nyuuzyou/wikis huggingface dataset:
Sample Space/Examples/Throwing a 6-Sided Die

== Example of Sample Space ==
Let $\EE$ be the experiment of throwing a standard $6$-sided die.
:The sample space of $\EE$ is $\Omega = \set {1, 2, 3, 4, 5, 6}$.
== Sources ==
* : $1$: Events and probabilities: $1.2$: Outcomes and events
* 
* 
Category:Examples of Sample Spaces
Sample Space is Union of All Distinct Simple Events

== Theorem ==
Let $\EE$ be an experiment.
Let $\Omega$ denote the sample space of $\EE$.
Then $\Omega$ is the union of the set of simple events in $\EE$.
== Proof ==
By Set is Subset of Itself:
:$\Omega \subseteq \Omega$
That is, $\Omega$ is itself an event in $\EE$.
The result as an application of Non-Trivial Event is Union of Simple Events.
== Sources ==
* : Chapter $2$: Probability and Discrete Sample Spaces: $2.2$ Sample spaces and events
Category:Events
Sampling Distribution/Examples/Normal Distribution

== Example of Sampling Distribution ==
Let $S$ be a random sample from a normal distribution $\Gaussian \mu {\sigma^2}$.
The sample mean $\overline x$ of $S$ is an observed value of a random variable $\overline X$, say, which has a probability distribution $\Gaussian \mu {\dfrac {\sigma^2} n}$
== Sources ==
* 
* 
Category:Examples of Sampling Distributions
Sampling Error/Examples/Normal Distribution

== Example of Sampling Error ==
Let $S = \set {x_1, x_2, \ldots, x_n}$ be a random sample of size $n$ from a normal distribution $\Gaussian \lambda {\rho^2}$ whose mean is $\lambda$ and whose standard deviation is $\rho$.
Then the mean $m$ of $S$ has a normal distribution whose mean is $\lambda$ and a standard deviation is $\dfrac \rho {\sqrt n}$.
If $\rho$ is unknown, then $\dfrac \rho {\sqrt n}$ is estimated by $\dfrac s {\sqrt n}$, where:
:$s^2 = \ds \sum_i \dfrac {\paren {x - m}^2} {n - 1}$
== Sources ==
* 
* 
Category:Examples of Sampling Errors
Sampling Frame/Examples/UK Register of Electors

== Example of Sampling Frame ==
The UK Register of Electors forms a sampling frame of all the people in each district who are qualified to vote in parliamentary elections.
Unfortunately, it rapidly becomes out of date.
People move in and out of the area, or they die, or they become of age, and so on.
== Sources ==
* 
* 
Category:Examples of Sampling Frames
Sampling Function is its own Fourier Transform

== Theorem ==
Consider the sampling function $\operatorname {III}: \R \to \R$.
Then:
:$\map \FF {\operatorname {III} } = \operatorname {III}$
where $\FF$ denotes the Fourier transform.
== Proof ==
== Sources ==
* : Preface to the Second Edition
* : Chapter $1$: Introduction
Category:Sampling Function
Category:Examples of Fourier Transforms
Sandwich Principle

== Theorem ==
Let $A$ be a class.
Let $g: A \to A$ be a mapping on $A$ such that:
:for all $x, y \in A$, either $\map g x \subseteq y$ or $y \subseteq x$.
Then:
:$\forall x, y \in A: x \subseteq y \subseteq \map g x \implies x = y \lor y = \map g x$
That is, there is no element $y$ of $A$ such that:
:$x \subset y \subset \map g x$
where $\subset$ denotes a proper subset.
=== Corollary 1 ===
=== Corollary 2 ===
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Sandwich Theorem or Sandwich Rule in the context of real analysis, also known as the Squeeze Theorem (preferred on )
== Linguistic Note ==
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 4$ A double induction principle and its applications: Lemma $4.9 \ (1)$
Category:Well-Orderings
Category:Named Theorems
Category:Sandwich Principle
Sandwich Principle/Corollary 1

== Theorem ==
Let $A$ be a class.
Let $g: A \to A$ be a mapping on $A$ such that:
:for all $x, y \in A$, either $\map g x \subseteq y$ or $y \subseteq x$.
Let:
:$x \subset y$
where $\subset$ denotes a proper subset.
Then:
:$\map g x \subseteq y$
== Proof ==
Let $x \subset y$.
By hypothesis, either $\map g x \subseteq y$ or $y \subseteq x$.
But because $x \subset y$, it follows that $y \subseteq x$ cannot be the case.
Hence the result.
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 4$ A double induction principle and its applications: Lemma $4.9 \ (2)$
Category:Sandwich Principle
Sandwich Principle/Corollary 2

== Theorem ==
Let $A$ be a class.
Let $g: A \to A$ be a mapping on $A$ such that:
:for all $x, y \in A$, either $\map g x \subseteq y$ or $y \subseteq x$.
Let $g$ be a progressing mapping.
Let $x \subseteq y$.
Then:
:$\map g x \subseteq \map g y$
== Proof ==
Let $x \subseteq y$.
Suppose $x = y$.
Then $\map g x \subseteq \map g y$ and the result holds.
Suppose that $x \ne y$.
Then $x \subset y$
It follows from Corollary 1 that:
:$\map g x \subseteq y$
As $g$ is a progressing mapping on $A$:
:$y \subseteq \map g y$
Hence by Subset Relation is Transitive:
:$\map g x \subseteq \map g y$
So in either case:
:$\map g x \subseteq \map g y$
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 4$ A double induction principle and its applications: Lemma $4.9 \ (3)$
Category:Sandwich Principle
Sandwich Principle/Linguistic Note

== Linguistic Note on Sandwich Principle ==
The name Sandwich Principle for this lemma appears to have been coined by and in their 
As they explain, when $x \subset y \subset z$, we may consider $y$ to be sandwiched between $x$ and $z$.
Thus the Sandwich Principle tells us that, given the conditions constraining $x$, $y$ and $\map g x$, there can be no such $y$ sandwiched between $x$ and $\map g x$.
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 4$ A double induction principle and its applications: Lemma $4.9$: Remarks
Category:Linguistic Notes
Sandwich Principle/Proof 1

== Theorem ==
== Proof ==
We are given that:
:for all $x, y \in A$, either $\map g x \subseteq y$ or $y \subseteq x$.
Let $x, y \in A$ such that:
:$x \subseteq y \subseteq \map g x$
 both $x \subset y$ and $y \subset \map g x$.
From $x \subset y$, it follows by definition of proper subset that:
:$\exists a \in y: a \notin x$
and so it is not the case that $y \subseteq x$.
From $y \subset \map g x$, it follows by definition of proper subset that:
:$\exists b \in \map g x: b \notin y$
and so it is not the case that $\map g x \subseteq y$.
Hence neither $y \subseteq x$ nor $\map g x \subseteq y$.
This contradicts the property of $g$.
Hence by Proof by Contradiction it follows that $y \subset \map g x$ and $x \subset y$ cannot both be true.
Therefore either $x = y$ or $y = \map g x$.
Hence the result.
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 4$ A double induction principle and its applications: Lemma $4.9$
Category:Sandwich Principle
Sandwich Principle/Proof 2

== Theorem ==
== Proof ==
We are given that:
:for all $x, y \in A$, either $\map g x \subseteq y$ or $y \subseteq x$.
Let $x, y \in A$ such that:
:$x \subseteq y \subseteq \map g x$
Then either we have:
:$\map g x \subseteq y$ and $y \subseteq \map g x$
in which case, by definition of set equality:
:$y = \map g x$
or we have that:
:$x \subseteq y$ and $y \subseteq x$
in which case, by definition of set equality:
:$x = y$
Thus either $y = \map g x$ or $x = y$ and the result follows.
Category:Sandwich Principle
Sandwich Principle for G-Towers

== Theorem ==
Let $M$ be a class.
Let $g: M \to M$ be a progressing mapping on $M$.
Let $M$ be a $g$-tower.
Let $x, y \in M$ be arbitrary.
Then it cannot be the case that:
:$x \subsetneqq y \subsetneqq \map g x$
=== Corollary 1 ===
=== Corollary 2 ===
== Proof ==
From Lemma $2$ of $g$-Tower is Nest we have that:
:$\forall x, y \in M: \map g x \subseteq y \lor y \subseteq x$
From the Sandwich Principle:
:$\forall x, y \in M: x \subseteq y \subseteq \map g x \implies x = y \lor y = \map g x$
That is, there is no element $y$ of $M$ such that:
:$x \subsetneqq y \subsetneqq \map g x$
== Also see ==
* Sandwich Principle
== Linguistic Note ==
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 2$ Superinduction and double superinduction: Theorem $2.4 \ (1)$
Category:G-Towers
Category:Sandwich Principle
Sandwich Principle for G-Towers/Corollary 1

== Theorem ==
Let $M$ be a class.
Let $g: M \to M$ be a progressing mapping on $M$.
Let $M$ be a $g$-tower.
Let $x, y \in M$ be arbitrary.
Then:
:$x \subsetneqq y \implies \map g x \subseteq y$
== Proof ==
From Lemma $2$ of $g$-Tower is Nest we have that:
:$\forall x, y \in M: \map g x \subseteq y \lor y \subseteq x$
From the Sandwich Principle: Corollary 1:
:$\forall x, y \in M: x \subseteq y \subseteq \map g x \implies x = y \lor y = \map g x$
That is, if:
:$x \subsetneqq y$
then:
:$\map g x \subseteq y$
== Also see ==
* Sandwich Principle
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 2$ Superinduction and double superinduction: Theorem $2.4 \ (2)$
Category:G-Towers
Category:Sandwich Principle
Sandwich Principle for G-Towers/Corollary 2

== Theorem ==
Let $M$ be a class.
Let $g: M \to M$ be a progressing mapping on $M$.
Let $M$ be a $g$-tower.
Let $x, y \in M$ be arbitrary such that $x \subseteq y$.
Then:
:$\map g x \subseteq \map g y$
== Proof ==
From Lemma $2$ of $g$-Tower is Nest we have that:
:$\forall x, y \in M: \map g x \subseteq y \lor y \subseteq x$
From the Sandwich Principle: Corollary 2:
:$\forall x, y \in M: x \subseteq y\implies \map g x \subseteq \map g y$
== Also see ==
* Sandwich Principle
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 2$ Superinduction and double superinduction: Theorem $2.4 \ (3)$
Category:G-Towers
Category:Sandwich Principle
Sandwich Principle for Minimally Closed Class

== Theorem ==
Let $N$ be a class which is closed under a progressing mapping $g$.
Let $b$ be an element of $N$ such that $N$ is minimally closed under $g$ with respect to $b$.
Then for all $x, y \in N$:
:$x \subseteq y \subseteq \map g x \implies x = y \lor y = \map g x$
== Proof ==
From Minimally Closed Class under Progressing Mapping induces Nest, we have that $N$ is a nest in which:
:$\forall x, y \in N: \map g x \subseteq y \lor y \subseteq x$
Thus the Sandwich Principle applies directly.
Category:Sandwich Principle
Category:Minimally Closed Classes
Category:Progressing Mappings
Sandwich Principle for Slowly Progressing Mapping

== Theorem ==
Let $M$ be a class.
Let $g: M \to M$ be a slowly progressing mapping on $M$.
Let $M$ be a minimally inductive class under $g$.
Then $x \subsetneqq y \subsetneqq \map g x$ can never hold.
=== Corollary ===
== Proof ==
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 9$ Supplement -- optional: Lemma $S_3 \ (1)$
Category:Minimally Inductive Classes
Category:Slowly Progressing Mappings
Category:Sandwich Principle
Category:Sandwich Principle for Slowly Progressing Mapping
Sandwich Principle for Slowly Progressing Mapping/Corollary

== Theorem ==
Let $M$ be a class.
Let $g: M \to M$ be a slowly progressing mapping on $M$.
Let $M$ be a minimally inductive class under $g$.
Let $N$ be a nest which is closed under $g$.
Then:
:$\forall x, y \in N: \map g x \subseteq y \lor y \subseteq x$
== Proof ==
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 9$ Supplement -- optional: Lemma $S_3 \ (2)$
Category:Sandwich Principle for Slowly Progressing Mapping
Satisfiability preserved in Supersignature

== Theorem ==
Let $\LL, \LL'$ be signatures for the language of predicate logic.
Let $\LL'$ be a supersignature of $\LL$.
Let $\Sigma$ be a set of $\LL$-sentences.
Then the following are equivalent:
:$\AA \models_{\mathrm{PL} } \Sigma$ for some $\LL$-structure $\AA$
:$\AA' \models_{\mathrm{PL} } \Sigma$ for some $\LL'$-structure $\AA'$
where $\models_{\mathrm{PL} }$ is the models relation.
That is to say, the notion of satisfiability is preserved in passing to a supersignature.
== Proof ==
== Sources ==
* : $\text{II}.8$ Further Semantic Notions: Lemma $\text{II}.8.15$
Category:Model Theory for Predicate Logic
Satisfiable Set Union Tautology is Satisfiable

== Theorem ==
Let $\LL$ be a logical language.
Let $\mathscr M$ be a formal semantics for $\LL$.
Let $\FF$ be an $\mathscr M$-satisfiable set of formulas from $\LL$.
Let $\phi$ be a tautology for $\mathscr M$.
Then $\FF \cup \set \phi$ is also $\mathscr M$-satisfiable.
== Proof ==
Since $\FF$ is $\mathscr M$-satisfiable, there exists some model $\MM$ of $\FF$:
:$\MM \models_{\mathscr M} \FF$
Since $\psi$ is a tautology, also:
:$\MM \models_{\mathscr M} \psi$
Therefore, we conclude that:
:$\MM \models_{\mathscr M} \FF \cup \set \phi$
that is, $\FF \cup \set \phi$ is satisfiable.
== Sources ==
* : $\S 2.5.2$: Theorem $2.45$
* : $\S 2.10$: Exercise $2.15$
Category:Formal Semantics
Satisfiable Set minus Formula is Satisfiable

== Theorem ==
Let $\LL$ be a logical language.
Let $\mathscr M$ be a formal semantics for $\LL$.
Let $\FF$ be an $\mathscr M$-satisfiable set of formulas from $\LL$.
Let $\phi \in \FF$.
Then $\FF \setminus \set \phi$ is also $\mathscr M$-satisfiable.
== Proof ==
This is an immediate consequence of Subset of Satisfiable Set is Satisfiable.
== Also see ==
* Subset of Satisfiable Set is Satisfiable
== Sources ==
* : $\S 2.5.2$: Theorem $2.44$
* : $\S 2.10$: Exercise $2.15$
Category:Formal Semantics
Satisfiable iff Negation is Falsifiable

== Theorem ==
Let $\mathbf A$ be a WFF of propositional logic.
Then $\mathbf A$ is satisfiable its negation $\neg \mathbf A$ is falsifiable.
== Proof ==
=== Necessary Condition ===
Let $\mathbf A$ be satisfiable.
Then there exists a boolean interpretation $v$ of $\mathbf A$ such that:
:$\map v {\mathbf A} = \T$
Hence, by definition of the boolean interpretation of negation:
:$\map v {\neg \mathbf A} = \F$
It follows that $\neg \mathbf A$ is falsifiable.
=== Sufficient Condition ===
Let $\neg \mathbf A$ be falsifiable.
Then there exists a boolean interpretation $v$ of $\neg \mathbf A$ such that:
:$\map v {\neg \mathbf A} = \F$
Hence, by definition of the boolean interpretation of negation:
:$\map v {\mathbf A} = \T$
It follows that $\mathbf A$ is satisfiable.
== Sources ==
* : $\S 2.5$: Theorem $2.39$
Category:Boolean Interpretations
Category:Formal Semantics
Saturated Implies Universal

== Theorem ==
Let $\kappa$ be an infinite cardinal.
Let $\MM$ be a model of the $\LL$-theory $T$.
If $\MM$ is $\kappa$-saturated, then it is $\kappa^+$-universal, where $\kappa^+$ is the successor cardinal of $\kappa$.
== Proof ==
The idea of the proof is that $\MM$ being saturated means that when we want to define an elementary map $\NN \to \MM$, we can find an image $y \in \MM$ for an element $x \in \NN$ by realizing the type made up of the formulas that such a $y$ would need to satisfy.
Let $\NN$ be a model of $T$ with universe of cardinality strictly less than $\kappa$.
We will construct an elementary embedding of $\NN$ into $\MM$ by transfinite recursion.
Since $\card \NN < \kappa$, we can write its elements as $n_\alpha$ for ordinals $\alpha < \kappa$.
For each ordinal $\alpha < \kappa$, let $A_\alpha$ be the subset $\set {n_\beta: \beta < \alpha}$ of the universe of $\NN$.
Note for clarity that $n_\alpha \in A_{\alpha + 1}$ but $n_\alpha \notin A_\alpha$.
*Base case $\alpha = 0$:
Define $f_0 = \O$.
Note that $f_0$ is trivially an elementary embedding from $A_0 = \O$ into $\MM$.
*Limit ordinals $\alpha$, assuming $f_\beta$ is defined and elementary $A_\beta \to \MM$ for all $\beta < \alpha$:
Let $\ds f_\alpha = \bigcup_{\beta \mathop < \alpha} f_\beta$.
If $\phi$ is an $\LL$-sentence with parameters from $A_\alpha$, then since it involves only finitely many such parameters, they must all be contained in some $A_\beta$ for $\beta < \alpha$. But $f_\alpha \restriction A_\beta = f_\beta$ is elementary, so $f_\alpha$ must be as well.
* Successor ordinals $\alpha = \beta + 1$, assuming $f_\beta$ is defined and elementary $A_\beta \to \MM$:
We need to extend $f_\beta$ to $A_\alpha = A_\beta \cup \set {n_\beta}$ so that truth of $\LL$-sentences with parameters from $A_\alpha$ is preserved.
Consider the subset $p = \set {\map \phi {v, \map {f_\beta} {\bar{a} } }: \bar{a} \text{ is a tuple from } A_\beta \text{ and } \NN \models \map \phi {n_\beta, \bar{a} } }$ of the set of $\LL$-formulas with one free variable and parameters from the image $\map {f_\beta} {A_\beta}$ of $A_\beta$ under $f_\beta$.
The set $p$ is a $1$-type over the image $\map {f_\beta} {A_\beta}$ in $\MM$.
Since $\card {A_\beta} < \kappa$ and by assumption $\MM$ is $\kappa$-saturated, this means that $p$ is realized in $\MM$ by some element $b$.
Thus, defining $f_\alpha$ to be $f_\beta \cup \set {(n_\beta, b)}$ makes it an elementary embedding $A_\alpha \to \MM$.
Now, define $\ds f = \bigcup_{\alpha \mathop < \kappa} f_\alpha$.
Then $f$ is an elementary embedding from $\NN$ to $\MM$ since $\ds \bigcup_{\alpha \mathop < \kappa} A_\alpha = \NN$, any finite set of parameters from $\NN$ must belong to one $A_\alpha$, and $f \restriction A_\alpha = f_\alpha$ is elementary.
Category:Model Theory for Predicate Logic
Saturation Under Equivalence Relation in Terms of Graph

== Theorem ==
Let $\RR \subset S \times S$ be an equivalence relation on a set $S$.
Let $\pr_1, \pr_2 : S \times S \to S$ denote the projections.
Let $T\subset S$ be a subset.
Let $\overline T$ denote its saturation.
Then the following hold:
:$\overline T = \map {\pr_1} {\RR \cap \map {\pr_2^{-1} } T}$
:$\overline T = \map {\pr_2} {\RR \cap \map {\pr_1^{-1} } T}$
== Proof ==
Let $s \in S$.
We have:
A similar reasoning proves the second identity.
Category:Equivalence Relations
Scalar-Valued Measure associated with Resolution of Identity is Positive Measure if Equal Vectors

== Theorem ==
Let $X$ be a topological space.
Let $\map \BB X$ be the Borel $\sigma$-algebra of $X$.
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.
Let $\map B {\HH}$ be the space of bounded linear transformations on $\HH$.
Let $\EE : \map \BB X \to \map B {\HH}$ be a resolution of the identity.
For each $x, y \in \HH$, define $\EE_{x, y} : \map \BB X \to \C$ by:
:$\map {\EE_{x, y} } A = \innerprod {\map \EE A x} y$
for each $A \in \map \BB X$.
If $x = y$, then $\EE_{x, y} := \EE_x$ is a measure with total variation $\norm x^2$. 
== Proof ==
Let $A \in \map \BB X$. 
We then have:
Hence $\EE_x$ is a measure.
Let $\norm {\EE_x}$ be the total variation of $\EE_x$.
Then:
Category:Resolutions of the Identity
Scalar Field (Physics)/Examples/Electric Charge Density

== Example of Scalar Field in the context of Physics ==
The electric charge density within a region of space is an example of a scalar field.
== Sources ==
* : Chapter $1$: Force and energy in electrostatics: $1.3$ Electric Fields in Matter: $1.3.1$ The atomic charge density
Category:Examples of Scalar Fields
Category:Electric Charge Density
Scalar Multiple of Cauchy Sequence in Topological Vector Space is Cauchy Sequence

== Theorem ==
Let $K$ be a topological field. 
Let $X$ be a topological vector space over $K$.
Let $\sequence {x_n}_{n \in \N}$ be a Cauchy sequence in $X$.
Let $\lambda \in K$.
Then $\sequence {\lambda x_n}_{n \in \N}$ is a Cauchy sequence in $X$.
== Proof ==
First consider the case $\lambda = 0_K$.
Then $\lambda x_n = {\mathbf 0}_X$ for each $n \in \N$. 
From Constant Sequence in Topological Space Converges, we have that $\lambda x_n \to {\mathbf 0}_X$ as $n \to \infty$.
From Convergent Sequence in Topological Vector Space is Cauchy, $\sequence {\lambda x_n}_{n \in \N}$ is therefore Cauchy.
Now take $\lambda \ne 0$. 
Let $U$ be an open neighborhood of ${\mathbf 0}_X$.
From Dilation of Open Set in Topological Vector Space is Open, $\lambda^{-1} U$ is an open neighborhood of ${\mathbf 0}_X$.
Since $\sequence {x_n}_{n \in \N}$ is Cauchy, there exists $N \in \N$ such that:
:$x_n - x_m \in \lambda^{-1} U$ for $n, m \ge N$.
We therefore have:
:$\lambda x_n - \lambda x_m \in U$ for $n, m \ge N$.
Since $U$ was an arbitrary open neighborhood of ${\mathbf 0}_X$, we have that $\sequence {\lambda x_n}_{n \in \N}$ is Cauchy.
Category:Cauchy Sequences in Topological Vector Spaces
Scalar Multiple of Convergent Net in Topological Vector Space is Convergent

== Theorem ==
Let $K$ be a topological field. 
Let $X$ be a topological vector space over $K$.
Let $\struct {\Lambda, \preceq}$ be a directed set.
Let $x \in X$ and $\mu \in K$.
Let $\family {x_\lambda}_{\lambda \in \Lambda}$ be a net converging to $x$.
Then $\family {\mu x_\lambda}_{\lambda \in \Lambda}$ converges to $\mu x$.
== Proof ==
For ease of reading, let $\succeq$ be the inverse relation of $\preceq$.
First consider the case $\mu = 0_K$.
Then $\mu x_n = {\mathbf 0}_X$ for each $n \in \N$. 
From Constant Net is Convergent, we have that $\mu x_\lambda$ converges to ${\mathbf 0}_X$. 
So $\sequence {\lambda x_n}_{n \in \N}$ converges to $\mu x$ for $\mu = 0_K$.
Now take $\mu \ne 0_K$.
Let $U$ be an open neighborhood of $\mu x$.
From Dilation of Open Set in Topological Vector Space is Open, $\mu^{-1} U$ is an open neighborhood of $x$. 
Since $\sequence {x_n}_{n \in \N}$ converges, there exists $\lambda_0 \in \Lambda$ such that:
:$x_n \in \mu^{-1} U$ for $\lambda \succeq \lambda_0$.
Then, we have:
:$\mu x_n \in U$ for $\lambda \succeq \lambda_0$. 
Since $U$ was an arbitrary open neighborhood of $\mu x$, we have that $\sequence {\mu x_\lambda}_{\lambda \in \Lambda}$ converges to $\mu x$ as required.
Category:Nets (Set Theory)
Category:Topological Vector Spaces
Scalar Multiple of Convergent Sequence in Topological Vector Space is Convergent

== Theorem ==
Let $K$ be a topological field. 
Let $X$ be a topological vector space over $K$.
Let $x \in X$.
Let $\sequence {x_n}_{n \in \N}$ be a convergent sequence in $X$ with:
:$x_n \to x$
Let $\lambda \in K$.
Then $\sequence {\lambda x_n}_{n \in \N}$ is a convergent sequence in $X$ with:
:$\lambda x_n \to \lambda x$. 
== Proof ==
First consider the case $\lambda = 0_K$.
Then $\lambda x_n = {\mathbf 0}_X$ for each $n \in \N$. 
From Constant Sequence in Topological Space Converges, we have that $\lambda x_n \to {\mathbf 0}_X = \lambda x$ as $n \to \infty$.
So $\sequence {\lambda x_n}_{n \in \N}$ converges for $\lambda = 0$. 
Now take $\lambda \ne 0_K$.
Let $U$ be an open neighborhood of $\lambda x$.
From Dilation of Open Set in Topological Vector Space is Open, $\lambda^{-1} U$ is an open neighborhood of $x$. 
Since $\sequence {x_n}_{n \in \N}$ converges, there exists $N \in \N$ such that:
:$x_n \in \lambda^{-1} U$ for $n \ge N$.
Then, we have:
:$\lambda x_n \in U$ for $n \ge N$. 
Since $U$ was an arbitrary open neighborhood of $\lambda x$, we have that $\sequence {\lambda x_n}_{n \in \N}$ converges to $\lambda x$ as required.
Category:Topological Vector Spaces
Scalar Multiple of Function of Exponential Order

== Theorem ==
Let $f: \R \to \F$ be a function, where $\F \in \set {\R, \C}$.
Let $\lambda$ be a complex constant.
Suppose $f$ is of exponential order $a$.
Then $\lambda f$ is also of exponential order $a$.
== Proof ==
If $\lambda = 0$, the theorem holds trivially.
Let $\lambda \ne 0$.
Category:Exponential Order
Scalar Multiple of Integrable Function is Integrable Function

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f: X \to \overline \R$ be a $\mu$-integrable function, and let $\lambda \in \R$.
Then $\lambda f: X \to \overline \R$, the pointwise $\lambda$-multiple of $f$, is also $\mu$-integrable.
That is, the space of integrable functions $\LL^1_{\overline \R}$ is closed under pointwise $\R$-scalar multiplication.
== Proof ==
== Sources ==
* : $10.4 \, \text{(i)}$
Category:Integrable Functions (Measure Space)‎
Scalar Multiple of Simple Function is Simple Function

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $f: X \to \R$ be a simple function, and let $\lambda \in \R$.
Then the pointwise scalar multiple $\lambda f: X \to \R$ of $f$ is also a simple function.
== Proof ==
Let $\Img f$ denote the image of $f$.
Let $\Img {\lambda f}$ denote the image of $\lambda f$.
Consider the surjection $l_\lambda: \Img f \to \Img {\lambda f}$ defined by:
:$\map {l_\lambda} {\map f x} := \lambda \map f x$
By Measurable Function is Simple Function iff Finite Image Set, $\card {\Img f}$ is finite.
Hence Cardinality of Codomain of Surjection yields that $\size {\Img {\lambda f} }$ is finite as well.
The result follows from a second application of Measurable Function is Simple Function iff Finite Image Set.
Category:Simple Functions