nyuuzyou/wikis huggingface dataset:
Image of Subset under Mapping/Examples/Image of -3 to 2 under x^4-1

== Example of Image of Element under Mapping ==
Let $f: \R \to \R$ be the mapping defined as:
:$\forall x \in \R: \map f x = x^4 - 1$
The image of the closed interval $\closedint {-3} 2$ is:
:$f \closedint {-3} 2 = \closedint {-1} {80}$
== Proof ==
Trivially, by differentiating $x^4 - 1$ $x$:
:$f' = 4 x^3$
and equating $f'$ to $0$, the minimum of $\Img f$ is seen to occur at $\map f 0 = -1$.
As $0 \in \closedint {-3} 2$, it can be seen that the minimum of $f \closedint {-3} 2$ is $-1$.
As $f$ is strictly increasing on $x > 0$ and strictly decreasing on $x < 0$, it suffices to inspect the images of the endpoints $-3$ and $2$.
Thus:
:$\map f {-3} = \paren {-3}^4 - 1 = 81 - 1 = 80$
:$\map f 2 = 2^4 - 1 = 16 - 1 = 15$
The result follows.
== Sources ==
* : Chapter $4$: Mappings: Exercise $3 \ \text {(ii)}$
Category:Examples of Images under Mappings
Image of Subset under Mapping/Examples/Image of 1 to 2 under x^2-x-2

== Example of Image of Subset under Mapping ==
Let $f: \R \to \R$ be the mapping defined as:
:$\forall x \in \R: \map f x = x^2 - x - 2$
The image of the closed interval $\closedint {-3} 2$ is:
:$f \closedint 1 2 = \closedint {-2} 0$
== Proof ==
Trivially, by differentiating $x^2 - x - 2$ $x$:
:$f' = 2 x - 1$
It is seen that, on $\closedint 1 2$, $f$ is strictly increasing.
Hence it suffices to inspect the images of the endpoints $1$ and $2$.
Thus:
:$\map f 1 = 1^2 - 1 - 1 = -2$
:$\map f 2 = 2^2 - 2 - 2 = 0$
The result follows.
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 6$: Functions
Category:Examples of Images under Mappings
Image of Subset under Mapping equals Union of Images of Elements

== Theorem ==
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping from $S$ to $T$.
Let $X \subseteq S$ be a subset of $S$.
Then:
:$\ds f \sqbrk X = \bigcup_{x \mathop \in X} \map f x$
where:
:$f \sqbrk X$ is the image of the subset $X$ under $f$
:$\map f x$ is the image of the element $x$ under $f$.
== Proof ==
By definition, a mapping is a relation.
Thus Image of Subset under Relation equals Union of Images of Elements applies.
== Sources ==
* : $\S 1.10$: Functions: Remark $10.8 \ \text{(d)}$
Category:Images
Category:Set Union
Image of Subset under Mapping is Subset of Image

== Corollary to Image of Subset under Relation is Subset of Image ==
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping from $S$ to $T$.
Let $A, B \subseteq S$ such that $A \subseteq B$.
Then the image of $A$ is a subset of the image of $B$:
:$A \subseteq B \implies f \sqbrk A \subseteq f \sqbrk B$
This can be expressed in the language and notation of direct image mappings as:
:$\forall A, B \in \powerset S: A \subseteq B \implies \map {f^\to} A \subseteq \map {f^\to} B$
== Proof ==
As $f: S \to T$ is a mapping, it is also a relation, and thus:
:$f \subseteq S \times T$
The result follows directly from Image of Subset under Relation is Subset of Image.
== Sources ==
* : $\S 2.2$: Homomorphisms: $\text{(ii)}$
* : $\S 1.10$: Functions: Exercise $5 \ \text{(c)}$
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.1: \ \text{(i)}$
* : $\S 21.3$: The image of a subset of the domain; surjections
* : $1$: Set Theory and Logic: $\S 2$: Functions: Exercise $2.2 \ \text{(e)}$
Category:Subsets
Category:Images
Image of Subset under Neighborhood of Diagonal is Neighborhood of Subset

== Theorem ==
Let $T = \struct{X, \tau}$ be a topological space.
Let $\tau_{X \times X}$ denote the product topology on the cartesian product $X \times X$.
Let $V$ be a neighborhood of the diagonal $\Delta_X$ of $X \times X$ in the product space $\struct {X \times X, \tau_{X \times X} }$.
Then:
:$\forall A \subseteq X : V \sqbrk A$ is a neighborhood of $A$ in $T$
== Proof ==
Let $A \subseteq X$.
From Image of Subset under Relation equals Union of Images of Elements:
:$V \sqbrk A = \ds \bigcup_{x \in A} \map V x$
From Subset of Union:
:$\forall x \in A : \map V x \subseteq V \sqbrk A$
From Image of Point under Neighborhood of Diagonal is Neighborhood of Point:
:$\forall x \in A : \map V x$ is a neighborhood of $x$ in $T$
From Superset of Neighborhood in Topological Space is Neighborhood:
:$\forall x \in A : \map V A$ is a neighborhood of $x$ in $T$
From Set is Neighborhood of Subset iff Neighborhood of all Points of Subset:
:$\map V A$ is a neighborhood of $A$ in $T$
Category:Neighborhoods
Category:Product Spaces
Category:Relations
Image of Subset under Open Neighborhood of Diagonal is Open Neighborhood of Subset

== Theorem ==
Let $T = \struct{X, \tau}$ be a topological space.
Let $\tau_{X \times X}$ denote the product topology on the cartesian product $X \times X$.
Let $U$ be an open neighborhood of the diagonal $\Delta_X$ of $X \times X$ in the product space $\struct {X \times X, \tau_{X \times X} }$.
Then:
:$\forall A \subseteq X : U \sqbrk A$ is an open neighborhood of $A$ in $T$
== Proof ==
Let $A \subseteq X$.
From Image of Subset under Neighborhood of Diagonal is Neighborhood of Subset:
:$U \sqbrk A$ is a neighborhood of $A$ in $T$
From Image of Subset under Relation equals Union of Images of Elements:
:$U \sqbrk A = \ds \bigcup_{x \in A} \map U x$
From Image of Point under Open Neighborhood of Diagonal is Open Neighborhood of Point:
:$\forall x \in A : \map U x$ is an open neighborhood of $x$ in $T$
By :
:$U \sqbrk A \in \tau$
Since $A$ was arbitrary:
:$\forall A \subseteq X : U \sqbrk A$ is an open neighborhood of $A$ in $T$
Category:Open Sets
Category:Product Spaces
Category:Relations
Image of Subset under Relation equals Union of Images of Elements

== Theorem ==
Let $S$ and $T$ be sets.
Let $\RR \subseteq S \times T$ be a relation on $S \times T$.
Let $X \subseteq S$ be a subset of $S$.
Then:
:$\ds \RR \sqbrk X = \bigcup_{x \mathop \in X} \map \RR x$
where:
:$\RR \sqbrk X$ is the image of the subset $X$ under $\RR$
:$\map \RR x$ is the image of the element $x$ under $\RR$.
== Proof ==
By definition:
:$\RR \sqbrk X = \set {y \in T: \exists x \in X: \tuple {x, y} \in \RR}$
:$\map \RR x = \set {y \in T:\tuple {x, y} \in \RR}$
First:
Then:
So:
:$\ds \bigcup_{x \mathop \in X} \map \RR x \subseteq \RR \sqbrk X$
and:
:$\ds \RR \sqbrk X \subseteq \bigcup_{x \mathop \in X} \map \RR x$
The result follows by definition of set equality.
Category:Relation Theory
Category:Subsets
Category:Set Union
Image of Subset under Relation is Subset of Image

== Theorem ==
Let $S$ and $T$ be sets.
Let $\RR \subseteq S \times T$ be a relation from $S$ to $T$.
Let $A, B \subseteq S$ such that $A \subseteq B$.
Then the image of $A$ is a subset of the image of $B$:
:$A \subseteq B \implies \RR \sqbrk A \subseteq \RR \sqbrk B$
In the notation of direct image mappings, this can be written:
:$A \subseteq B \implies \map {\RR^\to} A \subseteq \map {\RR^\to} B$
=== Corollary 1 ===
The same applies to the preimage, as follows.
=== Corollary 2 ===
The same applies for a mapping $f: S \to T$ and its inverse $f^{-1} \subseteq T \times S$, whether $f^{-1}$ is a mapping or not.
=== Corollary 3 ===
Similarly:
== Proof ==
Suppose $\RR \sqbrk A \nsubseteq \RR \sqbrk B$.
The result follows by the Rule of Transposition.
== Sources ==
* : $\S 1.12$: Set Inclusions for Image and Inverse Image Sets: Theorem $12.1$
Category:Subsets
Category:Relation Theory
Image of Subset under Relation is Subset of Image/Corollary 1

== Corollary to Image of Subset under Relation is Subset of Image ==
Let $S$ and $T$ be sets.
Let $\RR \subseteq S \times T$ be a relation from $S$ to $T$.
Let $C, D \subseteq T$.
Then:
:$C \subseteq D \implies \RR^{-1} \sqbrk C \subseteq \RR^{-1} \sqbrk D$
where $\RR^{-1} \sqbrk C$ is the preimage of $C$ under $\RR$.
== Proof ==
We have that $\RR^{-1}$ is itself a relation, by definition of inverse relation.
The result follows directly from Image of Subset under Relation is Subset of Image.
Category:Subsets
Category:Relation Theory
Image of Successor Mapping forms Peano Structure

== Theorem ==
Let $\left({P, s, 0}\right)$ be a Peano structure.
Let $P'$ be the set $s \left[{P}\right]$, that is:
:$P' = \left\{{s \left({n}\right): n \in P}\right\}$
Let $s'$ be the restriction of $s$ to $P'$.
Then $\left({P', s', s \left({0}\right)}\right)$ is also a Peano structure.
== Proof ==
We need to check that all of Peano's axioms hold for $\left({P', s', s \left({0}\right)}\right)$.
Now from Restriction of Injection is Injection, because $s$ is an injection then so is $s'$.
So Axiom $(P3)$ holds.
Suppose that for some $n \in P'$, we would have:
:$s' \left({n}\right) = s \left({0}\right)$
Then by definition of $s'$ as the restriction of $s$:
:$s \left({n}\right) = s \left({0}\right)$
However, since $0 \notin P'$, it follows that $n \ne 0$.
This contradicts the fact that $s$ is an injection.
Hence for all $n \in P'$:
:$s' \left({n}\right) \ne s \left({0}\right)$
So Axiom $(P4)$ holds as well.
Let $A \subseteq P'$ be such that $s \left({0}\right) \in A$ and:
:$\forall z \in A: s' \left({z}\right) \in A$
Let $A_0$ be defined as:
:$A_0 = A \cup \left\{{0}\right\}$
Then $0 \in A_0$, and since:
:$\forall z \in A: s' \left({z}\right) \in A$
:$s \left({0}\right) \in A$
it follows that:
:$\forall z \in A_0: s \left({z}\right) \in A_0$
Therefore, since $\left({P, s, 0}\right)$ is a Peano structure:
:$A_0 = P$
Therefore, since $s \left({n}\right) \ne 0$ for all $n \in P$:
:$\forall n \in P: s \left({n}\right) \in A$
Hence, by definition of $P'$:
:$P' \subseteq A$
By definition of set equality:
:$A = P'$
meaning that Axiom $(P5)$ holds as well.
Having verified all axioms, it follows that $\left({P', s', s \left({0}\right)}\right)$ is a Peano structure.
Category:Peano's Axioms
Image of Symmetric Set under Linear Transformation is Symmetric

== Theorem ==
Let $X$ and $Y$ be vector spaces over a subfield of $\C$.
Let $C$ be a symmetric subset of $X$.
Let $T : X \to Y$ be a linear transformation.
Then $\map T C$ is a symmetric subset of $Y$.
== Proof ==
Let $y \in \map T C$.
Then there exists $x \in C$ such that $y = T x$.
Then from linearity we have $-y = \map T {-x}$. 
Since $C$ is symmetric, we have $-x \in C$.
So $-y \in \map T C$.
So $\map T C$ is symmetric.
Category:Symmetric Subsets of Vector Spaces
Image of Translation of Set under Linear Transformation is Translation of Image

== Theorem ==
Let $K$ be a field.
Let $X$ and $Y$ be vector spaces over $K$. 
Let $T : X \to Y$ be a linear transformation.
Let $E \subseteq X$ be a non-empty set.
Let $x \in X$.
Then: 
:$\map T {E + x} = \map T E + T x$
where $E + x$ denotes the translation of $E$ by $x$.
== Proof ==
We have: 
:$y \in \map T {E + x}$
:
:$y = \map T {u + x}$ for some $u \in E$.
From the linearity of $T$, this is equivalent to: 
:$y = T u + T x$ for some $u \in E$.
This is equivalent to:
:$y \in \map T E + T x$
So by the definition of set equality we have:
:$\map T {E + x} = \map T E + T x$
Category:Linear Transformations
Category:Translation of Subsets of Vector Spaces
Image of Transpose of Linear Transformation is Annihilator of Kernel

== Theorem ==
Let $G$ and $H$ be $n$-dimensional vector spaces over a field.
Let $\map \LL {G, H}$ be the set of all linear transformations from $G$ to $H$.
Let $u \in \map \LL {G, H}$.
Let $u^\intercal$ be the transpose of $u$.
Then:
:The image of $u^\intercal$ is the annihilator of $\map \ker u$.
where $\map \ker u$ denotes the kernel of $u$.
== Proof ==
Let $x \in \map \ker u$.
Let $H^*$ be the algebraic dual of $H$.
Let $\innerprod x t$ be the evaluation linear transformation.
Then:
:$\forall y \in H^*: \innerprod x {\map {u^\intercal} y} = \innerprod {\map u x} y = \innerprod 0 y = 0$
So:
:$\map {u^\intercal} {H^*} \subseteq \paren {\map \ker u}^\circ$
From Rank Plus Nullity Theorem and Results Concerning Annihilator of Vector Subspace:
Hence:
:$\map {u^\intercal} {H^*} = \paren {\map \ker u}^\circ$
== Sources ==
* : Chapter $\text V$: Vector Spaces: $\S 28$. Linear Transformations: Theorem $28.12$
Category:Linear Algebra
Image of Ultrafilter is Ultrafilter

== Theorem ==
Let $X, Y$ be sets.
Let $f: X \to Y$ be a mapping.
Let $\FF$ be an ultrafilter on $X$.
Then the image filter $f \sqbrk \FF$ is an ultrafilter on $Y$.
== Proof 1 ==
== Proof 2 ==
Category:Image of Ultrafilter is Ultrafilter
Category:Ultrafilters
Category:Filter Theory
Image of Ultrafilter is Ultrafilter/Proof 1

== Theorem ==
== Proof ==
From Image Filter is Filter, we have that $\FF$ is a filter on $Y$.
Let $\GG$ be a filter on $Y$ such that $f \sqbrk \FF \subseteq \GG$.
We have to show that $f \sqbrk \FF = \GG$.
 there exists $U \in \GG$ such that $U \notin f \sqbrk \FF$.
By the definition of $f \sqbrk \FF$ this implies that $f^{-1} \sqbrk U \not \in \FF$.
By definition of ultrafilter:
:$V := X \setminus f^{-1} \sqbrk U \in \FF$
Because $V \subseteq f^{-1} \sqbrk {f \sqbrk V}$ it follows that:
:$f^{-1} \sqbrk {f \sqbrk V} \in \FF$
Thus also:
:$f \sqbrk V \in f \sqbrk \FF$
By assumption we have:
:$f \sqbrk \FF \subseteq \GG$
thus:
:$f \sqbrk V \in \GG$
But:
:$f \sqbrk V \cap U = f \sqbrk {X \setminus f^{-1} \sqbrk U} \cap U = \O \notin \GG$
Thus $\GG$ is not a filter, which is a contradiction to our assumptions.
Thus there can be no $U \in \GG$ such that $U \notin \FF$.
Therefore:
:$f \sqbrk \FF = \GG$
Hence $f \sqbrk \FF$ is an ultrafilter.
Category:Image of Ultrafilter is Ultrafilter
Image of Ultrafilter is Ultrafilter/Proof 2

== Theorem ==
== Proof ==
From Image Filter is Filter, we have that $\FF$ is a filter on $Y$.
In view of definition of ultrafilter, we need to show that:
:for every $A \subseteq Y$, either $A \in f \sqbrk \FF$ or $\relcomp Y A \in f \sqbrk \FF$
Let $A \subseteq Y$ be arbitrary.
Suppose $A \not \in f \sqbrk \FF$.
Then, by definition of image filter:
:$f^{-1} \sqbrk A \not \in \FF$
By definition of ultrafilter:
:$\relcomp X {f^{-1} \sqbrk A} \in \FF$
As Complement of Preimage equals Preimage of Complement:
:$\relcomp X {f^{-1} \sqbrk A} = f^{-1} \sqbrk {\relcomp Y A}$
Thus:
:$f^{-1} \sqbrk {\relcomp Y A} \in \FF$
That is, by definition of image filter:
:$\relcomp Y A \in f \sqbrk \FF$
Category:Image of Ultrafilter is Ultrafilter
Image of Union of Nest of Mappings is Union of Class of Images

== Theorem ==
Let $N$ be a nest of mappings.
Let $\bigcup N$ denote the union of $N$.
Then:
:$\Img {\bigcup N} = \ds \bigcup_{f \mathop \in N} \Img f$
where $\Img f$ denotes the image of $f$.
== Proof ==
== Sources ==
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 1$ A few preliminaries: Proposition $1.4$
Category:Image of Union of Nest of Mappings is Union of Class of Images
Category:Nests of Mappings
Image of Union of Nest of Mappings is Union of Class of Images/Proof

== Theorem ==
== Proof ==
From Union of Nest of Mappings is Mapping we have that $\bigcup N$ is a mapping.
Let $y \in \Img {\bigcup N}$.
Then by definition of mapping:
:$\exists \tuple {x, y} \in \bigcup N$
Then by definition of union of class:
:$\exists f \subseteq \bigcup N: \tuple {x, y} \in f$
Hence:
:$\exists f \subseteq \bigcup N: y \in \Img f$
That is:
:$y \in \ds \bigcup_{f \mathop \in N} \Img f$
That is:
:$\Img {\bigcup N} \subseteq \ds \bigcup_{f \mathop \in N} \Img f$
Let $y \in \ds \bigcup_{f \mathop \in N} \Img f$.
Then:
:$\exists f \subseteq \bigcup N: y \in \Img f$
Then by definition of mapping:
:$\exists f \subseteq \bigcup N: \tuple {x, y} \in f$
Thus by definition of union of class:
:$\exists \tuple {x, y} \in \bigcup N$
It follows that:
:$y \in \Img {\bigcup N}$
That is:
:$\ds \bigcup_{f \mathop \in N} \Img f \subseteq \Img {\bigcup N}$
Hence by definition of set equality:
:$\Img {\bigcup N} = \ds \bigcup_{f \mathop \in N} \Img f$
== Sources ==
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 1$ A few preliminaries: Exercise $1.1$
Category:Image of Union of Nest of Mappings is Union of Class of Images
Image of Union under Mapping

== Theorem ==
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Let $A$ and $B$ be subsets of $S$.
Then:
:$f \sqbrk {A \cup B} = f \sqbrk A \cup f \sqbrk B$
This can be expressed in the language and notation of direct image mappings as:
:$\forall A, B \in \powerset S: \map {f^\to} {A \cup B} = \map {f^\to} A \cup \map {f^\to} B$
=== General Result ===
=== Family of Sets ===
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Preimage of Union under Mapping
* Image of Intersection under Mapping
* Preimage of Intersection under Mapping
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.13 \ \text{(a)}$
* : Chapter $1$: Mappings: $\S 12 \alpha$
* : $\S 1.10$: Functions: Exercise $5 \ \text{(e)}$
* : $1$: Set Theory and Logic: $\S 2$: Functions: Exercise $2.2 \ \text{(f)}$
* : $\S 2$
Category:Images
Category:Set Union
Category:Image of Union under Mapping
Image of Union 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:
:$\ds f \sqbrk {\bigcup_{i \mathop \in I} S_i} = \bigcup_{i \mathop \in I} f \sqbrk {S_i}$
where $\ds \bigcup_{i \mathop \in I} S_i$ denotes the union of $\family {S_i}_{i \mathop \in I}$.
This can be expressed in the language and notation of direct image mappings as:
:$\ds \map {f^\to} {\bigcup_{i \mathop \in I} S_i} = \bigcup_{i \mathop \in I} \map {f^\to} {S_i}$
== Proof ==
As $f$, being a mapping, is also a relation, we can apply Image of Union under Relation: Family of Sets:
:$\ds \RR \sqbrk {\bigcup_{i \mathop \in I} S_i} = \bigcup_{i \mathop \in I} \RR \sqbrk {S_i}$
== Sources ==
* : $\S 10$: Inverses and Composites
* : $\S 1.3$: Functions and mappings. Images and preimages: Remark $2$
* : $\S 1.12$: Set Inclusions for Image and Inverse Image Sets: Theorem $12.6 \ \text{(a)}$
* : $\S 6$. Indexed families; partitions; equivalence relations: Exercise $1$
* : Chapter $1$: Theory of Sets: $\S 6$: Functions: Exercise $3 \ \text{(a})$
* : Notation and Terminology
* : Appendix $\text{A}.3$: Functions: Problem $\text{A}.3.1$
Category:Image of Union under Mapping
Image of Union 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:
:$\ds f \sqbrk {\bigcup \mathbb S} = \bigcup_{X \mathop \in \mathbb S} f \sqbrk X$
This can be expressed in the language and notation of direct image mappings as:
:$\ds \forall \mathbb S \subseteq \powerset S : \map {f^\to} {\bigcup \mathbb S} = \bigcup_{X \mathop \in \mathbb S} \map {f^\to} X$
== Proof ==
As $f$, being a mapping, is also a relation, we can apply Image of Union under Relation: General Result:
:$\ds \RR \sqbrk {\bigcup \mathbb S} = \bigcup_{X \mathop \in \mathbb S} \RR \sqbrk X$
== Sources ==
* : Chapter $0$: Functions
* : Part $1$: Not Entirely Naive Set Theory: Chapter $1$: Pairs, Relations, and Functions: Exercise $7 \ \text {(a)}$
* : $1$: Set Theory and Logic: $\S 2$: Functions: Exercise $2.3$
Category:Image of Union under Mapping
Image of Union under Mapping/Proof 1

== Theorem ==
== Proof ==
First we have:
Then:
Thus we have:
== Sources ==
* : $\S 1.3$: Functions and mappings. Images and preimages: Theorem $3$
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.1: \ \text{(ii)}$
* : $\S 21.4 \ \text{(ii)}$: The image of a subset of the domain; surjections
Category:Image of Union under Mapping
Image of Union under Mapping/Proof 2

== Theorem ==
== Proof ==
As $f$, being a mapping, is also a relation, we can apply Image of Union under Relation:
:$\RR \sqbrk {A \cup B} = \RR \sqbrk A \cup \RR \sqbrk B$
== Sources ==
* : Chapter $0$: Functions
Category:Image of Union under Mapping
Image of Union 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:
:$\RR \sqbrk {S_1 \cup S_2} = \RR \sqbrk {S_1} \cup \RR \sqbrk {S_2}$
That is, the image of the union of subsets of $S$ is equal to the union of their images.
=== General Result ===
=== Family of Sets ===
== Proof ==
== Also see ==
* Image of Intersection under Relation
* Preimage of Intersection under Relation
* Preimage of Union under Relation
== 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 $12.2 \ \text{(a)}$
Category:Relation Theory
Category:Set Union
Category:Image of Union under Relation
Image of Union 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:
:$\ds \RR \sqbrk {\bigcup_{i \mathop \in I} S_i} = \bigcup_{i \mathop \in I} \RR \sqbrk {S_i}$
where $\ds \bigcup_{i \mathop \in I} S_i$ denotes the union 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 $12.5 \ \text{(a)}$
Category:Image of Union under Relation
Category:Indexed Families
Image of Union 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:
:$\ds \RR \sqbrk {\bigcup \mathbb S} = \bigcup_{X \mathop \in \mathbb S} \RR \sqbrk X$
== Proof ==
Category:Image of Union under Relation
Image of Vector Subspace under Linear Transformation is Vector Subspace

== Theorem ==
Let $K$ be a field.
Let $X$ and $Y$ be vector spaces over $K$. 
Let $U$ be a vector subspace of $X$.
Let $T : X \to Y$ be a linear transformation.
Then $T \sqbrk U$ is a vector subspace of $Y$. 
== Proof ==
Since $U$ is a non-empty set, we can apply the One-Step Vector Subspace Test to $T \sqbrk U$.
Let $\lambda \in K$ and $u, v \in T \sqbrk U$.
Then there exists $x, y \in U$ such that: 
:$u = T x$
and:
:$v = T y$ 
Then: 
Since $U$ is a vector subspace, we have $\lambda x + y \in U$.
So $\map T {\lambda x + y} \in T \sqbrk U$.
That is, $\lambda u + v \in T \sqbrk U$.
So $T \sqbrk U$ is a vector subspace of $Y$.
Category:Vector Subspaces
Category:Linear Transformations
Image of Weakly Convergent Sequence under Compact Linear Transformation is Convergent

== Theorem ==
Let $\struct {X, \norm \cdot_X}$ and $\struct {Y, \norm \cdot_Y}$ be normed vector spaces.
Let $T : X \to Y$ be a compact linear transformation.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a weakly convergent sequence with:
:$x_n \weakconv x$
Then: 
:$T x_n \to T x$
in the strong sense.
== Proof ==
Let $X^\ast$ and $Y^\ast$ be the normed dual spaces of $X$ and $Y$ respectively.
Let $f \in Y^\ast$.
Then $f \circ T \in X^\ast$.
Since $x_n \weakconv x$, we have: 
:$\map f {T x_n} \to \map f {T x}$ as $n \to \infty$. 
So we have: 
:$T x_n \weakconv T x$ as $n \to \infty$. 
 that $\sequence {T x_n}_{n \in \N}$ does not converge to $T x$. 
Then there exists $\epsilon > 0$ such that: 
:$\norm {T x_{n_k} - T x}_Y \ge \epsilon$
for all $k \in \N$, for a subsequence $\sequence {x_{n_k} }_{k \in \N}$.
From Weakly Convergent Sequence in Normed Vector Space is Bounded, $\sequence {x_n}_{n \in \N}$ and hence $\sequence {x_{n_k} }_{k \in \N}$ is bounded.
Since $T$ is compact, there exists a subsequence $\sequence {T x_{n_{k_j} } }_{j \in \N}$ such that: 
:$T x_{n_{k_j} } \to z \in Y$
From Convergent Sequence in Normed Vector Space is Weakly Convergent, we have: 
:$T x_{n_{k_j} } \weakconv z$
Since: 
:$T x_n \weakconv T x$ 
we have that $z = T x$ from Weak Limit in Normed Vector Space is Unique.
But then there exists $N \in \N$ such that: 
:$\norm {T x_{n_{k_j} } - T x}_Y < \epsilon$
for $j \ge N$. 
This contradicts that: 
:$\norm {T x_{n_k} - T x}_Y \ge \epsilon$
for each $k \in \N$. 
So, we have $T x_n \to T x$. 
Category:Weak Convergence (Normed Vector Spaces)
Category:Compact Linear Transformations