nyuuzyou/wikis huggingface dataset:
Image Filter is Filter

== Theorem ==
Let $X, Y$ be sets.
Let $\powerset X$ and $\powerset Y$ be the power sets of $X$ and $Y$ respectively.
Let $f: X \to Y$ be a mapping.
Let $\FF \subset \powerset X$ be a filter on $X$.
Then the image filter of $\FF$ $f$:
:$f \sqbrk \FF := \set {U \subseteq Y: f^{-1} \sqbrk U \in \FF}$
is a filter on $Y$.
== Proof ==
From the definition of a filter we have to prove four things:
: $(1): \quad f \sqbrk \FF \subset \powerset Y$
: $(2): \quad Y \in f \sqbrk \FF, \O \notin f \sqbrk \FF$
: $(3): \quad U, V \in f \sqbrk \FF \implies U \cap V \in f \sqbrk \FF$
: $(4): \quad U \in f \sqbrk \FF, U \subseteq V \subseteq Y \implies V \in f \sqbrk \FF$
By construction we have:
:$f \sqbrk \FF \subseteq \powerset Y$
Since $f^{-1} \sqbrk \O = \O \notin \FF$ we know that:
:$\O \notin f \sqbrk \FF$
Therefore:
:$f \sqbrk \FF \ne \powerset Y$
which implies $(1)$.
Because $f^{-1} \sqbrk Y = X \in \FF$, we have:
:$Y \in f \sqbrk \FF$
Since we've already shown $\O \notin f \sqbrk \FF$, this implies $(2)$.
Let $U, V \in f \sqbrk \FF$.
From Preimage of Intersection under Mapping:
:$f^{-1} \sqbrk {U \cap V} = f^{-1} \sqbrk U \cap f^{-1} \sqbrk V \in \FF$
(since $\FF$ is a filter).
Thus $U \cap V \in f \sqbrk \FF$, and so $(3)$ holds.
Finally, let $U \in f \sqbrk \FF$ and $V \subseteq Y$ such that $U \subseteq V$.
Then from Preimage of Subset is Subset of Preimage:
:$f^{-1} \sqbrk U \subseteq f^{-1} \sqbrk V$
Since $f^{-1} \sqbrk U \in \FF$ and $\FF$ is a filter it follows that:
:$f^{-1} \sqbrk V \in \FF$
which implies:
:$V \in f \sqbrk \FF$
and thus $(4)$.
Category:Set Theory
Category:Filter Theory
Image is G-Module

== Theorem ==
Let $\struct {G, \cdot}$ be a group.
Let $f: \struct {V, \phi} \to \struct {V', \mu}$ be a $G$-Module Homomorphism.
Then $\Img f$ is a $G$-Submodule of $V'$.
== Proof ==
From $G$-Submodule Test it suffices to prove that $\map \mu {G, \Img f} \subseteq \Img f$.
In other words: for any $g \in G$ and $w \in \Img f$, it is to be shown that $\map \mu {g, w} \in \Img f$.
Assume that $g \in G$ and $w \in \Img f$.
Then:
:$\exists v \in V: \map f v = w$
By definition of homomorphism:
:$\map \mu {g, w} = \map \mu {g, \map f v} = \map f {\map \phi {g, v} }$
Hence:
:$\forall g \in G: \forall w \in \Img f: \map \mu {g, w} \in \Img f$
By $G$-Submodule Test, it follows that $\Img f$ is a $G$-Submodule of $V'$.
Category:Representation Theory
Image is Subset Iff Subset of Preimage

== Theorem ==
Let $S, T$ be sets.
Let $f: S \to T$ be a mapping.
Then:
:$\forall A \subseteq S, B \subseteq T : f \sqbrk A \subseteq B$ $A \subseteq f^{-1} \sqbrk B$ 
== Proof ==
=== Necessary Condition ===
We have:
=== Sufficient Condition ===
We have:
Category:Images
Category:Preimages
Image is Subset of Codomain

== Theorem ==
Let $\RR = S \times T$ be a relation.
For all subsets $A$ of the domain of $\RR$, the image of $A$ is a subset of the codomain of $\RR$:
:$\forall A \subseteq \Dom \RR: \RR \sqbrk A \subseteq T$
In the notation of direct image mappings, this can be written as:
:$\forall A \in \powerset S: \map {\RR^\to} A \in \powerset T$
=== Corollary 1 ===
These results also hold for mappings:
=== Corollary 2 ===
=== Corollary 3 ===
== Proof ==
Category:Relation Theory
Image is Subset of Codomain/Corollary 2

== Corollary of Image is Subset of Codomain ==
Let $f: S \to T$ be a mapping.
For all subsets $A$ of the domain $S$, the image of $A$ is a subset of the codomain of $f$:
:$\forall A \subseteq S: f \sqbrk A \subseteq T$
== Proof ==
As a mapping is by definition also a relation, the result follows immediately from Image is Subset of Codomain.
Category:Images
Image is Subset of Codomain/Corollary 3

== Corollary to Image is Subset of Codomain ==
Let $f: S \to T$ be a mapping.
The image of $f$ is a subset of the codomain of $f$:
:$\Img f \subseteq T$
== Proof ==
We have that a mapping is by definition also a relation.
The result follows from Image is Subset of Codomain: Corollary 1.
== Sources ==
* 
* 
Category:Images
Image of *-Algebra under *-Algebra Homomorphism is *-Subalgebra

== Theorem ==
Let $\struct {A, \ast}$ and $\struct {B, \square}$ be a $\ast$-algebras over $\C$.
Let $\phi : A \to B$ be a $\ast$-algebra homomorphism.
Then $\phi \sqbrk A$ is a $\ast$-subalgebra of $B$. 
== Proof ==
From Image of Algebra under Algebra Homomorphism is Subalgebra, $\phi \sqbrk A$ is a subalgebra of $B$. 
Further, for $a \in A$ we have:
:$\paren {\map \phi a}^\square = \map \phi {a^\ast} \in \phi \sqbrk A$
So:
:$x^\ast \in \phi \sqbrk A$ for $x \in \phi \sqbrk A$.
Hence $\phi \sqbrk A$ is a $\ast$-subalgebra of $B$.
Category:*-Algebras
Image of Absorbing Set under Surjective Linear Transformation is Absorbing

== Theorem ==
Let $\GF \in \set {\R, \C}$.
Let $X$ and $Y$ be vector spaces over $\GF$. 
Let $A \subseteq X$ be an absorbing set.
Let $T : X \to Y$ be a surjective linear transformation. 
Then $T \sqbrk A$ is an absorbing set.
== Proof ==
Let $y \in Y$.
Since $T$ is surjective, there exists $x \in X$ such that $y = T x$.
Since $A$ is absorbing, there exists $t \in \R_{> 0}$ such that:
:$x \in \alpha A$ for $\alpha \in \C$ with $\cmod \alpha \ge t$.
Then:
:$y = T x \in T \sqbrk {\alpha A}$ for $\alpha \in \C$ with $\cmod \alpha \ge t$. 
From Image of Dilation of Set under Linear Transformation is Dilation of Image, we have:
:$y \in \alpha T \sqbrk A$ for $\alpha \in \C$ with $\cmod \alpha \ge t$. 
So $T \sqbrk A$ is absorbing.
Category:Absorbing Sets
Image of Algebra under Algebra Homomorphism is Subalgebra

== Theorem ==
Let $R$ be a ring.
Let $A, B$ be $R$-algebras.
Let $\phi : A \to B$ be an $R$-algebra homomorphism.
Then $\phi \sqbrk A$ is a subalgebra of $B$. 
== Proof ==
From Image of Submodule under Linear Transformation is Submodule, $\phi \sqbrk A$ is a submodule of $B$. 
Let $a, b \in A$. 
We then have: 
:$\map \phi a \map \phi b = \map \phi {a b} \in \phi \sqbrk A$
since $\phi$ is a $R$-algebra homomorphism.
Hence $\phi \sqbrk A$ is a subalgebra of $B$.
Category:Algebras
Image of Balanced Set under Linear Transformation is Balanced

== Theorem ==
Let $\Bbb F \in \set {\R, \C}$.
Let $X$ and $Y$ be vector spaces over $\Bbb F$.
Let $E \subseteq X$ be balanced.
Let $T : X \to Y$ be a linear transformation.
Then $\map T E \subseteq Y$ is balanced.
== Proof ==
We aim to show that for all $s \in \R$ with $\cmod s \le 1$, we have: 
:$s \map T E \subseteq \map T E$
Let $y \in s \map T E$.
Then there exists $x \in E$ such that $y = s T x$.
From the linearity of $T$, we have: 
:$y = \map T {s x}$
Since $x \in E$, we have $s x \in s E$ from the definition of dilation of $E$ by $s$.
Since $E$ is balanced, we have $s E \subseteq E$ and so: 
:$s x \in E$
giving:
:$y = \map T {s x} \in \map T E$
Since $y \in s \map T E$ was arbitrary, we have $s \map T E \subseteq \map T E$, giving that $\map T E$ is balanced.
Category:Balanced Sets
Category:Linear Transformations
Image of Bijection on Set is Set with Axiom of Choice implies Axiom of Replacement

== Theorem ==
Let the Axiom of Choice be accepted.
Let it be accepted as axiomatic that:
:every class which can be put into one-to-one correspondence with a set.
Then the Axiom of Replacement holds:
== Proof ==
== Sources ==
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 3$ The axiom of substitution: Exercise $3.2$
Category:Axiom of Replacement
Category:Axiom of Choice
Image of Bounded Linear Transformation is Everywhere Dense iff Dual Operator is Injective

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $X$ and $Y$ be normed vector spaces over $\GF$. 
Let $X^\ast$ and $Y^\ast$ be the normed dual spaces of $X$ and $Y$ respectively. 
Let $T : X \to Y$ be a bounded linear transformation.
Let $T^\ast : Y^\ast \to X^\ast$ be the dual operator of $T$. 
Then $T \sqbrk X$ is everywhere dense in $Y$ $T^\ast$ is injective.
== Proof 1 ==
== Proof 2 ==
Category:Dual Operators
Category:Image of Bounded Linear Transformation is Everywhere Dense iff Dual Operator is Injective
Image of Bounded Linear Transformation is Everywhere Dense iff Dual Operator is Injective/Proof 1

== Theorem ==
== Proof ==
=== Necessary Condition ===
Suppose that $T \sqbrk X$ is everywhere dense in $Y$.
Let $f \in Y^\ast$ be such that $T^\ast f = 0$.
That is: 
:$\map f {T x} = 0$
for all $x \in X$. 
So:
:$\map f y = 0$ 
for all $y \in T \sqbrk X$.
Since $T \sqbrk X$ is everywhere dense in $Y$, $f$ is continuous and $\GF$ is Hausdorff, we have that: 
:$\map f y = 0$ for all $y \in Y$
from Continuous Mappings into Hausdorff Space coinciding on Everywhere Dense Set coincide.
So $T^\ast f = 0$ implies $f = 0$. 
So $T^\ast$ is injective from Linear Transformation is Injective iff Kernel Contains Only Zero.
=== Sufficient Condition ===
Suppose that $T \sqbrk X$ is not everywhere dense in $Y$.
Then $\map \cl {T \sqbrk X}$ is a proper closed linear subspace of $Y$.
From Existence of Distance Functional, there exists $f \in Y^\ast$ such that $f \ne 0$ and: 
:$\map f y = 0$ for all $y \in \map \cl {T \sqbrk X}$
In particular: 
:$\map f y = 0$ for all $y \in T \sqbrk X$
so that: 
:$\map f {T x} = 0$ for all $x \in X$.
Hence, we have: 
:$T^\ast f = 0$
while $f \ne 0$. 
So from Linear Transformation is Injective iff Kernel Contains Only Zero, we conclude that $T^\ast$ is not injective.
Hence by Proof by Contraposition, if $T^\ast$ is injective then $T \sqbrk X$ is everywhere dense in $Y$.
Category:Image of Bounded Linear Transformation is Everywhere Dense iff Dual Operator is Injective
Image of Bounded Linear Transformation is Everywhere Dense iff Dual Operator is Injective/Proof 2

== Theorem ==
== Proof ==
From Annihilator of Image of Bounded Linear Transformation is Kernel of Dual Operator, we have:
:$T \sqbrk X^\bot = \map \ker {T^\ast}$
where $T \sqbrk X^\bot$ denotes the annihilator of $T \sqbrk X$.
From Linear Transformation is Injective iff Kernel Contains Only Zero, we then have that $T^\ast$ is injective :
:$T \sqbrk X^\bot = \set { {\mathbf 0}_{Y^\ast} }$
From Annihilator of Subspace of Banach Space is Zero iff Subspace is Everywhere Dense, this is equivalent to:
:$T \sqbrk X$ is everywhere dense in $Y$.
This was the demand.
Category:Image of Bounded Linear Transformation is Everywhere Dense iff Dual Operator is Injective
Image of Canonical Injection is Kernel of Projection

== Theorem ==
Let $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$ be groups with identity elements $e_1$ and $e_2$ respectively.
Let $\struct {G_1 \times G_2, \circ}$ be the group direct product of $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$
Let:
:$\pr_1: \struct {G_1 \times G_2, \circ} \to \struct {G_1, \circ_1}$ be the first projection from $\struct {G_1 \times G_2, \circ}$ to $\struct {G_1, \circ_1}$
:$\pr_2: \struct {G_1 \times G_2, \circ} \to \struct {G_2, \circ_2}$ be the second projection from $\struct {G_1 \times G_2, \circ}$ to $\struct {G_2, \circ_2}$.
Let:
:$\inj_1: \struct {G_1, \circ_1} \to \struct {G_1 \times G_2, \circ}$ be the canonical injection from $\struct {G_1, \circ_1}$ to $\struct {G_1 \times G_2, \circ}$
:$\inj_2: \struct {G_2, \circ_2} \to \struct {G_1 \times G_2, \circ}$ be the canonical injection from $\struct {G_2, \circ_2}$ to $\struct {G_1 \times G_2, \circ}$.
Then:
:$(1): \quad \Img {\inj_1} = \map \ker {\pr_2}$
:$(2): \quad \Img {\inj_2} = \map \ker {\pr_1}$
That is:
:the image of the (first) canonical injection is the kernel of the second projection
:the image of the (second) canonical injection is the kernel of the first projection.
== Proof ==
The canonical injection $\inj_1: \struct {G_1, \circ_1} \to \struct {G_1 \times G_2, \circ}$ is defined as:
:$\forall x \in G_1: \map {\inj_1} x = \tuple {x, e_2}$
Thus:
:$\Img {\inj_1} = \set {\tuple {x, e_2}: x \in G_1}$
The second projection $\pr_2: \struct {G_1 \times G_2, \circ} \to \struct {G_2, \circ_2}$ is defined as:
:$\forall \tuple {x, y} \in G_1 \times G_2: \map {\pr_2} {x, y} = y$
Thus by definition of kernel:
:$\map \ker {\pr_2} = \map {\pr_2^{-1} } {e_2} = \set {\tuple {x, e_2}: x \in G_1}$
As can be seen:
:$\Img {\inj_1} = \map \ker {\pr_2}$
Similarly:
:$\Img {\inj_2} = \set {\tuple {e_1, y}: y \in G_2}$
:$\map \ker {\pr_1} = \map {\pr_1^{-1} } {e_1} = \set {\tuple {e_1, y}: y \in G_2}$
Hence the result.
== Sources ==
* : Chapter $\text{II}$: Groups: Direct Products
Category:Group Homomorphisms
Category:Group Direct Products
Image of Canonical Injection is Normal Subgroup

== Theorem ==
Let $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$ be groups with identity elements $e_1$ and $e_2$ respectively.
Let $\struct {G_1 \times G_2, \circ}$ be the group direct product of $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$.
Let:
:$\inj_1: \struct {G_1, \circ_1} \to \struct {G_1 \times G_2, \circ}$ be the canonical injection from $\struct {G_1, \circ_1}$ to $\struct {G_1 \times G_2, \circ}$
:$\inj_2: \struct {G_2, \circ_2} \to \struct {G_1 \times G_2, \circ}$ be the canonical injection from $\struct {G_2, \circ_2}$ to $\struct {G_1 \times G_2, \circ}$.
Then:
:$(1): \quad \Img {\inj_1} \lhd \struct {G_1 \times G_2, \circ}$
:$(2): \quad \Img {\inj_2} \lhd \struct {G_1 \times G_2, \circ}$
That is, the images of the canonical injections are normal subgroups of the group direct product of $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$.
== Proof ==
From Image of Canonical Injection is Kernel of Projection:
:$\Img {\inj_1} = \map \ker {\pr_2}$
:$\Img {\inj_2} = \map \ker {\pr_1}$
That is:
:the image of the (first) canonical injection is the kernel of the second projection
:the image of the (second) canonical injection is the kernel of the first projection.
The domain of the projections is $G_1 \times G_2$, by definition.
The result follows from Kernel is Normal Subgroup of Domain.
== Sources ==
* : Chapter $\text{II}$: Groups: Direct Products
Category:Normal Subgroups
Category:Group Homomorphisms
Category:Group Direct Products
Image of Cauchy Sequence in Topological Vector Space is von Neumann-Bounded

== Theorem ==
Let $\struct {X, \tau}$ be a topological vector space. 
Let $\sequence {x_n}_{n \in \N}$ be a Cauchy sequence. 
Let:
:$E = \set {x_n : n \in \N}$
Then $E$ is von Neumann-bounded. 
== Proof ==
Let $W$ be an open neighborhood of ${\mathbf 0}_X$.
From Open Neighborhood of Origin in Topological Vector Space contains Balanced Open Neighborhood, take a balanced open neighborhood of ${\mathbf 0_X}$, $V \subseteq W$. 
From , there exists a symmetric open neighborhood $U$ of ${\mathbf 0}_X$ such that: 
:$U + U \subseteq V$
From Open Neighborhood of Origin in Topological Vector Space contains Balanced Open Neighborhood, we can take $U$ to be balanced.
Since $\sequence {x_n}_{n \in \N}$ is Cauchy, there exists $N \in \N$ such that: 
:$x_n \in x_N + U$
for $n \ge N$.
From Topological Vector Space as Union of Dilations of Open Neighborhood of Origin, there exists $M \in \N$ such that: 
:$x_N \in k U$
Then we have: 
:$x_n \in x_N + U \subseteq k U + U \subseteq k \paren {U + U} \subseteq k V$
for $n \ge N$. 
Applying Topological Vector Space as Union of Dilations of Open Neighborhood of Origin again, for each $1 \le j \le N - 1$ we have $r_j \in \N$ such that $x_j \in r_j V$. 
Taking $K = \max \set {r_1, \ldots, r_{N - 1}, k}$, we have: 
:$x_n \in K V$ for all $n \in \N$.
Since $V$ is balanced, we have that: 
:$x_n \in s V$ for all $s > K$. 
That is: 
:$x_n \in s W$ for all $s > K$. 
Category:Cauchy Sequences in Topological Vector Spaces
Category:Von Neumann-Bounded Subsets of Topological Vector Spaces
Image of Class under Mapping is Image of Restriction of Mapping to Class

== Theorem ==
Let $V$ be a basic universe
Let $f: V \to V$ be a mapping.
Let $A$ be a class.
Let $f \sqbrk A$ denote the image of $A$ under $f$.
Then $f \sqbrk A$ is the image of the restriction of $f$ to $A$:
:$f \sqbrk A = \Img {f {\restriction} A}$
== Proof ==
By definition, $f {\restriction} A$ is class of all ordered pairs $\tuple {a, \map f a}$, where $a \in A$.
The result follows from Restriction of Mapping is Subset of Cartesian Product.
== Sources ==
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 1$ A few preliminaries
Category:Images
Category:Restrictions
Image of Closed Real Interval is Bounded

== Theorem ==
Let $f$ be a real function which is continuous on the closed interval $\closedint a b$.
Then $f$ is bounded on $\closedint a b$.
== Proof ==
Suppose $f$ is not bounded on $\closedint a b$.
Then from the corollary to Limit of Sequence to Zero Distance Point, there exists a sequence $\sequence {x_n}$ in $\closedint a b$ such that $\size {\map f {x_n} } \to +\infty$ as $n \to \infty$.
Since $\closedint a b$ is a closed interval, from Convergent Subsequence in Closed Interval, $\sequence {x_n}$ has a subsequence $\sequence {x_n}$ which converges to some $\xi \in \closedint a b$.
Because $f$ is continuous on $\closedint a b$, it follows from Limit of Image of Sequence that $\map f {x_{n_r} } \to \map f \xi$ as $r \to \infty$.
But this contradicts our supposition that there exists a sequence $\sequence {x_n}$ in $\closedint a b$ such that $\size {\map f {x_n} } \to +\infty$ as $n \to \infty$.
The result follows.
== Sources ==
* : $5$: Compact spaces: $5.1$: Motivation: Proposition $5.1.1$
* : $\S 9.11$
* 
Category:Analysis
Image of Closed Set under Linear Isometry from Banach Space to Normed Vector Space is Closed

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a Banach space over $\GF$. 
Let $\struct {Y, \norm {\, \cdot \,}_Y}$ be a normed vector space over $\GF$. 
Let $T : X \to Y$ be a linear isometry.
Let $A$ be a closed subset of $X$.
Then $T \sqbrk A$ is closed in $X$.
== Proof ==
Let $\sequence {y_n}_{n \mathop \in \N}$ be a convergent sequence in $T \sqbrk A$ with:
:$y_n \to y$ as $n \to \infty$. 
From Convergent Sequence is Cauchy Sequence, $\sequence {y_n}_{n \mathop \in \N}$ is a Cauchy sequence. 
Then there exists $x_n \in X$ with $y_n = T x_n$.
Let $\epsilon > 0$.
Let $N \in \N$ be such that:
:$\norm {y_n - y_m}_Y = \norm {\map T {x_n - x_m} } < \epsilon$ for $n, m \ge N$.
Since $T$ is a linear isometry, we therefore have:
:$\norm {x_n - x_m} < \epsilon$ for $n, m \ge N$.
So $\sequence {x_n}_{n \mathop \in \N}$ is a Cauchy sequence in $X$.
Hence there exists $x \in X$ such that $x_n \to x$.
Since $A$ is closed in $X$, we further have $x \in A$ from Subset of Metric Space contains Limits of Sequences iff Closed.
Since $T$ is continuous, we have $T x_n \to T x$. 
From Limit in Normed Vector Space is Unique, we have $y = T x$ with $x \in A$.
Hence $y \in T \sqbrk A$. 
Hence every convergent sequence in $T \sqbrk A$ has its limit in $T \sqbrk A$.
Hence $T \sqbrk A$ is closed.
Category:Linear Isometries
Category:Banach Spaces
Image of Closure Operator Inherits Infima

== Theorem ==
Let $L = \struct {S, \preceq}$ be an ordered set.
Let $f$ be a closure operator on $L$.
Then $R = \struct {f \sqbrk S, \precsim}$ inherits infima,
where
:$\mathord \precsim = \mathord \preceq \cap \paren {f \sqbrk S \times f \sqbrk S}$
:$f \sqbrk S$ denotes the image of $f$.
== Proof ==
Let $X$ be subset of $f \sqbrk S$ such that
:$X$ admits an infimum in $L$.
By Closure Operator does not Change Infimum of Subset of Image:
:$\map f {\inf_L X} = \inf_L X$
By definition of image of mapping:
:$\inf_L X \in f \sqbrk S$
Thus by Infimum in Ordered Subset:
:$X$ admits an infimum in $R$ and $\inf_R X = \inf_L X$
== Sources ==
* 
* 
Category:Closure Operators
Image of Compact Subset under Directed Suprema Preserving Closure Operator

== Theorem ==
Let $L = \struct {S, \preceq}$ be a bounded below algebric lattice.
Let $c: S \to S$ be a closure operator that preserves directed suprema.
Then:
:$c \sqbrk {\map K L} = \map K {\struct {c \sqbrk S, \precsim} }$
where
:$\map K L$ denotes the compact subset of $L$,
:$c \sqbrk S$ denotes the image of $S$ under $c$,
:$\mathord \precsim = \mathord \preceq \cap \paren {c \sqbrk S \times c \sqbrk S}$
== Proof ==
We will prove that:
:$\map K {\struct {c \sqbrk S, \precsim} } \subseteq c \sqbrk {\map K L}$
By Image of Compact Subset under Directed Suprema Preserving Closure Operator is Subset of Compact Subset:
:$c \sqbrk {\map K L} \subseteq \map K {\struct {c \sqbrk S, \precsim} }$
Thus the result by definition of set equality.
== Sources ==
* 
* 
Category:Continuous Lattices
Category:Way Below Relation
Image of Compact Subset under Directed Suprema Preserving Closure Operator is Subset of Compact Subset

== Theorem ==
Let $L = \struct {S, \preceq}$ be an algebric lattice.
Let $c: S \to S$ be a closure operator that preserves directed suprema.
Then:
:$c \sqbrk {\map K L} \subseteq \map K {\struct {c \sqbrk S, \precsim} }$
where:
:$\map K L$ denotes the compact subset of $L$
:$c \sqbrk S$ denotes the image of $S$ under $c$
:$\mathord \precsim = \mathord \preceq \cap \paren {c \sqbrk S \times \map c S}$
== Proof ==
Let $x \in c \sqbrk {\map K L}$.
By definition of image of set:
:$\exists y \in \map K L: x = \map c y$
and
:$x \in c \sqbrk S$
By definition of compact subset:
:$y$ is compact in $L$.
By definition of compact element:
:$y \ll y$
where $\ll$ denotes the way below relation.
Define $P = \struct {c \sqbrk S, \precsim}$ as an ordered subset of $L$.
We will prove that:
:for every directed subset $D$ of $c \sqbrk S$: $x \precsim \sup_P D \implies \exists d \in D: d \precsim x$
Let $D$ be a directed subset of $c \sqbrk S$.
By definition of ordered subset:
:$D$ is directed in $L$.
By definition of algebraic ordered set:
:$L$ is up-complete.
By definition of up-complete:
:$D$ admits a supremum in $L$.
By definition of image of set:
:$\sup_L D \in c \sqbrk S$
By Supremum in Ordered Subset:
:$\sup_L D = \sup_P D$
By definition of ordered subset:
:$x \preceq \sup_L D$
By definition of closure operator/inflationary:
:$y \preceq x$
By definition of transitivity:
:$y \preceq \sup_L D$
By definition of way below relation:
:$\exists d \in D: y \preceq d$
By definition of subset:
:$d \in c \sqbrk S$
By definition of closure operator/idempotent:
:$d = \map c d$
By definition of closure operator/increasing:
:$x = \map c y \preceq \map c d = d$
Thus by definition of ordered subset:
:$\exists d \in D: x \precsim d$
By definition of way below relation:
:$x \ll_P x$
By definition:
:$x$ is a compact element in $P$.
Thus by definition of compact subset:
:$x \in \map K P$
== Sources ==
* 
* 
Category:Continuous Lattices
Category:Way Below Relation
Image of Complex Exponential Function

== Theorem ==
The image of the complex exponential function is $\C \setminus \left\{ {0}\right\}$.
== Proof ==
Let $z \in \C \setminus \left\{ {0}\right\}$.
Let $r = \cmod z$ be the modulus of $z$, and let $\theta = \arg \left({z}\right)$ be the argument of $z$.
Then $r>0$.
Let $\ln$ denote the real natural logarithm, and let $e$ denote the real exponential function.
Then:
Hence, $z \in \operatorname{Im} \left({\exp}\right)$.
Suppose instead that $z=0$.
Let $z_0 = r_0 \left({\cos \theta_0 + i \sin \theta_0 }\right) \in \C$.
From Exponential Tends to Zero and Infinity, it follows that $e^{ r_0 } \ne 0$.
As $\cmod {\cos \theta_0 + i \sin \theta_0} = 1$, it follows that $\cos \theta_0 + i \sin \theta_0 \ne 0$.
Then this equation has no solutions:
:$0 = \exp z_0 = e^{ r_0 } \left({\cos \theta_0 + i \sin \theta_0 }\right)$
Hence, $\operatorname{Im} \left({\exp}\right) = \C \setminus \left\{ {0}\right\}$-
== Also see ==
* Exponential of Real Number is Strictly Positive
== Sources ==
* : $\S 1.5$
Category:Exponential Function