nyuuzyou/wikis huggingface dataset:
Resonance Frequency is less than Natural Frequency

== Theorem ==
Consider a physical system $S$ whose behaviour is defined by the second order ODE:
:$(1): \quad \dfrac {\d^2 y} {\d x^2} + 2 b \dfrac {\d y} {\d x} + a^2 x = K \cos \omega x$
where:
:$K \in \R: k > 0$
:$a, b \in \R_{>0}: b < a$
Then the resonance frequency of $S$ is smaller than the natural frequency of the associated second order ODE:
:$(2): \quad \dfrac {\d^2 y} {\d x^2} + 2 b \dfrac {\d y} {\d x} + a^2 x = 0$
== Proof ==
From Natural Frequency of Underdamped System, the natural frequency of $(2)$ is:
:$\nu = \dfrac {\sqrt {a^2 - b^2} } {2 \pi}$
From Condition for Resonance in Forced Vibration of Underdamped System, the resonance frequency of $S$ is:
:$\omega = \sqrt {a^2 - 2 b^2}$
:$\nu_R = \dfrac {\sqrt {a^2 - 2 b^2} } {2 \pi}$
We have that:
Thus $\nu_R < \nu$ $a^2 - 2 b^2 > 0$.
But from Condition for Resonance in Forced Vibration of Underdamped System, in order for there to be resonance, $a^2 - 2 b^2 > 0$.
The two conditions are compatible, and so hence the result.
== Sources ==
* : $\S 3.20$: Problem $(1)$
Category:Mathematical Physics
Restricted Dipper Operation is Associative

== Theorem ==
Let $m, n \in \N_{>0}$ be non-zero natural numbers.
Let $\N_{< \paren {m \mathop + n} }$ denote the initial segment of the natural numbers:
:$\N_{< \paren {m \mathop + n} } := \set {0, 1, \ldots, m + n - 1}$
Let $\N^*_{< \paren {m \mathop + n} }$ denote the set defined as $\N_{< \paren {m \mathop + n} } \setminus \set 0$:
:$\N^*_{< \paren {m \mathop + n} } := \set {1, 2, \ldots, m + n - 1}$
The restricted dipper operation $+^*_{m, n}$ on $\N^*_{< \paren {m \mathop + n} }$ is associative.
== Proof ==
By definition, $+^*_{m, n}$ is the restriction of the dipper relation $+_{m, n}$ to $\N_{>0}$.
We have from Dipper Operation is Associative that $+_{m, n}$ is an associative operation.
The result follows from Restriction of Associative Operation is Associative.
== Also see ==
* Restricted Dipper Operation is Commutative
Category:Restricted Dipper Operations
Category:Examples of Associative Operations
Restricted Dipper Operation is Commutative

== Theorem ==
Let $m, n \in \N_{>0}$ be non-zero natural numbers.
Let $\N_{< \paren {m \mathop + n} }$ denote the initial segment of the natural numbers:
:$\N_{< \paren {m \mathop + n} } := \set {0, 1, \ldots, m + n - 1}$
Let $\N^*_{< \paren {m \mathop + n} }$ denote the set defined as $\N_{< \paren {m \mathop + n} } \setminus \set 0$:
:$\N^*_{< \paren {m \mathop + n} } := \set {1, 2, \ldots, m + n - 1}$
The restricted dipper operation $+^*_{m, n}$ on $\N^*_{< \paren {m \mathop + n} }$ is commutative.
== Proof ==
By definition, $+^*_{m, n}$ is the restriction of the dipper relation $+_{m, n}$ to $\N_{>0}$.
We have from Dipper Operation is Associative that $+_{m, n}$ is a commutative operation.
The result follows from Restriction of Commutative Operation is Commutative.
== Also see ==
* Restricted Dipper Operation is Associative
Category:Restricted Dipper Operations
Category:Examples of Commutative Operations
Restricted Dipper Relation is Congruence for Addition

== Theorem ==
Let $m, n \in \N_{>0}$ be non-zero natural numbers.
Let $\RR^*_{m, n}$ be the restricted dipper relation on $\N$:
:$\forall x, y \in \N_{>0}: x \mathrel {\RR^*_{m, n} } y \iff \begin {cases} x = y \\ m \le x < y \text { and } n \divides \paren {y - x} \\ m \le y < x \text { and } n \divides \paren {x - y} \end {cases}$
Then $\RR^*_{m, n}$ is a congruence relation for addition.
== Proof ==
By definition, $\RR^*_{m, n}$ is the restriction of the dipper relation $\RR_{m, n}$ to $\N_{>0}$.
We have from Dipper Relation is Congruence for Addition that $\RR_{m, n}$ is a congruence relation for addition.
The result follows from Restriction of Congruence Relation is Congruence.
== Also see ==
* Restricted Dipper Relation is Congruence for Multiplication
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.7 \ \text {(a)}$
Category:Restricted Dipper Relations
Category:Examples of Congruence Relations
Restricted Dipper Relation is Congruence for Multiplication

== Theorem ==
Let $m, n \in \N_{>0}$ be non-zero natural numbers.
Let $\RR^*_{m, n}$ be the restricted dipper relation on $\N$:
:$\forall x, y \in \N_{>0}: x \mathrel {\RR^*_{m, n} } y \iff \begin {cases} x = y \\ m \le x < y \text { and } n \divides \paren {y - x} \\ m \le y < x \text { and } n \divides \paren {x - y} \end {cases}$
Then $\RR^*_{m, n}$ is a congruence relation for multiplication.
== Proof ==
By definition, $\RR^*_{m, n}$ is the restriction of the dipper relation $\RR_{m, n}$ to $\N_{>0}$.
We have from Dipper Relation is Congruence for Multiplication that $\RR_{m, n}$ is a congruence relation for multiplication.
The result follows from Restriction of Congruence Relation is Congruence.
== Also see ==
* Restricted Dipper Relation is Congruence for Addition
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.7 \ \text {(a)}$
Category:Restricted Dipper Relations
Category:Examples of Congruence Relations
Restricted Dipper Relation is Equivalence Relation

== Theorem ==
Let $m, n \in \N_{>0}$ be non-zero natural numbers.
Let $\RR^*_{m, n}$ be the restricted dipper relation on $\N$:
:$\forall x, y \in \N_{>0}: x \mathrel {\RR^*_{m, n} } y \iff \begin {cases} x = y \\ m \le x < y \text { and } n \divides \paren {y - x} \\ m \le y < x \text { and } n \divides \paren {x - y} \end {cases}$
Then $\RR^*_{m, n}$ is an equivalence relation.
== Proof ==
By definition, $\RR^*_{m, n}$ is the restriction of the dipper relation $\RR_{m, n}$ to $\N_{>0}$.
We have from Dipper Relation is Equivalence Relation that $\RR_{m, n}$ is an equivalence relation.
The result follows from Restriction of Equivalence Relation is Equivalence.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.7 \ \text {(a)}$
Category:Restricted Dipper Relations
Category:Examples of Equivalence Relations
Restricted Measure is Measure

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\Sigma'$ be a sub-$\sigma$-algebra of $\Sigma$.
Then the restricted measure $\mu \restriction_{\Sigma'}$ is a measure on the measurable space $\struct {X, \Sigma'}$.
== Proof ==
Verify the axioms for a measure in turn for $\mu \restriction_{\Sigma'}$:
=== Axiom $(1)$ ===
The statement of axiom $(1)$ for $\mu \restriction_{\Sigma'}$ is:
:$\forall E' \in \Sigma': \map {\mu \restriction_{\Sigma'} } {E'} \ge 0$
Now, for every $E' \in \Sigma'$, compute:
=== Axiom $(2)$ ===
Let $\sequence {E'_n}_{n \mathop \in \N}$ be a sequence of pairwise disjoint sets in $\Sigma'$.
Then the statement of axiom $(2)$ for $\mu \restriction_{\Sigma'}$ is:
:$\ds \map {\mu \restriction_{\Sigma'} } {\bigcup_{n \mathop \in \N} E'_n} = \sum_{n \mathop \in \N} \map {\mu \restriction_{\Sigma'} } {E'_n}$
One can show this by means of the following computation:
=== Axiom $(3')$ ===
The statement of axiom $(3')$ for $\mu \restriction_{\Sigma'}$ is:
:$\map {\mu \restriction_{\Sigma'} } \O = 0$
By Sigma-Algebra Contains Empty Set:
:$\O \in \Sigma'$
Hence:
:$\map {\mu \restriction_{\Sigma'} } \O = \map \mu \O = 0$
because $\mu$ is a measure.
Having verified the measure axioms, it follows that $\mu \restriction_{\Sigma'}$ is a measure.
== Sources ==
* : $\S 4$: Problem $14 \ \text{(i)}$
Category:Restricted Measures
Restricted P-adic Metric is Metric

== Theorem ==
Let $p \in \N$ be a prime.
Let $d^\Z_p$ be the $p$-adic metric on $\Z$:
:$\forall x, y \in \Z: \map {d^\Z_p} {x, y} = \norm {x - y}_p$
where $\norm {x - y}_p$ denotes the $p$-adic norm.
Then $d^\Z_p$ is a metric.
== Proof ==
From $p$-adic Metric is Metric, the $p$-adic metric on $\Q$:
:$\forall x, y \in \Q: \map {d_p} {x, y} = \norm {x - y}_p$
forms a metric space $\struct {\Q, d_p}$.
The mapping:
:$\forall x, y \in \Z: \map {d^\Z_p} {x, y} = \norm {x - y}_p$
is the restriction of $d_p$ to the integers.
Hence the $p$-adic metric on $\Z$ is a metric subspace $\struct {\Z, d^\Z_p}$ of $\struct {\Q, d_p}$.
The result follows from Subspace of Metric Space is Metric Space.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 2$: Metric Spaces: Exercise $8$
Category:P-adic Metrics
Restricted P-adic Valuation is Valuation

== Theorem ==
Let $\nu_p^\Z: \Z \to \Z \cup \set {+\infty}$ be the $p$-adic valuation restricted to the integers.
Then $\nu_p^\Z$ is a valuation.
== Proof ==
To prove that $\nu_p^\Z$ is a valuation it is necessary to demonstrate:
=== Axiom $(\text V 1)$ ===
Let $m, n \in \Z$.
Let $m = 0$ or $n = 0$.
Then:
:$\map {\nu_p^\Z} m = +\infty$
or:
:$\map {\nu_p^\Z} n = +\infty$
Also:
:$n m = 0$
and hence:
Let $n m \ne 0$.
Then by definition of the restricted $p$-adic valuation:
:$p^{\map {\nu_p^\Z} n} \divides n$
:$p^{\map {\nu_p^\Z} n + 1} \nmid n$
Also:
:$p^{\map {\nu_p^\Z} m} \divides m$
:$p^{\map {\nu_p^\Z} m + 1} \nmid m$
Hence:
:$p^{\map {\nu_p^\Z} n + \map {\nu_p^\Z} m} \divides n m$
:$p^{\map {\nu_p^\Z} n + \map {\nu_p^\Z} m + 1} \nmid n m$
So:
:$\map {\nu_p^\Z} {n m} = \map {\nu_p^\Z} n + \map {\nu_p^\Z} m$
=== Axiom $(\text V 2)$ ===
By definition of the restricted $p$-adic valuation:
:$\forall n \in \Z: \map {\nu_p^\Z} n = +\infty \iff n = 0$
=== Axiom $(\text V 3)$ ===
Let $m, n \in \Z$.
 let:
:$p^\alpha \divides n$
:$p^\beta \divides m$
where $\alpha \ge \beta$.
Then $\exists t \in \Z, k \in \Z$ such that:
Thus:
:$p^\beta \divides \paren {m + n}$
Hence by the definition of $\nu_p^\Z$:
:$\map {\nu_p^\Z} {m + n} \ge \min \set {\map {\nu_p^\Z} m, \map {\nu_p^\Z} n}$
Category:P-adic Valuations
Restricted Tukey's Theorem/Strong Form

== Theorem ==
Let $X$ be a set.
Let $\AA$ be a non-empty set of subsets of $X$.
Let $'$ be a unary operation on $X$.
Let $\AA$ have finite character.
For all $A \in \AA$ and all $x \in X$, let either:
: $A \cup \set x \in \AA$
or:
:$A \cup \set {x'} \in \AA$
Then for each $A \in \AA$ there exists a $C \in \AA$ such that:
:$A \subseteq C$
and:
:for all $x \in X$, either $x \in C$ or $x' \in C$.
== Proof ==
Let $A \in \AA$.
Let:
:$\BB = \set {B: B \subseteq X \text{ and } A \cup B \in \AA}$
It is to be shown that $\BB$ has finite character:
First suppose that $B \in \BB$ and $F$ is a Definition:Finite Subset subset of $B$.
Then since $B \in \BB$, $B \subseteq X$ and $A \cup B \in \AA$.
We wish to show that $F \in \BB$.
Since $F \subseteq B \subseteq X$:
:$F \subseteq X$
It remains to be shown that:
:$A \cup F \in \AA$.
$A \cup F \subseteq A \cup B$.
Let $G$ be a finite subset of $A \cup F$.
Then $G$ is a finite subset of $A \cup B$.
Since $A \cup B \in \AA$ and $\AA$ has finite character, $G \in \AA$.
Thus every finite subset of $A \cup F$ is in $\AA$.
Since $\AA$ has finite character, $A \cup F \in \AA$.
Thus $F \in \BB$.
Suppose instead that $B \subseteq X$ and every finite subset of $B$ is an element of $\BB$.
We wish to show that $B \in \BB$.
In order to do this, we must show that $A \cup B \in \AA$.
Let $F$ be a finite subset of $A \cup B$.
$\AA$ has finite character, $F \cap A \in \AA$.
Since $F \cap B$ is a finite subset of $B$, $F \cap B \in \BB$ by assumption.
Then by the definition of $\BB$:
:$\paren {F \cap B} \cup A \in \AA$
But $F$ is a finite subset of $\paren {F \cap B} \cup A \in \AA$.
Since $\AA$ has finite character, $F \in \AA$.
As this holds for all finite subsets of $A \cup B$ and $\AA$ has finite character:
:$A \cup B \in \AA$
Let $B \in \BB$ and $x \in X$.
Then:
:$B \cup A \in \AA$
so either $B \cup A \cup \set x$ or $B \cup A \cup \set {x'}$ is in $\AA$.
But then $B \cup \set x$ or $B \cup \set {x'}$ is in $\BB$ by the definition of $\BB$.
Thus $\BB$ satisfies the premises of the weak form of the Restricted Tukey's Theorem.
Thus there is a $B \in \BB$ such that for all $x \in X$, either $x \in B$ or $x' \in B$.
Let $C = A \cup B$.
Then since $B \subseteq C$, if $x \in X$ then either $x \in C$ or $x' \in C$.
But since $B \in \BB$:
:$C = A \cup B \in \AA$
== Sources ==
* 
Category:Tukey's Lemma
Restricted Tukey's Theorem/Weak Form

== Theorem ==
Let $X$ be a set.
Let $\AA$ be a non-empty set of subsets of $X$.
Let $'$ be a unary operation on $X$.
Let $\AA$ have finite character.
For all $A \in \AA$ and all $x \in X$, let either:
:$A \cup \set x \in \AA$
or:
:$A \cup \set {x'} \in \AA$
Then there exists a $B \in \AA$ such that for all $x \in X$, either $x \in B$ or $x' \in B$.
== Proof ==
== Sources ==
* 
Category:Tukey's Lemma
Restricting Measure Preserves Finiteness

== Theorem ==
Let $\left({X, \Sigma, \mu}\right)$ be a measure space.
Let $\mu$ be a finite measure.
Let $\Sigma'$ be a sub-$\sigma$-algebra of $\Sigma$.
Then the restricted measure $\mu \restriction_{\Sigma'}$ is also a finite measure.
== Proof ==
By Restricted Measure is Measure, $\mu \restriction_{\Sigma'}$ is a measure.
Now by definition of $\mu \restriction_{\Sigma'}$, have:
:$\mu \restriction_{\Sigma'} \left({X}\right) = \mu \left({X}\right) < \infty$
as $\mu$ is a finite measure.
Hence $\mu \restriction_{\Sigma'}$ is also a finite measure.
== Sources ==
* : $\S 4$: Problem $14 \ \text{(ii)}$
Category:Restricted Measures
Category:Finite Measures
Restriction/Mapping/Examples/Bijective Restriction of Real Sine Function

== Example of Restriction of Mapping ==
Let $f: \R \to \R$ be the mapping defined as:
:$\forall x \in \R: f \paren x = \sin x$
Then a bijective restriction $g$ of $f$ can be defined as:
:$g: S \to T: \forall x \in S: g \paren x = \sin x$
where:
:$S = \closedint {-\dfrac \pi 2} {\dfrac \pi 2}$
:$T = \closedint {-1} 1$
== Proof ==
We note that $f$ is neither injective nor surjective:
From Shape of Sine Function:
From Graph of Sine Function this is apparent:
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 9$: Inverse Functions, Extensions, and Restrictions
* : $\S 23$: Restriction of a Mapping
Category:Examples of Restrictions of Mappings
Restriction/Mapping/Examples/Restriction of Square Function on Natural Numbers

== Example of Restriction of Mapping ==
Let $f: \N \to \N$ be the mapping defined as:
:$\forall n \in \N: \map f n = n^2$
Let $S = \set {x \in \N: \exists y \in \N_{>0}: x = 2 y} = \set {2, 4, 6, 8, \ldots}$
Let $g: S \to \N$ be the mapping defined as:
:$\forall n \in \N: \map g n = n^2$
Then $g$ is a restriction of $f$.
== Proof ==
== Sources ==
* : $\S 2$: Sets and functions: Graphs and functions
Category:Examples of Restrictions of Mappings
Restriction is Subset of Relation

== Theorem ==
Let $\RR \subseteq S \times T$ be a relation.
Let $X \subseteq S$.
Then the restriction of $\RR$ to $X$ is a subset of $\RR$.
== Proof ==
From the definition of restriction:
:$\forall x \in X: \map {\RR \restriction_X} x = \map \RR x$
Thus:
:$\forall x \in X: \exists t \in T: \tuple {x, t} \in \RR \restriction_X$
But $\tuple {x, t}$ is also (by definition) in $\RR$.
It follows that:
:$\RR \restriction_X \subseteq \RR$
Category:Restrictions
Restriction of Antireflexive Relation is Antireflexive

== Theorem ==
Let $S$ be a set.
Let $\RR \subseteq S \times S$ be an antireflexive relation on $S$.
Let $T \subseteq S$ be a subset of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the restriction of $\RR$ to $T$.
Then $\RR {\restriction_T}$ is an antireflexive relation on $T$.
== Proof ==
Suppose $\RR$ is antireflexive on $S$.
Then by definition of antireflexive:
:$\forall x \in S: \tuple {x, x} \notin \RR$
We are given $T$ is a subset of $S$, so:
:$\forall x \in T: \tuple {x, x} \notin \RR$
We are given $\RR {\restriction_T} \subseteq T \times T$, so:
:$\forall x \in T: \tuple {x, x} \notin \RR \restriction_T$
Hence by definition, $\RR {\restriction_T}$ is antireflexive on $T$.
== Also see ==
* Properties of Restriction of Relation‎ for other similar properties of the restriction of a relation.
Category:Antireflexive Relations
Category:Restrictions
Restriction of Antisymmetric Relation is Antisymmetric

== Theorem ==
Let $S$ be a set.
Let $\RR \subseteq S \times S$ be an antisymmetric relation on $S$.
Let $T \subseteq S$ be a subset of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the restriction of $\RR$ to $T$.
Then $\RR {\restriction_T}$ is an antisymmetric relation on $T$.
== Proof ==
Suppose $\RR$ is antisymmetric on $S$.
Then:
Thus $\RR {\restriction_T}$ is antisymmetric on $T$.
== Also see ==
* Properties of Restriction of Relation‎ for other similar properties of the restriction of a relation.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings
Category:Antisymmetric Relations
Category:Restrictions
Restriction of Antitransitive Relation is Antitransitive

== Theorem ==
Let $S$ be a set.
Let $\RR \subseteq S \times S$ be an antitransitive relation on $S$.
Let $T \subseteq S$ be a subset of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the restriction of $\RR$ to $T$.
Then $\RR {\restriction_T}$ is an antitransitive relation on $T$.
== Proof ==
Suppose $\RR$ is antitransitive on $S$.
Then by definition:
:$\set {\tuple {x, y}, \tuple {y, z} } \subseteq \RR \implies \tuple {x, z} \notin \RR$
So:
Therefore, if $x, y, z \in T$, it follows that:
:$\set {\tuple {x, y}, \tuple {y, z} } \subseteq \RR {\restriction_T} \implies \tuple {x, z} \notin \RR {\restriction_T}$
and so by definition $\RR {\restriction_T}$ is an antitransitive relation on $T$.
== Also see ==
* Properties of Restriction of Relation‎ for other similar properties of the restriction of a relation.
Category:Antitransitive Relations
Category:Restrictions
Restriction of Associative Operation is Associative

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
Let $T \subseteq S$.
Let $\circ_T$ denote the restriction of $\circ$ to $T$.
Let the operation $\circ$ be associative on $\struct {S, \circ}$.
Then $\circ_T$ is associative on $\struct {T, \circ_T}$.
== Proof ==
== Sources ==
* : $\S 5.1$. Subsets closed to an operation
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets
* : Chapter $\text{II}$: Groups: Subgroups
* : Chapter $4$: Subgroups
Category:Associativity
Category:Semigroups
Restriction of Asymmetric Relation is Asymmetric

== Theorem ==
Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a asymmetric relation on $S$.
Let $T \subseteq S$ be a subset of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the restriction of $\RR$ to $T$.
Then $\RR {\restriction_T}$ is a asymmetric relation on $T$.
== Proof ==
Suppose $\RR$ is asymmetric on $S$.
Then:
and so $\RR {\restriction_T}$ is asymmetric on $T$.
== Also see ==
* Properties of Restriction of Relation‎ for other similar properties of the restriction of a relation.
Category:Asymmetric Relations
Category:Restrictions
Restriction of Canonical Surjection to Restricted Dipper Semigroup is Isomorphism

== Theorem ==
Let $m, n \in \N_{>0}$ be non-zero natural numbers.
Let $\N_{< \paren {m \mathop + n} }$ denote the initial segment of the natural numbers:
:$\N_{< \paren {m \mathop + n} } := \set {0, 1, \ldots, m + n - 1}$
Let $\N^*_{< \paren {m \mathop + n} }$ denote the set defined as $\N_{< \paren {m \mathop + n} } \setminus \set 0$:
:$\N^*_{< \paren {m \mathop + n} } := \set {1, 2, \ldots, m + n - 1}$
Let $+_{m, n}$ be the dipper operation on $\N_{< \paren {m \mathop + n} }$:
:$\forall a, b \in \N_{< \paren {m \mathop + n} }: a +_{m, n} b = \begin{cases} a + b & : a + b < m \\ a + b - k n & : a + b \ge m \end{cases}$
where $k$ is the largest integer satisfying:
:$m + k n \le a + b$
Let $+^*_{m, n}$ be the restricted dipper operation on $\N^*_{< \paren {m \mathop + n} }$ defined as:
:$\forall a, b \in \N^*_{< \paren {m \mathop + n} }: a +^*_{m, n} b = \begin{cases} a + b & : a + b < m \\ a + b - k n & : a + b \ge m \end{cases}$
where $k$ is the largest integer satisfying:
:$m + k n \le a + b$
Consider the algebraic structure $\struct {\N^*_{< \paren {m \mathop + n} }, +^*_{m, n} }$.
Let $\phi^*_{m, n}$ be the restriction of the canonical surjection from $\N_{>0}$ to the restricted dipper semigroup $\struct {\map {D^*} {m, n}, \oplus^*_{m, n} }$.
Then $\phi^*_{m, n}$ is an isomorphism from $\struct {\N^*_{< \paren {m \mathop + n} }, +^*_{m, n} }$ to $\struct {\map {D^*} {m, n}, \oplus^*_{m, n} }$.
== Proof ==
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.7 \ \text {(b)}$
Category:Restricted Dipper Semigroups
Restriction of Commutative Operation is Commutative

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
Let $T \subseteq S$.
Let the operation $\circ$ be commutative on $\struct {S, \circ}$.
Then the restriction $\circ {\restriction_T}$ of $\circ$ to $T$ is also commutative.
== Proof ==
== Sources ==
* : $\S 5.1$. Subsets closed to an operation
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets
Category:Commutativity
Restriction of Composition is Composition of Restriction

== Theorem ==
Let $X, Y, Z$ be sets.
Let $f: X \to Y$ and $g: Y \to Z$ be mappings.
Let $S \subseteq X$.
Then:
:$\paren {g \circ f} \restriction S = g \circ \paren {f \restriction S}$
== Proof ==
By definitions of composition of mappings and restriction of mapping:
:$\paren {g \circ f} \restriction S: S \to Z$ and $g \circ \paren {f \restriction S}: S \to Z$
Let $s \in S$.
By definition of restriction of mapping:
:$\map {\paren {\paren {g \circ f} \restriction S} } s = \map {\paren {g \circ f} } s$
Thus
== Sources ==
* 
Category:Mapping Theory
Restriction of Congruence Relation is Congruence

== Definition ==
Let $\struct {S, \circ}$ be an algebraic structure.
Let $\RR$ be a congruence relation for $\circ$ on $S$.
Let $T \subseteq S$ be a subset of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the restriction of $\RR$ to $T$.
Then $\RR {\restriction_T}$ is a congruence relation for $\circ_T$ on $T$.
== Proof ==
Hence the result by definition of congruence relation.
Category:Congruence Relations
Category:Restrictions
Restriction of Connected Relation is Connected

== Theorem ==
Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a connected relation on $S$.
Let $T \subseteq S$ be a subset of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the restriction of $\RR$ to $T$.
Then $\RR {\restriction_T}$ is a connected relation on $T$.
== Proof ==
Suppose $\RR$ is connected on $S$.
That is:
:$\forall a, b \in S: a \ne b \implies \tuple {a, b} \in \RR \lor \tuple {b, a} \in \RR$
So:
and so $\RR {\restriction_T}$ is connected on $T$.
== Also see ==
* Properties of Restriction of Relation‎ for other similar properties of the restriction of a relation.
Category:Connected Relations
Category:Restrictions
Restriction of Continuous Mapping is Continuous/Metric Spaces

== Theorem ==
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $S \subseteq M_1$ be a subset of $M_1$.
Let $f: M_1 \to M_2$ be a mapping which is continuous at a point $\alpha \in S$.
Let $f \restriction_S = g: S \to M_2$ be the restriction of $f$ to $S$.
Then $g$ is continuous at $\alpha$.
== Proof ==
Let $\sequence {z_n}$ be a sequence in $S$ such that:
:$\ds \lim_{n \mathop \to \infty} z_n = \alpha$
Since $\sequence {z_n}$ and $\alpha$ both lie in $S$:
:$\ds \lim_{n \mathop \to \infty} \map f {z_n} \to \alpha$
But:
:$\forall n \in \N: \map g {z_n} = \map f {z_n}$
and also:
:$\map g \alpha = \map f \alpha$
The result follows from Limit of Function by Convergent Sequences.
Category:Metric Spaces
Category:Continuous Mappings on Metric Spaces
Restriction of Continuous Mapping is Continuous/Topological Spaces

== Theorem ==
Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $M_1 \subseteq S_1$ be a subset of $S_1$.
Let $f: S_1 \to S_2$ be a mapping which is continuous.
Let $M_2 \subseteq S_2$ be a subset of $S_2$ such that $f \sqbrk {M_1} \subseteq M_2$.
Let $f \restriction_{M_1 \times M_2}: M_1 \to M_2$ be the restriction of $f$ to $M_1 \times M_2$.
Then $f \restriction_{M_1 \times M_2}$ is continuous, where $M_1$ and $M_2$ are equipped with the respective subspace topologies.
== Proof 1 ==
== Proof 2 ==
Category:Topology
Category:Continuous Mappings