bounds:
:$\map f {\sup D}$ is upper bound for $f \sqbrk D$.
Thus by definition of supremum:
:$\map \sup {f \sqbrk D} \precsim \map f {\sup D}$
== Sources ==
* 
Category:Order Theory
Subset equals Image of Preimage iff Mapping is Surjection

== Theorem ==
Let $f: S \to T$ be a mapping.
Let $f^\to: \powerset S \to \powerset T$ be the direct image mapping of $f$.
Similarly, let $f^\gets: \powerset T \to \powerset S$ be the inverse image mapping of $f$.
Then:
:$\forall B \in \powerset T: B = \map {\paren {f^\to \circ f^\gets} } B$
 $f$ is a surjection.
== Proof ==
=== Sufficient Condition ===
Let $f$ be such that:
:$\forall B \in \powerset T: B = \map {\paren {f^\to \circ f^\gets} } B$
From Subset equals Image of Preimage implies Surjection, $f$ is a surjection.
=== Necessary Condition ===
Let $f$ be a surjection.
Then by Image of Preimage of Subset under Surjection equals Subset:
:$B = \map {f^\to} {\map {f^\gets} B}$
== Sources ==
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.8$
* : Appendix $\text{A}.5$: Identity, One-one, and Onto Functions: Proposition $\text{A}.5.1: 2 \ \text{(d)}$
Category:Surjections
Subset equals Image of Preimage implies Surjection

== Theorem ==
Let $f: S \to T$ be a mapping.
Let:
:$\forall B \subseteq T: B = \paren {f \circ f^{-1} } \sqbrk B$
where $f \sqbrk B$ denotes the image of $B$ under $f$.
Then $f$ is a surjection.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Image of Preimage of Subset under Surjection equals Subset
* Subset equals Image of Preimage iff Mapping is Surjection
Category:Surjections
Category:Subset equals Image of Preimage implies Surjection
Subset equals Image of Preimage implies Surjection/Proof 1

== Theorem ==
== Proof ==
Let $f$ be such that:
:$\forall B \subseteq T: B = \paren {f \circ f^{-1} } \sqbrk B$
In particular, it holds for $T$ itself.
Hence:
So:
:$T \subseteq \Img f \subseteq T$
and so by definition of set equality:
:$\Img f = T$
So, by definition, $f$ is a surjection.
== Sources ==
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.8$
Category:Subset equals Image of Preimage implies Surjection
Subset equals Image of Preimage implies Surjection/Proof 2

== Theorem ==
== Proof ==
Suppose $f$ is not a surjection.
$T$ must have at least two elements for this to be the case.
Let one of these two elements not be the image of any element of $S$.
That is, let $b_1, b_2 \in T$ such that:
:$\exists a \in S: f \paren a = b_1$
:$\nexists x \in S: f \paren x = b_2$
Let $B = \set {b_1, b_2}$.
Then:
So by the Rule of Transposition:
:$\forall B \subseteq S: B = \paren {f \circ f^{-1} } \sqbrk B$
implies that $f$ is an surjection.
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 6$: Functions: Exercise $2 \ \text{(b})$
Category:Subset equals Image of Preimage implies Surjection
Subset equals Preimage of Image iff Mapping is Injection

== Theorem ==
Let $f: S \to T$ be a mapping.
Let $f^{-1}$ denote the inverse of $f$.
Then:
:$\forall A \subseteq S: A = \paren {f^{-1} \circ f} \sqbrk A$ $f$ is an injection
where:
:$f \sqbrk A$ denotes the image of $A$ under $f$
:$f^{-1} \circ f$ denotes the composition of $f^{-1}$ and $f$.
This can be expressed in the language and notation of direct image mappings and inverse image mappings as:
:$\forall A \in \powerset S: A = \map {\paren {f^\gets \circ f^\to} } A$ $f$ is an injection
== Proof ==
=== Sufficient Condition ===
Let $g$ be such that:
:$\forall A \subseteq S: A = \paren {f^{-1} \circ f} \sqbrk A$
Then by Subset equals Preimage of Image implies Injection, $f$ is an injection.
=== Necessary Condition ===
Let $f$ be an injection.
Then by Preimage of Image of Subset under Injection equals Subset:
:$\forall A \subseteq S: A = \paren {f^{-1} \circ f} \sqbrk A$
== Sources ==
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.8$
* : Appendix $\text{A}.5$: Identity, One-one, and Onto Functions: Proposition $\text{A}.5.1: 1 \ \text{(c)}$
Category:Injections
Category:Subsets
Category:Composite Mappings
Category:Preimages under Mappings
Subset equals Preimage of Image implies Injection

== Theorem ==
Let $f: S \to T$ be a mapping.
Let $f^\to: \powerset S \to \powerset T$ be the direct image mapping of $f$.
Similarly, let $f^\gets: \powerset T \to \powerset S$ be the inverse image mapping of $f$.
Let:
:$\forall A \in \powerset S: A = \map {\paren {f^\gets \circ f^\to} } A$
Then $f$ is an injection.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Preimage of Image of Subset under Injection equals Subset
* Subset equals Preimage of Image iff Mapping is Injection
Category:Injections
Category:Subsets
Category:Subset equals Preimage of Image implies Injection
Subset equals Preimage of Image implies Injection/Proof 1

== Theorem ==
== Proof ==
Let $f$ be such that:
:$\forall A \in \powerset S: A = \map {\paren {f^\gets \circ f^\to} } A$
In particular, it holds for all subsets of $A$ which are singletons.
Now, consider any $x, y \in A$.
We have:
So $f$ is an injection.
== Sources ==
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.8$
Category:Subset equals Preimage of Image implies Injection
Subset equals Preimage of Image implies Injection/Proof 2

== Theorem ==
== Proof ==
Suppose that $f$ is not an injection.
Then two elements of $S$ map to the same one element of $T$.
That is:
:$\exists a_1, a_2 \in S, b \in T: \map f {a_1} = \map f {a_2} = b$
Let $A = \set {a_1}$.
Then:
So by the Rule of Transposition:
:$\forall A \in \powerset S: A = \map {\paren {f^\gets \circ f^\to} } A$
implies that $f$ is an injection.
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 6$: Functions: Exercise $2 \ \text{(a})$
Category:Subset equals Preimage of Image implies Injection
Subset has 2 Conjugates then Normal Subgroup

== Theorem ==
Let $G$ be a group.
Let $S$ be a subset of $G$.
Let $S$ have exactly two conjugates in $G$.
Then $G$ has a proper non-trivial normal subgroup.
== Proof ==
Consider the centralizer $\map {C_G} S$ of $S$ in $G$.
From Centralizer of Group Subset is Subgroup, $\map {C_G} S$ is a subgroup of $G$.
If $\map {C_G} S = G$, then $S$ has no conjugate but itself.
So, in order for $S$ to have exactly two conjugates in $G$, it is necessary for $\map {C_G} S$ to be a proper subgroup.
Let $e$ be the identity of $G$.
If $\map {C_G} S = \set e$, then for there to be exactly two conjugates of $S$: 
:$\forall a \ne b \in G \setminus \set e: b x b^{-1} = a x a^{-1}$
But:
This implies either that $\map {C_G} S$ is actually nontrivial, or that $a^{-1}b = e \iff a = b$, a contradiction.
Thus $\map {C_G} S$ is a nontrivial proper subgroup of $G$.
We have that there are exactly $2$ conjugacy classes of $S$.
These are in one-to-one correspondence with cosets of $S$.
Thus the index $\index G {\map {C_G} S}$ of the centralizer is:
:$\index G {\map {C_G} S} = 2$
From Subgroup of Index 2 is Normal:
:$\map {C_G} S$ is a proper nontrivial normal subgroup of $G$.
== Sources ==
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 49 \alpha$
Category:Conjugacy
Category:Normal Subgroups
Subset implies Cardinal Inequality

== Theorem ==
Let $S$ and $T$ be sets such that $S \subseteq T$.
Furthermore, let:
:$T \sim \card T$
where $\card T$ denotes the cardinality of $T$.
Then:
:$\card S \le \card T$
== Proof ==
For the proof:
:the ordering relation $\le$ for ordinals
and
:the subset relation $\subseteq$
shall be used interchangeably.
Let $f: T \to \card T$ be a bijection.
It follows that $f \restriction_S : S \to \card T$ is an injection.
The image of $S$ under $f$ is a subset of $\card T$ and thus is a subset of an ordinal.
By Unique Isomorphism between Ordinal Subset and Unique Ordinal, there is a unique mapping $\phi$ and a unique ordinal $x$ such that $\phi: x \to f \sqbrk S$ is an order isomorphism.
It follows that $S \sim x$ by the definition of order isomorphism.
Furthermore, $\phi$ is a strictly increasing mapping from ordinals to ordinals.
Therefore, $y \in x \implies y \in \card T$ and $x \le \card T$ by the definition of subset.
But $\card S \le x$ by Cardinal Number Less than Ordinal.
So $\card S \le \card T$ by the fact that Subset Relation is Transitive.
== Also see ==
* Set Equivalent to Cardinal, which requires the axiom of choice.
== Sources ==
* : $\S 10.22$
* 
Category:Cardinals
Subset in Neighborhood Space is Neighborhood iff it contains Open Set

== Theorem ==
Let $\struct {S, \NN}$ be a neighborhood space.
Let $x \in S$ be a point of $S$.
Let $N \subseteq S$ be a subset of $S$.
Then $N$ is a neighborhood of $x$ there exists an open set $U$ of $\struct {S, \NN}$ such that $x \in U \subseteq N$.
== Proof ==
=== Necessary Condition ===
Let $N$ be a neighborhood of $x$.
Then by neighborhood space axiom $\text N 5$, $N$ contains a neighborhood $U$ of $x$ such that $U$ is a neighborhood of each of its points.
By neighborhood space axiom $\text N 2$, $x \in U$.
=== Sufficient Condition ===
Let $N$ be such that:
:$\exists U \in \NN: x \in U \subseteq N$
By definition of open set, $U$ is a neighborhood of $x$.
By neighborhood space axiom $\text N 3$, $N$ is then a neighborhood of $x$.
== Sources ==
* : Chapter $3$: Topological Spaces: $\S 3$: Neighborhoods and Neighborhood Spaces: Lemma $3.7$
Category:Neighborhood Spaces
Subset in Subsets

== Theorem ==
Let $S, B$ be sets.
Let $A$ be subset of $S$.
Then:
:$A \subseteq B \iff \forall x \in S: x \in A \implies x \in B$
== Proof ==
=== Sufficient Condition ===
Follows by definition of subset.
=== Necessary Condition ===
Let:
:$\forall x \in S: x \in A \implies x \in B$
Let $x \in A$.
By definition of subset:
:$x \in S$
Thus by assumption:
:$x \in B$
The result follows by definition of subset.
Category:Subsets
Subset is Compatible with Ordinal Addition

== Theorem ==
Let $x, y, z$ be ordinals.
Then:
:$(1): \quad x \le y \implies \paren {z + x} \le \paren {z + y}$
:$(2): \quad x \le y \implies \paren {x + z} \le \paren {y + z}$
== Proof ==
The result follows from Subset is Left Compatible with Ordinal Addition and Subset is Right Compatible with Ordinal Addition.
Category:Ordinal Arithmetic
Subset is Compatible with Ordinal Multiplication

== Theorem ==
Let $x, y, z$ be ordinals.
Then:
:$(1): \quad x \le y \implies \paren {z \cdot x} \le \paren {z \cdot y}$
:$(2): \quad x \le y \implies \paren {x \cdot z} \le \paren {y \cdot z}$
== Proof ==
The result follows from Subset is Left Compatible with Ordinal Multiplication and Subset is Right Compatible with Ordinal Multiplication.
Category:Ordinal Arithmetic
Subset is Compatible with Ordinal Successor

== Theorem ==
Let $x$ and $y$ be ordinals and let $x^+$ denote the successor set of $x$.
Let $x \in y$.
Then:
:$x^+ \in y^+$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
Category:Ordinals
Category:Subset is Compatible with Ordinal Successor
Subset is Compatible with Ordinal Successor/Proof 2

== Theorem ==
== Proof ==
First note that by Successor Set of Ordinal is Ordinal, $x^+$ and $y^+$ are ordinals.
Let $x \in y$.
We wish to show that $x^+ \in y^+$.
By Ordinal Membership is Trichotomy:
:Either $x^+ \in y^+$, $y^+ \in x^+$, or $x^+ = y^+$.
 $y^+ = x^+$.
Then $y \in x$ or $y = x$ by the definition of successor set.
If $y = x$ then $x \in x$, contradicting the fact that Ordinal is not Element of Itself.
If $y \in x$ then since an ordinal is transitive, $y \in y$, again contradicting Ordinal is not Element of Itself.
Thus $y^+ ≠ x^+$.
 $y^+ \in x^+$.
By definition of successor set, $y^+ \in x$ or $y^+ = x$.
If $y^+ \in x$, then since $y^+$ and $x$ are both ordinals, $y^+ \subsetneqq x$.
Then $y \in x$.
Since $y$ is transitive, $y \in y$, contradicting Ordinal is not Element of Itself.
If instead $y^+ = x$, then $x \in y \in y^+ = x$, so the same contradiction arises because $x$ is transitive.
Thus $y^+ \notin x^+$.
So the only remaining possibility, that $x^+ \in y^+$, must hold.
Category:Subset is Compatible with Ordinal Successor