nyuuzyou/wikis huggingface dataset:
Continuous Image of Connected Space is Connected/Corollary 3

== Corollary to Continuous Image of Connected Space is Connected ==
Let $I = \closedint a b$ be a closed real interval.
Let $f: I \to \R$ be a continuous mapping.
Then $f$ has the intermediate value property.
== Proof ==
From Subset of Real Numbers is Interval iff Connected, $I$ is connected.
By Continuous Image of Connected Space is Connected: Corollary 2, $f \sqbrk I$ is also connected.
So from Subset of Real Numbers is Interval iff Connected, $f \sqbrk I$ is a real interval.
The result follows by definition of an real interval and the intermediate value property.
== Sources ==
* : $6.2$: Connectedness: Corollary $6.2.14$
Category:Continuous Functions
Category:Continuous Image of Connected Space is Connected
Continuous Image of Connected Space is Connected/Proof 1

== Theorem ==
== Proof ==
Let $i: f \sqbrk {T_1} \to T_2$ be the inclusion mapping.
Let $g: T_1 \to f \sqbrk {T_1}$ be the surjective restriction of $f$.
Then $f = i \circ g$.
Hence, by Continuity of Composite with Inclusion: Inclusion on Mapping, it follows that $g$ is continuous.
We use a Proof by Contradiction.
Suppose that $A \mid B$ is a partition of $f \sqbrk {T_1}$.
Then it follows that $g^{-1} \sqbrk A \mid g^{-1} \sqbrk B$ is a partition of $T_1$.
== Sources ==
* : $6.2$: Connectedness: Proposition $6.2.10$
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness: Functions and Products
Category:Continuous Image of Connected Space is Connected
Continuous Image of Connected Space is Connected/Proof 2

== Theorem ==
== Proof ==
Suppose that $S_2 = f \sqbrk {S_1}$ is not connected in $T_2$.
Then by definition there exist open sets $U_2$ and $V_2$ in $T_2$ such that:
:$S_2 \subseteq U_2 \cup V_2$
:$U_2 \cap V_2 \cap S_2 = \O$
:$U_2 \cap S_2 \ne \O$
:$V_2 \cap S_2 \ne \O$
We have that $f: T_1 \to T_2$ is continuous.
Thus $U_1 = f^{-1} \sqbrk {U_2}$ and $V_1 = f^{-1} \sqbrk {V_2}$ are open in $T_1$.
We have that:
:$U_2 \cap S_2 \ne \O$
Therefore:
:$\exists x \in S_1: \map f x \in U_2$
Then:
:$x \in f^{-1} \sqbrk {U_2} = U_1$
and:
:$x \in S_1$
so:
:$U_1 \cap S_1 \ne \O$
Similarly:
:$V_1 \cap S_1 \ne \O$
Suppose there exists $x \in S_1$ such that $x \in U_1 \cap V_1 \cap S_1$.
Then:
:$\map f x \in U_2 \cap V_2 \cap S_2$
which is a contradiction.
It follows that:
:$U_1 \cap V_1 \cap S_1 = \O$
Thus by definition $S_1$ is not connected in $T_1$.
The result follows by the Rule of Transposition.
Category:Continuous Image of Connected Space is Connected
Continuous Image of Everywhere Dense Set is Everywhere Dense

== Theorem ==
Let $A_T = \struct {A, \tau_A}$ and $B_T = \struct {B, \tau_B}$ be topological spaces.
Let $f : A \to B$ be an everywhere continuous surjection.
Let $S \subseteq A$ be everywhere dense in $A_T$.
Then, $f \sqbrk S$ is everywhere dense in $B_T$.
== Proof ==
As $\paren {f \sqbrk S}^- \subseteq B$, by set equality:
:$\paren {f \sqbrk S}^- = B$
Therefore, $f \sqbrk S$ is everywhere dense in $B$ by definition.
Category:Denseness
Continuous Image of Path-Connected Set is Path-Connected

== Theorem ==
Let $\struct {T_1, \tau_1}, \struct {T_2, \tau_2}$ be topological spaces.
Let $f: T_1 \to T_2$ be a continuous mapping.
Let $S \subseteq T_1$ be a subset of $T_1$.
Let $S$ be path-connected in $\struct {T_1, \tau_1}$.
Then $f \sqbrk S$ is path-connected in $\struct {T_2, \tau_2}$.
=== Metric Space ===
== Proof ==
Let $\map f s, \map f {s'} \in f \sqbrk S$, for some $s, s' \in S$.
Let $\mathbb I$ denote the closed unit interval:
:$\mathbb I = \closedint 0 1$
Let $p: \mathbb I \to S$ be a continuous mapping such that:
:$\map p 0 = s, \map p 1 = s'$
Such a $p$ exists since $S$ is path-connected in $\struct {T_1, \tau_1}$.
Now define $q: \mathbb I \to f \sqbrk S$ by:
:$\map q t := f \circ \map p t$
Composite of Continuous Mappings is Continuous shows that $q$ is a continuous mapping, and:
:$\map q 0 = \map f {\map p 0} = \map f s$
:$\map q 1 = \map f {\map p 1} = \map f {s'}$
Thus $\map f s$ and $\map f {s'}$ are connected by a path in $f \sqbrk S$.
Since these points were arbitrary in $f \sqbrk S$, the result follows.
== Sources ==
* : $6.3$: Path-connectedness (in passing)
Category:Path-Connected Spaces
Category:Topology
Continuous Image of Path-Connected Set is Path-Connected/Metric Space

== Theorem ==
Let $M_1, M_2$ be metric spaces whose metrics are $d_1, d_2$ respectively.
Let $f: M_1 \to M_2$ be a continuous mapping.
Let $S \subseteq M_1$ be a path-connected subspace of $M_1$.
Then $f \sqbrk S$ is a path-connected subspace of $M_2$.
== Proof ==
Let $\map f s, \map f {s'} \in f \sqbrk S$, for some $s, s' \in S$.
Let $\mathbb I$ denote the closed unit interval:
:$\mathbb I = \closedint 0 1$
Let $p: \mathbb I \to S$ be a continuous mapping such that:
:$\map p 0 = s, \map p 1 = s'$
Such a $p$ exists since $S$ is a path-connected subspace of $M_1$.
Now define $q: \mathbb I \to f \sqbrk S$ by:
:$\map q t := f \circ \map p t$
Then $q$ is a continuous mapping, being the composition of such, and:
:$\map q 0 = \map f {\map p 0} = \map f s$
:$\map q 1 = \map f {\map p 1} = \map f {s'}$
Thus $\map f s$ and $\map f {s'}$ are connected by a path in $f \sqbrk S$.
Since these points were arbitrary in $f \sqbrk S$, the result follows.
== Sources ==
* : Chapter $\text{III}$: Metric Spaces: Path-Connectedness
Category:Path-Connected Spaces
Category:Metric Spaces
Continuous Image of Separable Space is Separable

== Definition ==
Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $f: T_1 \to T_2$ be a continuous mapping.
If $T_1$ is separable, then so is the image $f \sqbrk {T_1}$.
That is, separability is a continuous invariant.
== Proof ==
By definition, $T_1 = \struct {S_1, \tau_1}$ is separable there exists a countable subset $D \subset S_1$ which is (everywhere) dense in $T_1$.
We need to show that if there exists a continuous mapping $f: T_1 \to T_2$, then $f \sqbrk {T_1}$ is also separable.
That is, that there exists a countable subset of $f \sqbrk {S_1}$ which is dense in $T_2$.
Let $x_2$ be any point in the image $f \sqbrk {S_1}$ of $S_1$ under $f$.
Let $U \in \tau_2$ be an arbitrary open set such that $x_2 \in U$.
By definition of image set, there exists some $x_1 \in S_1$ with $\map f {x_1} = x_2$.
Since $f$ is continuous, $f^{-1} \sqbrk U$ is open in $T_1$.
By definition of preimage, $x_1$ is in this set.
We are given that $D$ is a countable subset of $S_1$ which is dense in $T_1$.
By definition of dense, $D \cap f^{-1} \sqbrk U \ne \O$
Thus there exists some $d \in D$ such that $d \in f^{-1} \sqbrk U$.
Therefore $\map f d \in U$.
Since $U$ was arbitrary, it follows that $f \sqbrk D$ is dense in $T_2$.
By Image of Countable Set under Mapping is Countable, $f \sqbrk D$ is countable.
Hence $T_2$ is separable.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Functions
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Invariance Properties
Category:Separable Spaces
Category:Continuous Invariants
Continuous Implies Locally Bounded

== Theorem ==
Let $X$ be a topological space.
Let $M$ be a metric space.
Let $f: X \to M$ be continuous.
Then $f$ is locally bounded.
== Proof ==
Let $x \in X$.
Let $U = \map {f^{-1} } {\map B {\map f x, 1} }$.
By continuity, $U$ is a neighborhood of $x$.
Because $\map f U \subset \map B {\map f x, 1}$, $f$ is bounded on $U$.
Thus $f$ is locally bounded.
Category:Boundedness
Continuous Injection of Interval is Strictly Monotone

== Theorem ==
Let $I$ be a real interval.
Let $f: I \to \R$ be an injective continuous real function.
Then $f$ is strictly monotone.
== Proof ==
 $f$ is not strictly monotone.
That is, there exist $x, y, z \in I$ with $x < y < z$ such that either:
:$\map f x \le \map f y$ and $\map f y \ge \map f z$
or:
:$\map f x \ge \map f y$ and $\map f y \le \map f z$
Suppose $\map f x \le \map f y$ and $\map f y \ge \map f z$.
If $\map f x = \map f y$, or $\map f y = \map f z$, or $\map f x = \map f z$, $f$ is not injective, which is a contradiction.
Thus, $\map f x < \map f y$ and $\map f y > \map f z$.
Suppose $\map f x < \map f z$.
That is:
:$\map f x < \map f z < \map f y$
As $f$ is continuous on $I$, the Intermediate Value Theorem can be applied.
Hence there exists $c \in \openint x y$ such that $\map f c = \map f z$.
As $z \notin \openint x y$, we have $c \ne z$.
So $f$ is not injective, which is a contradiction.
Suppose instead $\map f x > \map f z$.
That is:
:$\map f z < \map f x < \map f y$
Again, as $f$ is continuous on $I$, the Intermediate Value Theorem can be applied.
Then, there exists $c \in \openint y z$ such that $\map f c = \map f x$.
So $f$ is not injective, which is a contradiction.
If we suppose $\map f x \ge \map f y$ and $\map f y \le \map f z$, we reach a similar contradiction.
By Proof by Contradiction, $f$ is strictly monotone.
Category:Continuous Functions
Continuous Inverse Theorem

== Theorem ==
Let $f$ be a real function defined on an interval $I$.
Let $f$ be strictly monotone and continuous on $I$.
Let $g$ be the inverse mapping to $f$.
Let $J := f \left[{I}\right]$ be the image of $I$ under $f$.
Then $g$ is strictly monotone and continuous on $J$.
== Proof ==
From Strictly Monotone Real Function is Bijective, $f$ is a bijection.
From Inverse of Strictly Monotone Function, $f^{-1} : J \to I$ exists and is strictly monotone.
From Surjective Monotone Function is Continuous, $f^{-1}$ is continuous.
Hence the result.
== Sources ==
* : $\S 5.6$: Theorem $5$
Category:Real Analysis
Category:Continuity
Category:Named Theorems
Continuous Involution is Homeomorphism

== Theorem ==
Let $\struct {S, \tau}$ be a topological space.
Let $f: S \to S$ be a continuous involution.
Then $f$ is a homeomorphism.
== Proof ==
From Involution is Permutation, $f$ is a permutation and so a bijection.
Since $f$ is continuous, it suffices to verify that its inverse is also continuous.
Now recall $f$ is an involution, that is, $f^{-1} = f$.
Thus $f^{-1}$ is also continuous.
Hence $f$ is a homeomorphism.
Category:Continuous Mappings
Category:Homeomorphisms (Topological Spaces)
Category:Involutions
Continuous Lattice Subframe of Algebraic Lattice is Algebraic Lattice

== Theorem ==
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a bounded below algebraic lattice.
Let $P = \struct {T, \precsim}$ be a continuous lattice subframe of $L$.
Then $P$ is algebraic lattice.
== Proof ==
By definition of algebraic ordered set:
:$L$ is up-complete.
By Up-Complete Lower Bounded Join Semilattice is Complete:
:$L$ is a complete lattice.
By definition:
:$P$ is closure system of $L$.
By Image of Operator Generated by Closure System is Set of Closure System
:$\map {\operatorname {operator} } P \sqbrk S = T$
By Closure Operator Preserves Directed Suprema iff Image of Closure Operator Inherits Directed Suprema
:$\map {\operatorname {operator} } P$ preserves directed suprema.
Thus by Image of Directed Suprema Preserving Closure Operator is Algebraic Lattice:
:$P$ is an algebraic lattice.
== Sources ==
* 
* 
Category:Continuous Lattices
Continuous Lattice and Way Below implies Preceding implies Preceding

== Theorem ==
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a continuous complete lattice.
Let $a, b \in S$.
Let:
:$\forall c \in S: c \ll a \implies c \preceq b$
Then $a \preceq b$
== Proof ==
By definition of way below closure:
:$\forall c \in a^\ll: c \preceq b$
By definition:
:$b$ is an upper bound for $a^\ll$
By definition of supremum:
:$\map \sup {a^\ll} \preceq b$
By definition of continuous:
:$L$ satisfies the axiom of approximation.
Thus by the axiom of approximation:
:$a = \map \sup {a^\ll} \preceq b$
== Sources ==
* 
* 
Category:Continuous Lattices
Continuous Lattice iff Auxiliary Approximating Relation is Superset of Way Below Relation

== Theorem ==
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a complete lattice.
Let $\map {\operatorname{App} } L$ be the set of all auxiliary approximating relation on $S$.
Then
:$L$ is continuous
:$\forall \RR \in \map {\operatorname{App} } L: \ll \subseteq \RR$ and $\ll$ is an approximating relation
== Proof ==
=== Sufficient Condition ===
Let $L$ be continuous.
Let $\RR \in \map {\operatorname{App} } L$
Let $\tuple {a, b} \in S \times S$ such that:
:$a \ll b$
By definition of way below closure:
:$a \in b^\ll$
where $b^\ll$ denotes the way below closure of $b$.
By Complete Lattice is Bounded and Continuous Lattice is Meet-Continuous:
$L$ is a bounded below meet-continuous lattice.
By Intersection of Relation Segments of Approximating Relations equals Way Below Closure:
:$\ds \bigcap \set {b^\RR: \RR \in \map {\operatorname{App} } L} = b^\ll$
where $b^\RR$ denotes the $\RR$-segment of $b$.
By Intersection is Subset/General Result:
:$b^\ll \subseteq b^\RR$
By definition of subset:
:$a \in b^\RR$
By definition of $\RR$-segment:
:$\tuple {a, b} \in \RR$
Thus by definition of subset:
:$\ll \subseteq \RR$
Thus by Way Below is Approximating Relation:
:$\ll$ is an approximating relation.
=== Necessary Condition ===
Let
:$\forall \RR \in \map {\operatorname{App} } L: \ll \subseteq \RR$ and $\ll$ is an approximating relation
Define $\RR := \mathord \ll$.
By definition of approximating relation:
:$\forall x \in S: x = \map \sup {x^\RR}$
By definitions of way below closure and $\RR$-segment:
:$\forall x \in S: x = \map \sup {x^\ll}$
By definition:
:$L$ satisfies the axiom of approximation.
By Way Below Closure is Directed in Bounded Below Join Semilattice:
:$\forall x \in S: x^\ll$ is directed
By definition of complete lattice:
:$L$ is up-complete.
Hence $L$ is continuous.
== Sources ==
* 
* 
Category:Approximating Relations
Category:Continuous Lattices
Continuous Lattice is Meet-Continuous

== Theorem ==
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a bounded below continuous lattice.
Then $L$ is meet-continuous.
== Proof ==
Let $x \in S$, $D$ be a directed subset of $S$ such that
:$x \preceq \sup D$
By Way Below Closure is Directed in Bounded Below Join Semilattice:
:$x^\ll$ is directed.
By definition of continuous:
:$L$ is up-complete and satisfies the axiom of approximation.
By definition of up-complete:
:$x^\ll$ admits a supremum.
By Lower Closure of Element is Ideal and definition of ideal:
:$\paren {\sup \set {x \wedge d: d \in D} }^\preceq$ is directed.
By definition of up-complete:
:$\paren {\sup \set {x \wedge d: d \in D} }^\preceq$ admits a supremum.
We will prove that
:$x^\ll \subseteq \paren {\sup \set {x \wedge d: d \in D} }^\preceq$
Let $a \in x^\ll$
By definition of way below closure:
:$a \ll x$
By Way Below implies Preceding:
:$a \preceq x$
By definition of way below relation:
:$\exists d \in D: a \preceq d$
By Meet is Idempotent and Meet Semilattice is Ordered Structure:
:$a = a \wedge a \preceq x \wedge d \in \set {x \wedge y: y \in D}$
By definition of supremum:
:$\sup \set {x \wedge y: y \in D}$ is upper bound for $\set {x \wedge y: y \in D}$
By definition of upper bound:
:$x \wedge d \preceq \sup \set {x \wedge y: y \in D}$
By definition of transitivity:
:$a \preceq \sup \set {x \wedge y: y \in D}$
Thus by definition of lower closure of element:
:$a \in \paren {\sup \set {x \wedge d: d \in D} }^\preceq$
By Supremum of Subset:
:$\map \sup {x^\ll} \preceq \map \sup {\paren {\sup \set {x \wedge d: d \in D} }^\preceq}$
By the axiom of approximation and Supremum of Lower Closure of Element:
:$x \preceq \sup \set {x \wedge d: d \in D}$
By definition:
:$x$ is upper bound for $\set {x \wedge d: d \in D}$
By definition of supremum:
:$\sup \set {x \wedge d: d \in D} \preceq x$
Thus by definition of antisymmetry:
:$x = \sup \set {x \wedge d: d \in D}$
Thus by Meet-Continuous iff if Element Precedes Supremum of Directed Subset then Element equals Supremum of Meet of Element by Directed Subset:
:$L$ is meet-continuous.
== Sources ==
* 
* 
Category:Meet-Continuous Lattices
Category:Continuous Lattices
Continuous Linear Operator over Finite Dimensional Vector Space is Invertible

== Theorem ==
Let $\struct {X, \norm {\, \cdot\,}_X}$ be a normed vector space.
Let $\map {CL} X := \map {CL} {X, X}$ be a continuous linear transformation space.
Let $I \in \map {CL} X$ be the identity element.
Let $S, T \in \map {CL} X$.
Suppose the dimension of $X$ is finite:
:$d = \dim X < \infty$
Suppose $T \circ S = I$ where $\circ$ denotes the composition of mappings.
Then $T$ and $S$ are invertible.
== Proof ==
Let $x \in X$.
Then:
Let $\mathbf 0 \in X$ be the zero vector of $X$.
Suppose $\map S x = \mathbf 0$.
Then:
Hence:
:$\paren {\map S x = \mathbf 0} \implies \paren{x = \mathbf 0} \paren \star$
In other words:
:$\map \ker S = \set {\mathbf 0}$
where $\ker$ denotes the kernel.
Let $\set {v_1, \ldots v_d}$ be the basis for $X$.
Suppose $\forall k \in \N_{> 0} : k \le d : \alpha_k \in \mathbb F$ are scalars such that:
:$\ds \sum_{k \mathop = 1}^d \alpha_k \map S {v_k} = \mathbf 0$
where $\mathbb F$ is a number field.
By definition of linear transformation on vector space:
:$\ds \map S {\sum_{k \mathop = 1}^d \alpha_k v_k} = \mathbf 0$
By Eq. $\paren \star$:
:$\ds \sum_{k \mathop = 1}^d \alpha_k v_k = \mathbf 0$
Since $\set {v_1, \ldots v_d}$ was arbitrary, we have that:
:$\forall k \in \N_{> 0} : k \le d : \alpha_k = 0$
By definition of linearly independent sequence of vectors, $\set {\map S {v_1}, \ldots \map S {v_d} }$ is a basis for $X$.
Therefore:
:$\ds \forall x \in X : \forall k \in \N_{> 0} : k \le d : \exists \beta_k \in \mathbb F : x = \sum_{k \mathop = 1}^d \beta_k \map S {v_k}$
Thus:
All in all:
:$\paren {T \circ S = I} \implies \paren {T \circ S = S \circ T = I}$
== Sources ==
* : Chapter $\S 2.4$: Composition of continuous linear transformations
Category:Continuous Linear Transformations
Category:Inverse Mappings
Continuous Linear Operator over Infinite Dimensional Vector Space is not necessarily Invertible

== Theorem ==
Let $\struct {X, \norm {\, \cdot\,}_X}$ be a normed vector space.
Let $\map {CL} X := \map {CL} {X, X}$ be a continuous linear transformation space.
Let $I \in \map {CL} X$ be the identity element.
Let $S, T \in \map {CL} X$.
Suppose the dimension of $X$ is finite:
:$d = \dim X = \infty$
Suppose $T \circ S = I$ where $\circ$ denotes the composition of mappings.
Then $T$ and $S$ are not necessarily invertible.
== Proof ==
Let $X, Y$ be $2$-sequence spaces.
Let $T = L : X \to Y$ be the left shift operator.
Let $S = R : X \to Y$ be the right shift operator.
Let $x := \tuple {a_1, a_2, \ldots} \in X$.
We have that:
Hence:
:$L \circ R = I$
However:
Therefore, $T \circ S = I$ but $S \circ T \ne I$.
== Sources ==
* : Chapter $\S 2.4$: Composition of continuous linear transformations
Category:Continuous Linear Transformations
Category:Inverse Mappings
Continuous Linear Transformation Algebra has Two-Sided Identity

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a normed vector space.
Let $\map {CL} X := \map {CL} {X, X}$ be a continuous linear transformation space.
Let $\struct {\map {CL} X, *}$ be an associative algebra.
Then there exists an identity element $I \in \map {CL} X$ such that:
:$\forall x \in X : \map I x = x$
== Proof ==
Let $X$ be normed vector space over $K$.
Let $\map \LL {X, X}$ denote the set of all linear transformations from $X$ to itself.
Let $\map C {X, X}$ denote the continuous mapping space from $X$ to itself.
Suppose:
:$\map {CL} X := \map {CL} {X, X}$
Since $\struct {\map {CL} X, *}$ is an associative algebra, by definition of the continuous linear transformation space from $X$ to itself, $\map {CL} {X, X}$, we have:
:$\map {CL} {X, X} := \map C {X, X} \cap \map \LL {X, X}$
By Identity Mapping is Continuous, we have that $I \in \map C {X, X}$.
By Identity Mapping on Normed Vector Space is Bounded Linear Operator, we have that $I \in \map \LL {X, X}$.
Hence $I \in \map {CL} X $ by definition of identity element.
== Sources ==
* : Chapter $\S 2.4$: Composition of continuous linear transformations
Continuous Linear Transformation Algebra with Supremum Operator Norm is Normed Algebra

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a normed vector space.
Let $\map {CL} X := \map {CL} {X, X}$ be a continuous linear transformation space.
Let $\struct {\map {CL} X, *}$ be an associative algebra.
Let $\norm {\, \cdot \,}$ be the supremum operator norm.
Then $\struct {\struct {\map {CL} X, *}, \norm {\, \cdot \,}}$ is a normed algebra.
== Proof ==
We need to show:
$\norm {a \ast b} \le \norm a \norm b$
== Sources ==
* : Chapter $\S 2.4$: Composition of continuous linear transformations
Continuous Linear Transformation Space as Algebra

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Let $\map {CL} X := \map {CL} {X, X}$ be a continuous linear transformation space.
Let $* : \map {CL} X \times \map {CL} X \mapsto \map {CL} X$ be a bilinear mapping.
Then $\struct {\map {CL} X, *}$ is an associative algebra.
== Proof ==
== Sources ==
* : Chapter $\S 2.4$: Composition of continuous linear transformations
Continuous Linear Transformation Space as Banach Algebra

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a normed vector space.
Let $\map {CL} X := \map {CL} {X, X}$ be a continuous linear transformation space.
Let $*_X : X \times X \to X$ and $* : \map {CL} X \times \map {CL} X \to \map {CL} X$ be bilinear mappings.
Suppose $\struct {\struct {X, \norm {\, \cdot \,}_X}, *_X}$ is a Banach algebra.
Let $\norm {\, \cdot \,}$ be the supremum operator norm.
Then $\struct {\struct {\map {CL} X, \norm {\, \cdot \,}}, *}$ is a Banach algebra.
== Proof ==
== Sources ==
* : Chapter $\S 2.4$: Composition of continuous linear transformations
Continuous Linear Transformation between Topological Vector Spaces is Bounded

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $X$ and $Y$ be topological vector spaces over $\GF$. 
Let $T : X \to Y$ be a continuous linear transformation.
Then $T$ is bounded.
== Proof ==
Let $E$ be a von Neumann-bounded subset of $X$. 
We want to show that $T \sqbrk E$ is a von Neumann-bounded subset of $Y$. 
Let $W$ be a open neighborhood of ${\mathbf 0}_Y$ in $Y$.
Since $T$ is continuous, it is continuous at $\mathbf 0_X$.
Since $\map T {\mathbf 0_X} = \mathbf 0_Y$, there exists an open neighborhood $V$ of $\mathbf 0_X$ in $X$ such that: 
:$T \sqbrk V \subseteq W$
Since $E$ is von Neumann-bounded subset, there exists $s > 0$ such that: 
:$E \subseteq t V$ for $t > s$.
Then we have: 
for $t > s$. 
Since $W$ was an arbitrary open neighborhood of ${\mathbf 0}_Y$ in $Y$, we have that $T \sqbrk E$ is von Neumann-bounded.
So $T$ is bounded.
== Sources ==
* : $1.32$: Theorem
Category:Bounded Linear Transformations (Topological Vector Spaces)
Continuous Linear Transformations form Subspace of Linear Transformations

== Theorem ==
The space of continuous linear transformations is a subspace of the space of linear transformations.
== Proof ==
Let $\struct {X, \norm \cdot }$ and $\struct {Y, \norm \cdot }$ be normed vector spaces.
=== Closure under vector addition ===
Let $S, T \in \map {CL} {X, Y}$.
By Continuity of Linear Transformation between Normed Vector Spaces:
:$\exists M_S \in \R : M_S > 0 : \forall x \in X : \norm {S x} \le M_S \norm x$
:$\exists M_T \in \R : M_T > 0 : \forall x \in X : \norm {T x} \le M_T \norm x$
Furthremore:
By Continuity of Linear Transformation between Normed Vector Spaces:
:$S + T \in \map {CL} {X, Y}$
=== Closure under scalar multiplication ===
Let $\alpha \in \R$.
Let $T \in \map {CL} {X, Y}$.
By Continuity of Linear Transformation between Normed Vector Spaces:
:$\exists M \in \R_{> 0} : \forall x \in X : \norm {T x} \le M \norm x$
Hence:
By Continuity of Linear Transformation between Normed Vector Spaces:
:$\alpha T \in \map {CL} {X, Y}$
=== Existence of identity element under vector addition ===
Let $\mathbf 0 : X \to Y$ be the zero mapping.
Then:
Hence:
:$\mathbf 0 \in \map {CL} {X, Y}$
== Sources ==
* : Chapter $\S 2.3$: The normed space $\map {CL} {X, Y}$. Operator norm and the normed space $\map {CL} {X, Y}$
Category:Linear Transformations
Category:Continuous Linear Transformations
Category:Continuous Mappings
Category:Examples of Vector Subspaces