nyuuzyou/wikis huggingface dataset:
Image of Group of Units in Unital Algebra under Unital Algebra Homomorphism/Corollary

== Corollary ==
Let $R$ be a ring.
Let $A, B$ be unital $R$-algebras.
Let $\phi : A \to B$ be an unital algebra isomorphism.
Let $\map G A$ and $\map G B$ the groups of units of $A$ and $B$ respectively. 
Then:
:$\phi \sqbrk {\map G A} = \map G B$
== Proof ==
From Image of Group of Units in Unital Algebra under Unital Algebra Homomorphism, we have:
:$\phi \sqbrk {\map G A} \subseteq \map G B$
From Inverse of Algebraic Structure Isomorphism is Isomorphism: General Result, $\phi^{-1} : B \to A$ is a unital algebra isomorphism.
Hence applying Image of Group of Units in Unital Algebra under Unital Algebra Homomorphism again we have:
:$\phi^{-1} \sqbrk {\map G B} \subseteq \map G A$
Hence we have:
:$\phi \sqbrk {\phi^{-1} \sqbrk {\map G B} } \subseteq \phi \sqbrk {\map G A}$
Since $\phi$ is a surjection, from Image of Preimage of Subset under Surjection equals Subset we have:
:$\phi \sqbrk {\phi^{-1} \sqbrk {\map G B} } = \map G B$
Hence we have:
:$\map G B \subseteq \phi \sqbrk {\map G A}$
We conclude that:
:$\phi \sqbrk {\map G A} = \map G B$
Category:Image of Group of Units in Unital Algebra under Unital Algebra Homomorphism
Image of Idempotent and Directed Suprema Preserving Mapping is Complete Lattice

== Theorem ==
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a complete lattice.
Let $f:S \to S$ be a mapping that is idempotent and preserves directed suprema.
Let $F = \struct {f \sqbrk S, \precsim}$ be an ordered subset of $L$.
Then $F$ inherits directed suprema and is complete lattice.
== Proof ==
We will prove that
:$F$ inherits directed suprema.
Let $D$ be a directed subset of $f \sqbrk S$ such that
:$D$ admits a supremum in $L$.
By definition of ordered subset:
:$D$ is directed in $L$.
By definition of mapping preserves directed suprema:
:$\map {\sup_L} {f \sqbrk D} = \map f {\sup_L D}$
By definition of idempotent mapping:
:$f \sqbrk D = D$
Thus by definition of image of set:
:$\sup_L D \in f \sqbrk S$
By definition of idempotent mapping:
:$f \sqbrk S = \set {x \in S: \map f x = x}$
By Directed Suprema Preserving Mapping is Increasing:
:$f$ is an increasing mapping.
Thus by Image under Increasing Mapping equal to Special Set is Complete Lattice:
:$F$ is complete lattice.
== Sources ==
* 
Category:Complete Lattices
Image of Interior of Set under Homeomorphism is Interior of Image

== Theorem ==
Let $X$ and $Y$ be topological spaces.
Let $f : X \to Y$ be a homeomorphism.
Let $A \subseteq X$.
Then:
:$\paren {f \sqbrk A}^\circ = f \sqbrk {A^\circ}$
where:
:$A^\circ$ is the interior of $A$ in $X$
:$\paren {f \sqbrk A}^\circ$ is the interior of $f \sqbrk A$ in $Y$.
== Proof ==
First, we show that $f \sqbrk {A^\circ} \subseteq \paren {f \sqbrk A}^\circ$. 
Let $y \in f \sqbrk {A^\circ}$.
Then there exists $x \in A^\circ$ such that $y = \map f x$.
Since $x \in A^\circ$, there exists an open set $U \subseteq A$ in $X$ such that $x \in U$.
Then $y = \map f x \in f \sqbrk U \subseteq f \sqbrk A$.
By definition of a homeomorphism, $f$ is an open map.
So $f \sqbrk U$ is an open set in $Y$ contained in $f \sqbrk A$ and containing $y$.
So from the definition of interior, we have $y \in \paren {f \sqbrk A}^\circ$. 
Hence $y \in f \sqbrk {A^\circ}$ implies that $y \in \paren {f \sqbrk A}^\circ$.
So $f \sqbrk {A^\circ} \subseteq \paren {f \sqbrk A}^\circ$.
Now we show that $\paren {f \sqbrk A}^\circ \subseteq f \sqbrk {A^\circ}$. 
Let $y \in \paren {f \sqbrk A}^\circ$.
Then there exists an open set $V \subseteq f \sqbrk A$ in $Y$ such that $y \in V$.
We then have $\map {f^{-1} } y \in f^{-1} \sqbrk V$.
Further, from Preimage of Image of Subset under Injection equals Subset and Image of Subset under Mapping is Subset of Image, we have $f^{-1} \sqbrk V \subseteq A$. 
Since $f$ is continuous, $f^{-1} \sqbrk V$ is open in $X$, is contained in $A$ and contains $\map {f^{-1} } y$.
So we have $\map {f^{-1} } y \in A^\circ$.
Since $y = \map f {\map {f^{-1} } y}$, we obtain $y \in f \sqbrk {A^\circ}$.
Hence $y \in \paren {f \sqbrk A}^\circ$ implies $y \in f \sqbrk {A^\circ}$.
So $\paren {f \sqbrk A}^\circ \subseteq f \sqbrk {A^\circ}$. 
Hence we obtain $\paren {f \sqbrk A}^\circ = f \sqbrk {A^\circ}$.
Category:Homeomorphisms (Topological Spaces)
Category:Set Interiors
Image of Intersection under Injection

== Theorem ==
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Then:
:$\forall A, B \subseteq S: f \sqbrk {A \cap B} = f \sqbrk A \cap f \sqbrk B$
 $f$ is an injection.
This can be expressed in the language and notation of direct image mappings as:
:$\forall A, B \subseteq S: \map {f^\to} {A \cap B} = \map {f^\to} A \cap \map {f^\to} B$
=== General Result ===
=== Family of Sets ===
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 10$: Inverses and Composites: Exercise $\text{(ii)}$
* : Chapter $1$: Mappings: $\S 12 \alpha$
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Exercise $2$
* : Chapter $4$: Mappings: Exercise $7$
* : $1$: Set Theory and Logic: $\S 2$: Functions: Exercise $2.2 \ \text{(g)}$
Category:Injections
Category:Set Intersection
Category:Image of Intersection under Injection
Image of Intersection under Injection/Family of Sets

== Theorem ==
Let $S$ and $T$ be sets.
Let $\family {S_i}_{i \mathop \in I}$ be a family of subsets of $S$.
Let $f: S \to T$ be a mapping.
Then:
:$\ds f \sqbrk {\bigcap_{i \mathop \in I} S_i} = \bigcap_{i \mathop \in I} f \sqbrk {S_i}$
 $f$ is an injection.
This can be expressed in the language and notation of direct image mappings as:
:$\ds \map {f^\to} {\bigcap_{i \mathop \in I} S_i} = \bigcap_{i \mathop \in I} \map {f^\to} {S_i}$
== Proof ==
An injection is a type of one-to-one relation, and therefore also a one-to-many relation.
Therefore Image of Intersection under One-to-Many Relation: Family of Sets applies:
:$\ds \RR \sqbrk {\bigcap_{i \mathop \in I} S_i} = \bigcap_{i \mathop \in I} \RR \sqbrk {S_i}$
 $\RR$ is a one-to-many relation.
We have that $f$ is a mapping and therefore a many-to-one relation.
So $f$ is a one-to-many relation $f$ is also an injection.
It follows that:
:$\ds f \sqbrk {\bigcap_{i \mathop \in I} S_i} = \bigcap_{i \mathop \in I} f \sqbrk {S_i}$
 $f$ is an injection.
== Sources ==
* : $\S 6$. Indexed families; partitions; equivalence relations: Exercise $1$
* : Chapter $1$: Theory of Sets: $\S 6$: Functions: Exercise $3 \ \text{(c})$
Category:Image of Intersection under Injection
Image of Intersection under Injection/General Result

== Theorem ==
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Let $\powerset S$ be the power set of $S$.
Then:
:$\ds \forall \mathbb S \subseteq \powerset S: f \sqbrk {\bigcap \mathbb S} = \bigcap_{X \mathop \in \mathbb S} f \sqbrk X$ 
 $f$ is an injection.
This can be expressed in the language and notation of direct image mappings as:
:$\ds \forall \mathbb S \subseteq \powerset S : \map {f^\to} {\bigcap \mathbb S} = \bigcap_{X \mathop \in \mathbb S} \map {f^\to} X$
== Proof ==
An injection is a type of one-to-one relation, and therefore also a one-to-many relation.
Therefore Image of Intersection under One-to-Many Relation applies:
:$\ds \forall \mathbb S \subseteq \powerset S: \RR \sqbrk {\bigcap \mathbb S} = \bigcap_{X \mathop \in \mathbb S} \RR \sqbrk {\mathbb S}$
 $\RR$ is a one-to-many relation.
We have that $f$ is a mapping and therefore a many-to-one relation.
So $f$ is a one-to-many relation $f$ is also an injection.
It follows that:
:$\ds \forall \mathbb S \subseteq \powerset S: f \sqbrk {\bigcap \mathbb S} = \bigcap_{X \mathop \in \mathbb S} f \sqbrk X$ 
 $f$ is an injection.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $1$: Pairs, Relations, and Functions: Exercise $7 \ \text {(b)}$
Category:Image of Intersection under Injection
Image of Intersection under Injection/Proof 1

== Theorem ==
== Proof ==
An injection is a type of one-to-one relation, and therefore also a one-to-many relation.
Therefore Image of Intersection under One-to-Many Relation applies:
:$\forall A, B \subseteq S: \RR \sqbrk {A \cap B} = \RR \sqbrk A \cap \RR \sqbrk B$
 $\RR$ is a one-to-many relation.
We have that $f$ is a mapping and therefore a many-to-one relation.
So $f$ is a one-to-many relation $f$ is also an injection.
It follows that:
:$\forall A, B \subseteq S: f \sqbrk {A \cap B} = f \sqbrk A \cap f \sqbrk B$
 $f$ is an injection.
Category:Image of Intersection under Injection
Image of Intersection under Injection/Proof 2

== Theorem ==
== Proof ==
From Image of Intersection under Mapping:
:$f \sqbrk {A \cap B} \subseteq f \sqbrk A \cap f \sqbrk B$
which holds for all mappings.
It remains to be shown that:
:$f \sqbrk A \cap f \sqbrk B \subseteq f \sqbrk {A \cap B}$
 $f$ is an injection.
=== Sufficient Condition ===
Suppose that:
:$\forall A, B \subseteq S: f \sqbrk {A \cap B} = f \sqbrk A \cap f \sqbrk B$
If $S$ is singleton, then $f$ is a fortiori an injection from Mapping from Singleton is Injection.
So, assume $S$ is not singleton.
 $f$ is specifically not an injection.
Then:
:$\exists x, y \in S: \exists z \in T: \tuple {x, z} \in T, \tuple {y, z} \in T, x \ne y$
and of course $\set x \subseteq S, \set y \subseteq S$.
So:
:$z \in f \sqbrk {\set x}$
:$z \in f \sqbrk {\set y}$
and so by definition of intersection:
:$z \in f \sqbrk {\set x} \cap f \sqbrk {\set y}$
But:
:$\set x \cap \set y = \O$
Thus from a corollary to Image of Empty Set is Empty Set:
:$f \sqbrk {\set x \cap \set y} = \O$
and so:
:$f \sqbrk {\set x \cap \set y} \ne f \sqbrk {\set x} \cap f \sqbrk {\set y}$
But by hypothesis:
:$\forall A, B \subseteq S: f \sqbrk {A \cap B} = f \sqbrk A \cap f \sqbrk B$
From this contradiction it follows that $f$ is an injection.
=== Necessary Condition ===
Let $f$ be an injection.
It is necessary to show:
:$f \sqbrk {S_1} \cap f \sqbrk {S_2} \subseteq f \sqbrk {S_1 \cap S_2}$
Let $t \in f \sqbrk {S_1} \cap f \sqbrk {S_2}$.
Then:
So if $f$ is an injection, it follows that:
:$f \sqbrk {S_1 \cap S_2} = f \sqbrk {S_1} \cap f \sqbrk {S_2}$
Putting the results together, we see that:
:$f \sqbrk {S_1 \cap S_2} = f \sqbrk {S_1} \cap f \sqbrk {S_2}$ $f$ is an injection.
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 6$: Functions: Exercise $2 \ \text{(c})$
* : $2.1$
Category:Image of Intersection under Injection
Image of Intersection under Mapping

== Theorem ==
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Let $S_1$ and $S_2$ be subsets of $S$.
Then:
:$f \sqbrk {S_1 \cap S_2} \subseteq f \sqbrk {S_1} \cap f \sqbrk {S_2}$
This can be expressed in the language and notation of direct image mappings as:
:$\forall S_1, S_2 \in \powerset S: \map {f^\to} {S_1 \cap S_2} \subseteq \map {f^\to} {S_1} \cap \map {f^\to} {S_2}$
That is, the image of the intersection of subsets of a mapping is a subset of the intersection of their images.
=== General Result ===
=== Family of Sets ===
== Proof 1 ==
== Proof 2 ==
== Examples ==
== Also see ==
* Image of Intersection under Injection: equality happens $f$ is an injection 
* Image of Intersection under One-to-Many Relation
* Preimage of Intersection under Mapping
* Image of Union under Mapping
* Preimage of Union under Mapping
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.13 \ \text{(a)}$
* : $\S 1.3$: Functions and mappings. Images and preimages: Remark $1$
* : $\S 1.10$: Functions: Exercise $5 \ \text{(f)}$
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.1: \ \text{(iii)}$
* : $1$: Set Theory and Logic: $\S 2$: Functions: Exercise $2.2 \ \text{(g)}$
* : $\S 2$
* : $\S 2$: Problem $4 \ \text{(i)}$
Category:Images
Category:Set Intersection
Category:Image of Intersection under Mapping
Image of Intersection under Mapping/Examples/First Projection on Subsets of Cartesian Natural Number Space

== Example of Image of Intersection under Mapping ==
Let $\pr_1: \N \times \N \to \N$ denote the first projection from the cartesian space $\N \times \N$ of the natural numbers.
Let:
First note that we have:
Then:
while:
As can be seen, the inclusion is proper, that is:
:$\pr_1 \sqbrk {S_1 \cap S_2} \ne \pr_1 \sqbrk {S_1} \cap \pr_1 \sqbrk {S_2}$
== Sources ==
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Exercise $1$
Category:Image of Intersection under Mapping
Image of Intersection under Mapping/Examples/Square Function

== Example of Image of Intersection under Mapping ==
Let:
:$S_1 = \set {x \in \Z: x \le 0}$
:$S_2 = \set {x \in \Z: x \ge 0}$
:$f: \Z \to \Z: \forall x \in \Z: \map f x = x^2$
We have:
:$f \sqbrk {S_1} = \set {0, 1, 4, 9, 16, \ldots} = f \sqbrk {S_2}$
Then:
:$f \sqbrk {S_1} \cap f \sqbrk {S_2} = \set {0, 1, 4, 9, 16, \ldots}$
but:
:$f \sqbrk {S_1 \cap S_2} = f \sqbrk {\set 0} = \set 0$
As can be seen, the inclusion is proper, that is:
:$f \sqbrk {S_1 \cap S_2} \ne f \sqbrk {S_1} \cap f \sqbrk {S_2}$
== Also see ==
* Image of Intersection under Injection: equality holds $f$ is an injection.
== Sources ==
* : Chapter $1$: Mappings: $\S 12 \alpha$
* : $\S 21$: The image of a subset of the domain; surjections: Postscript to $\S 21.4 \ \text{(i)}$
Category:Image of Intersection under Mapping
Image of Intersection under Mapping/Family of Sets

== Theorem ==
Let $S$ and $T$ be sets.
Let $\family {S_i}_{i \mathop \in I}$ be a family of subsets of $S$.
Let $f: S \to T$ be a mapping.
Then:
:$\ds f \sqbrk {\bigcap_{i \mathop \in I} S_i} \subseteq \bigcap_{i \mathop \in I} f \sqbrk {S_i}$
where $\ds \bigcap_{i \mathop \in I} S_i$ denotes the intersection of $\family {S_i}_{i \mathop \in I}$.
== Proof ==
As $f$, being a mapping, is also a relation, we can apply Image of Intersection under Relation: Family of Sets:
:$\ds \RR \sqbrk {\bigcap_{i \mathop \in I} S_i} \subseteq \bigcap_{i \mathop \in I} \RR \sqbrk {S_i}$
== Sources ==
* : $\S 10$: Inverses and Composites
* : $\S 1.12$: Set Inclusions for Image and Inverse Image Sets: Theorem $12.6 \ \text{(b)}$
* : $\S 6$. Indexed families; partitions; equivalence relations: Exercise $1$
* : Chapter $1$: Theory of Sets: $\S 6$: Functions: Exercise $3 \ \text{(b})$
* : Notation and Terminology
* : Appendix $\text{A}.3$: Functions: Problem $\text{A}.3.1$
Category:Image of Intersection under Mapping
Image of Intersection under Mapping/General Result

== Theorem ==
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Let $\powerset S$ be the power set of $S$.
Let $\mathbb S \subseteq \powerset S$.
Then:
:$\ds f \sqbrk {\bigcap \mathbb S} \subseteq \bigcap_{X \mathop \in \mathbb S} f \sqbrk X$
== Proof ==
As $f$, being a mapping, is also a relation, we can apply Image of Intersection under Relation: General Result:
:$\ds \RR \sqbrk {\bigcap \mathbb S} \subseteq \bigcap_{X \mathop \in \mathbb S} \RR \sqbrk X$
== Sources ==
* : Chapter $0$: Functions
* : Part $1$: Not Entirely Naive Set Theory: Chapter $1$: Pairs, Relations, and Functions: Exercise $7 \ \text {(b)}$
* : $1$: Set Theory and Logic: $\S 2$: Functions: Exercise $2.3$
Category:Image of Intersection under Mapping
Image of Intersection under Mapping/Proof 2

== Theorem ==
== Proof ==
As $f$, being a mapping, is also a relation, we can apply Image of Intersection under Relation:
:$\RR \sqbrk {S_1 \cap S_2} \subseteq \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$
== Sources ==
* : Chapter $0$: Functions
Category:Image of Intersection under Mapping
Image of Intersection under One-to-Many Relation

== Theorem ==
Let $S$ and $T$ be sets.
Let $\RR \subseteq S \times T$ be a relation.
Then:
:$\forall S_1, S_2 \subseteq S: \RR \sqbrk {S_1 \cap S_2} = \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$ 
 $\RR$ is one-to-many.
=== General Result ===
=== Family of Sets ===
== Proof ==
=== Sufficient Condition ===
Suppose that:
:$\forall S_1, S_2 \subseteq S: \RR \sqbrk {S_1 \cap S_2} = \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$
If $S$ is singleton, the result follows immediately as $\RR$ would have to be one-to-many.
So, assume $S$ is not singleton.
Suppose $\RR$ is specifically not one-to-many.
So:
:$\exists x, y \in S: \exists z \in T: \tuple {x, z} \in T, \tuple {y, z} \in T, x \ne y$
and of course $\set x \subseteq S, \set y \subseteq S$.
So:
:$z \in \RR \sqbrk {\set x}$
:$z \in \RR \sqbrk {\set y}$
and so by definition of intersection:
:$z \in \RR \sqbrk {\set x} \cap \RR \sqbrk {\set y}$
But:
:$\set x \cap \set y = \O$
Thus from Image of Empty Set is Empty Set:
:$\RR \sqbrk {\set x \cap \set y} = \O$
and so:
:$\RR \sqbrk {\set x \cap \set y} \ne \RR \sqbrk {\set x} \cap \RR \sqbrk {\set y}$
=== Necessary Condition ===
Let $\RR$ be one-to-many.
From Image of Intersection under Relation, we already have:
:$\RR \sqbrk {S_1 \cap S_2} \subseteq \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$
So we just need to show:
:$\RR \sqbrk {S_1} \cap \RR \sqbrk {S_2} \subseteq \RR \sqbrk {S_1 \cap S_2}$
Let $t \in \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$.
Then:
So if $\RR$ is one-to-many, it follows that:
:$\RR \sqbrk {S_1 \cap S_2} = \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$
Putting the results together, we see that:
:$\RR \sqbrk {S_1 \cap S_2} = \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$ $\RR$ is one-to-many.
Category:One-to-Many Relations
Category:Set Intersection
Category:Image of Intersection under One-to-Many Relation
Image of Intersection under One-to-Many Relation/Family of Sets

== Theorem ==
Let $S$ and $T$ be sets.
Let $\RR \subseteq S \times T$ be a relation.
Then $\RR$ is a one-to-many relation :
:$\ds \RR \sqbrk {\bigcap_{i \mathop \in I} S_i} = \bigcap_{i \mathop \in I} \RR \sqbrk {S_i}$ 
where $\family {S_i}_{i \mathop \in I}$ is any family of subsets of $S$.
== Proof ==
=== Sufficient Condition ===
Suppose:
:$\ds \RR \sqbrk {\bigcap_{i \mathop \in I} S_i} = \bigcap_{i \mathop \in I} \RR \sqbrk {S_i}$ 
where $\family {S_i}_{i \mathop \in I}$ is any family of subsets of $S$.
Then by definition of $\family {S_i}_{i \mathop \in I}$:
:$\forall i, j \in I: \RR \sqbrk {S_i \cap S_j} = \RR \sqbrk {S_i} \cap \RR \sqbrk {S_j}$
and the sufficient condition applies for Image of Intersection under One-to-Many Relation.
So $\RR$ is one-to-many.
=== Necessary Condition ===
Suppose $\RR$ is one-to-many.
From Image of Intersection under Relation: Family of Sets, we already have:
:$\ds \RR \sqbrk {\bigcap_{i \mathop \in I} S_i} \subseteq \bigcap_{i \mathop \in I} \RR \sqbrk {S_i}$
so we just need to show:
:$\ds \bigcap_{i \mathop \in I} \RR \sqbrk {S_i} \subseteq \RR \sqbrk {\bigcap_{i \mathop \in I} S_i}$
Let:
:$\ds t \in \bigcap_{i \mathop \in I} \RR \sqbrk {S_i}$
Then:
So if $\RR$ is one-to-many, it follows that:
:$\ds \RR \sqbrk {\bigcap_{i \mathop \in I} S_i} = \bigcap_{i \mathop \in I} \RR \sqbrk {S_i}$ 
Putting the results together:
$\RR$ is one-to-many :
:$\ds \RR \sqbrk {\bigcap_{i \mathop \in I} S_i} = \bigcap_{i \mathop \in I} \RR \sqbrk {S_i}$
where $\family {S_i}_{i \mathop \in I}$ is any family of subsets of $S$.
Category:Image of Intersection under One-to-Many Relation
Category:Indexed Families
Image of Intersection under One-to-Many Relation/General Result

== Theorem ==
Let $S$ and $T$ be sets.
Let $\RR \subseteq S \times T$ be a relation.
Let $\powerset S$ be the power set of $S$.
Then:
:$\ds \forall \mathbb S \subseteq \powerset S: \RR \sqbrk {\bigcap \mathbb S} = \bigcap_{X \mathop \in \mathbb S} \RR \sqbrk X$ 
 $\RR$ is one-to-many.
== Proof ==
=== Sufficient Condition ===
Suppose:
:$\ds \RR \sqbrk {\bigcap \mathbb S} = \bigcap_{X \mathop \in \mathbb S} \RR \sqbrk X$ 
where $\mathbb S$ is any subset of $\powerset S$.
Then by definition of $\mathbb S$:
:$\forall S_1, S_2 \in \mathbb S: \RR \sqbrk {S_1 \cap S_2} = \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$
and the sufficient condition applies for Image of Intersection under One-to-Many Relation.
So $\RR$ is one-to-many.
=== Necessary Condition ===
Suppose $\RR$ is one-to-many.
From Image of Intersection under Relation: General Result, we already have:
:$\ds \RR \sqbrk {\bigcap \mathbb S} \subseteq \bigcap_{X \mathop \in \mathbb S} \RR \sqbrk X$
so we just need to show:
:$\ds \forall \mathbb S \subseteq \powerset S: \bigcap_{X \mathop \in \mathbb S} \RR \sqbrk X \subseteq \RR \sqbrk {\bigcap \mathbb S}$
Let:
:$\ds t \in \bigcap_{X \mathop \in \mathbb S} \RR \sqbrk X$
Then:
So if $\RR$ is one-to-many, it follows that:
:$\ds \forall \mathbb S \subseteq \powerset S: \RR \sqbrk {\bigcap \mathbb S} = \bigcap_{X \mathop \in \mathbb S} \RR \sqbrk X$ 
Putting the results together:
$\RR$ is one-to-many :
:$\ds \RR \sqbrk {\bigcap \mathbb S} = \bigcap_{X \mathop \in \mathbb S} \RR \sqbrk X$ 
where $\mathbb S$ is any subset of $\powerset S$.
Category:Image of Intersection under One-to-Many Relation
Image of Intersection under Relation

== Theorem ==
Let $S$ and $T$ be sets.
Let $\RR \subseteq S \times T$ be a relation.
Let $S_1$ and $S_2$ be subsets of $S$.
Then:
:$\RR \sqbrk {S_1 \cap S_2} \subseteq \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$
That is, the image of the intersection of subsets of $S$ is a subset of the intersection of their images.
=== General Result ===
=== Family of Sets ===
== Proof ==
== Also see ==
* Image of Union under Relation
* Preimage of Union under Relation
* Preimage of Intersection under Relation
== Also see ==
* Image of Intersection under One-to-Many Relation, which shows that, for the general relation $\RR$, equality holds $\RR$ is one-to-many.
== Sources ==
* : Chapter $0$: Relations: Theorem $5 \ \text{(c)}$
* : Chapter $\text{I}$: Sets and Functions: Problem $\text{AA}$: Relations
* : $\S 1.12$: Set Inclusions for Image and Inverse Image Sets: Theorem $12.2 \ \text{(b)}$
Category:Relation Theory
Category:Set Intersection
Category:Image of Intersection under Relation
Image of Intersection under Relation/Family of Sets

== Theorem ==
Let $S$ and $T$ be sets.
Let $\family {S_i}_{i \mathop \in I}$ be a family of subsets of $S$.
Let $\RR \subseteq S \times T$ be a relation.
Then:
:$\ds \RR \sqbrk {\bigcap_{i \mathop \in I} S_i} \subseteq \bigcap_{i \mathop \in I} \RR \sqbrk {S_i}$
where $\ds \bigcap_{i \mathop \in I} S_i$ denotes the intersection of $\family {S_i}_{i \mathop \in I}$.
== Proof ==
== Sources ==
* : Chapter $0$: Relations: Theorem $5 \ \text{(d)}$
* : $\S 1.12$: Set Inclusions for Image and Inverse Image Sets: Theorem $12.5 \ \text{(b)}$
Category:Image of Intersection under Relation
Category:Indexed Families
Image of Intersection under Relation/General Result

== Theorem ==
Let $S$ and $T$ be sets.
Let $\RR \subseteq S \times T$ be a relation.
Let $\powerset S$ be the power set of $S$.
Let $\mathbb S \subseteq \powerset S$.
Then:
:$\ds \RR \sqbrk {\bigcap \mathbb S} \subseteq \bigcap_{X \mathop \in \mathbb S} \RR \sqbrk X$
== Proof ==
Category:Image of Intersection under Relation
Image of Interval by Continuous Function is Interval/Proof 1

== Theorem ==
== Proof ==
Let $J$ be the image of $f$.
By definition of real interval, it suffices to show that:
:$\forall y_1, y_2 \in J: \forall \lambda \in \R: y_1 \le \lambda \le y_2 \implies \lambda \in J$
So suppose $y_1, y_2 \in J$, and suppose $\lambda \in \R$ is such that $y_1 \le \lambda \le y_2$.
Consider these subsets of $I$:
:$S = \set {x \in I: \map f x \le \lambda}$
:$T = \set {x \in I: \map f x \ge \lambda}$
As $y_1 \in S$ and $y_2 \in T$, it follows that $S$ and $T$ are both non-empty.
Also, $I = S \cup T$.
So from Interval Divided into Subsets, a point in one subset is at zero distance from the other.
So, suppose that $s \in S$ is at zero distance from $T$.
From Limit of Sequence to Zero Distance Point, we can find a sequence $\sequence {t_n}$ in $T$ such that $\ds \lim_{n \mathop \to \infty} t_n = s$.
Since $f$ is continuous on $I$, it follows from Limit of Image of Sequence that:
:$\ds \lim_{n \mathop \to \infty} \map f {t_n} = \map f s$
But:
:$\forall n \in \N_{> 0}: \map f {t_n} \ge \lambda$
Therefore by Lower and Upper Bounds for Sequences:
:$\map f s \ge \lambda$
We already have that:
:$\map f s \le \lambda$
Therefore:
:$\map f s = \lambda$
and so:
:$\lambda \in J$
A similar argument applies if a point of $T$ is at zero distance from $S$.
== Sources ==
* : $4.23$
* : $\S 9.9$
Category:Image of Interval by Continuous Function is Interval