nyuuzyou/wikis huggingface dataset:
Restriction of Continuous Mapping is Continuous/Topological Spaces/Proof 1

== Theorem ==
== Proof ==
Let $V \subseteq M_2$ be an open set with respect to the subspace topology of $M_2$.
By definition of subspace topology, $V = U \cap M_2$ for an open set $U \in \tau_2$.
We have that:
By definition of continuous mapping, $f^{-1} \sqbrk U$ is open in $T_1$.
By definition of subspace topology, $f^{-1} \sqbrk U \cap M_1$ is open with respect to the subspace topology of $M_1$.
As $f^{-1} \sqbrk U \cap M_1 = {f \restriction_{M_1 \times M_2} }^{-1} \sqbrk V$ it follows that ${f \restriction_{M_1 \times M_2} }^{-1} \sqbrk V$ is open with respect to the subspace topology of $M_1$.
Since $V \subseteq M_2$ was an arbitrary open set, $f$ is continuous.
Restriction of Continuous Mapping is Continuous/Topological Spaces/Proof 2

== Theorem ==
== Proof ==
Consider first the restriction $f \restriction_{M_1}$.
From Composition of Mapping with Inclusion is Restriction:
:$f \restriction_{M_1} = f \circ i_{M_1}$
where $i_{M_1}$ is the inclusion of $M_1$ into $S_1$.
From Continuity of Composite with Inclusion: Mapping on Inclusion, it follows that $f \circ i_{M_1} = f \restriction_{M_1}$ is continuous.
Consider now a mapping $\tilde f : S_1 \to M_2$ where $f \sqbrk {S_1} \subseteq M_2$
Then:
$f = i_{M_2} \circ \tilde f$
where $i_{M_2}$ is the inclusion of $M_2$ into $S_2$.
From Continuity of Composite with Inclusion: Inclusion on Mapping, it follows that $\tilde f$ is continuous $i_{M_2} \circ \tilde f = f$ is continuous.
Thus $\tilde f$ is continuous, since $f$ is continuous by assumption.
The result follows from combining these partial results.
== Sources ==
* : $\S 3$: New Spaces From Old: Subspaces
Restriction of Dipper Operation to Non-Zero Initial Segment is Closed

== 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 $+_{m, n}$ denote 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 $\N^*_{< \paren {m \mathop + n} }$ denote the set defined as $\N_{< \paren {m \mathop + n} } \setminus \set 0$:
:$\N^*_{< \paren {m \mathop + n} } := \set {1, 2, \ldots, m + n - 1}$
Then $+_{m, n}$ is closed on $\N^*_{< \paren {m \mathop + n} }$.
== Proof ==
From Dipper Operation is Closed on Initial Segment, $+_{m, n}$ is closed on $\N_{< \paren {m \mathop + n} }$.
That is:
:$\forall a, b \in \N_{< \paren {m \mathop + n} }: a +_{m, n} b < m + n$
It remains to be shown that:
:$\forall a, b \in \N^*_{< \paren {m \mathop + n} }: a +_{m, n} b > 0$
Let $a + b < m$.
Then as $a > 0$ and $b > 0$ it follows that $a + b < 0$.
That is:
:$a +_{m, n} b > 0$
Let $a + b > m$.
Then:
:$a +_{m, n} b = a + b - k n$
where $k$ is the largest integer satisfying $m + k n \le a + b$.
That is:
:$m < a + b - k n$
But as $m > 0$ we must have that:
:$0 < a +_{m, n} b$
and the result follows.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.7 \ \text {(b)}$
Category:Restricted Dipper Operations
Category:Dipper Operations
Category:Algebraic Closure
Restriction of Equivalence Relation is Equivalence

== Theorem ==
Let $S$ be a set.
Let $\RR \subseteq S \times S$ be an equivalence 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 equivalence relation on $T$.
== Proof ==
Let $\RR$ be an equivalence relation on $S$.
Then by definition:
:$\RR$ is a reflexive relation on $S$
:$\RR$ is a symmetric relation on $S$
:$\RR$ is a transitive relation on $S$.
Then:
:from Restriction of Reflexive Relation is Reflexive, $\RR {\restriction_T}$ is a reflexive relation on $T$
:from Restriction of Symmetric Relation is Symmetric, $\RR {\restriction_T}$ is a symmetric relation on $T$
:from Restriction of Transitive Relation is Transitive, $\RR {\restriction_T}$ is a transitive relation on $T$
and so it follows by definition that $\RR {\restriction_T}$ is an equivalence relation on $T$.
== Also see ==
* Restriction of Ordering is Ordering
* 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
* : $1$: Set Theory and Logic: $\S 3$: Relations: Exercise $3.2$
Category:Equivalence Relations
Category:Restrictions
Restriction of Homeomorphism is Homeomorphism

== Theorem ==
Let $T_1 = \struct {S_1, \tau_1}$, $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $f: S_1 \to S_2$ be a homeomorphism between $T_1$ and $T_2$.
Let $S$ be a subset of $S_1$.
Let $f {\restriction_{S \times f \sqbrk S} } : S \to f \sqbrk S$ be the restriction of $f$ to $S \times f \sqbrk S$.
Let $S$ and $f \sqbrk S$ bear their respective subspace topologies.
Then $f {\restriction_{S \times f \sqbrk S} }$ is a homeomorphism.
== Proof ==
By Restriction of Continuous Mapping is Continuous, $f {\restriction_{S \times f \sqbrk S} }$ is continuous.
By Restriction of Inverse is Inverse of Restriction, $\left({f {\restriction_{S \times f \sqbrk S} } }\right)^{-1}$ is well-defined and equal to $f^{-1} {\restriction_{f \sqbrk S \times S} }$.
By Restriction of Continuous Mapping is Continuous, $f^{-1} {\restriction_{f \sqbrk S \times S} }$ is continuous.
Since $\paren {f {\restriction_{S \times f \sqbrk S} } }^{-1} = f^{-1} {\restriction_{f \sqbrk S \times S} }$, $\paren {f {\restriction_{S \times f \sqbrk S} } }^{-1}$ is continuous.
Hence, $f {\restriction_{S \times f \sqbrk S} }$ is a homeomorphism.
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9 : 17$
Category:Homeomorphisms (Topological Spaces)
Restriction of Homomorphism is Homomorphism

== Theorem ==
Let $\struct {S, \circ}$ and $\struct {T, \odot}$ be algebraic structures.
Let $\phi: S \to T$ be a homomorphism.
Let $A \subseteq S$ be a subset of $S$.
Then the restriction of $\phi$ to $A \times \Img A$ is also a homomorphism.
== Proof ==
Category:Homomorphisms (Abstract Algebra)
Category:Restrictions
Restriction of Homomorphism to Image is Epimorphism

== Theorem ==
Let $S$ and $T$ be algebraic structures.
Let $\phi: S \to T$ be a homomorphism.
Then a surjective restriction of $\phi$ can be produced by limiting the codomain of $\phi$ to its image $\Img \phi$.
== Proof ==
Let $\phi: S \to T$ be a homomorphism.
Let $\Img \phi = T'$
By Morphism Property Preserves Closure, $T'$ is closed.
From Restriction of Mapping to Image is Surjection, $\phi \to \Img \phi$ is a surjection.
Thus $\phi: S \to T$ is an epimorphism.
Therefore, by suitably restricting the codomain of a homomorphism, it is possible to regard it as an epimorphism.
Category:Epimorphisms (Abstract Algebra)
Restriction of Idempotent Operation is Idempotent

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
Let $T \subseteq S$.
Let the operation $\circ$ be idempotent.
Then $\circ$ is also idempotent upon restriction to $\struct {T, \circ \restriction_T}$.
== Proof ==
Category:Abstract Algebra
Category:Idempotence
Restriction of Injection is Injection

== Theorem ==
Let $f: S \to T$ be an injection.
Let $X \subseteq S$ be a subset of $S$.
Let $f \sqbrk X$ denote the image of $X$ under $f$.
Let $Y \subseteq T$ be a subset of $T$ such that $f \sqbrk X \subseteq Y$.
The restriction $f \restriction_{X \times Y}$ of $f$ to $X \times Y$ is an injection from $X$ to $Y$.
== Proof ==
First we show that $f \restriction_{X \times Y}$ is a mapping from $X$ to $Y$.
By Restriction of Mapping is Mapping, $f \restriction_{X \times T}$ is a mapping from $X$ to $T$.
If $x \in X$, then by the definition of image:
:$\map f x \in f \sqbrk X$
Since $f \sqbrk X \subseteq Y$, $f \restriction_{X \times Y}$ is a mapping from $X$ to $Y$.
By definition of an injection:
:$\forall s_1, s_2 \in S: \map f {s_1} = \map f {s_2} \implies s_1 = s_2$.
 $f \restriction_{X \times Y}: X \to Y$ were not an injection.
Then:
:$\exists x_1, x_2 \in X: x_1 \ne x_2, \map f {x_1} = \map f {x_2}$
But then:
:$\exists x_1, x_2 \in S: x_1 \ne x_2, \map f {x_1} = \map f {x_2}$
So $f: S \to T$ would not then be an injection.
Hence the result.
Category:Injections
Restriction of Mapping is Mapping

== Theorem ==
Let $f: S \to T$ be a mapping.
Let $X \subseteq S$.
Let $f \restriction_X$ be the restriction of $f$ to $X$.
Then $f \restriction_X: X \to T$ is a mapping:
: whose domain is $X$
: whose preimage is $X$.
=== General Result ===
== Proof ==
As $f: S \to T$ is a mapping, we have that:
:$\forall x \in S: \tuple {x, y_1} \in f \land \tuple {x, y_2} \in f \implies y_1 = y_2$
From the definition of a subset, $x \in X \implies x \in S$, and so:
:$\forall x \in X: \tuple {x, y_1} \in f \restriction_X \land \tuple {x, y_2} \in f \restriction_X \implies y_1 = y_2$
Also from the definition of a mapping:
:$\forall x \in S: \exists y \in T: \tuple {x, y} \in f$
Again from the definition of a subset, $x \in X \implies x \in S$, and so:
:$\forall x \in X: \exists y \in T: \tuple {x, y} \in f \restriction_X$
So $f \restriction_X: X \to T$ is a mapping.
The fact that the domain of $f \restriction_X$ is $X$ follows from the definition of domain.
The fact that the preimage of $f \restriction_X$ is also $X$ follows from Preimage of Mapping equals Domain.
== Sources ==
* : $\S 1.6$
Category:Restrictions
Restriction of Mapping is Mapping/General Result

== Theorem ==
Let $f: S \to T$ be a mapping.
Let $X \subseteq S$.
Let $f \sqbrk X \subseteq Y \subseteq T$.
Let $f \restriction_{X \times Y}$ be the restriction of $f$ to $X \times Y$.
Then $f \restriction_{X \times Y}: X \to Y$ is a mapping:
: whose domain is $X$
: whose preimage is $X$
: whose codomain is $Y$.
== Proof ==
=== $f \restriction_{X \times Y}$ is Many-to-one ===
We have:
=== $f \restriction_{X \times Y}$ is Left-total ===
We have:
Hence by definition, $f \restriction_{X \times Y}: X \to Y$ is a mapping.
By definition of domain, the domain of $f \restriction_{X \times Y}$ is $X$.
By definition of codomain, the codomain of $f \restriction_{X \times Y}$ is $Y$.
From Preimage of Mapping equals Domain, the preimage of $f \restriction_{X \times Y}$ is also $X$.
Category:Restrictions
Restriction of Mapping is Subclass of Cartesian Product

== 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 $f$.
Let $f {\restriction} A$ denote the restriction of $f$ to $A$.
Then $f {\restriction} A$ is a subclass of the cartesian product of $A$ with its image:
:$f {\restriction} A \subseteq A \times f \sqbrk A$
== Proof ==
Follows directly from:
:the definition of restriction
:the definition of mapping.
== Sources ==
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 1$ A few preliminaries
Category:Images
Category:Cartesian Product
Restriction of Mapping is its Intersection with Cartesian Product of Subset with Image

== Theorem ==
Let $f: S \to T$ be a mapping.
Let $X \subseteq S$.
Let $f {\restriction_X}$ be the restriction of $f$ to $X$.
Then:
:$f {\restriction_X} = f \cap \paren {X \times \Img f}$
where:
:$\Img f$ denotes the image of $f$, defined as:
::$\Img f = \set {t \in T: \exists s \in S: t = \map f s}$
:$X \times \Img f$ denotes the cartesian product of $X$ with $\Img f$.
== Proof ==
By Restriction of Mapping is Mapping, $f {\restriction_X}: X \to T$ is a mapping.
We have:
Then:
Thus we have:
== Sources ==
* : $\S 1$: Naive Set Theory: $\S 1.6$: Functions: Exercise $1.6.4 \ \text{(ii)}$
Category:Restrictions
Restriction of Mapping to Image is Surjection

== Theorem ==
Let $f: S \to T$ be a mapping.
Let $g: S \to \Img f$ be the restriction of $f$ to $S \times \Img f$.
Then $g$ is a surjective restriction of $f$.
== Proof ==
From Image is Subset of Codomain:
:$\Img g \subseteq T$
Furthermore, by definition of image, we have:
:$\forall s \in S: g \paren s \in \Img g$
Therefore, $g$ may be viewed as a mapping $g: S \to \Img g$.
Thus, by definition, $g$ is a surjection.
== Comment ==
Thus, for any mapping $f: S \to T$ which is not surjective, by restricting its codomain to its image, it can be considered as a surjection.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 5$: Composites and Inverses of Functions
* : $\S 23$: Restriction of a Mapping
Category:Restrictions
Category:Surjections
Restriction of Mapping to Small Class is Small

== Theorem ==
Let $F$ be a mapping.
Let $A$ be a small class.
Then the restriction $F {\restriction_A}$ is a small class.
== Proof ==
The domain of $F {\restriction_A}$ is a subset of $A$.
By Axiom of Subsets Equivalents, the domain is a small class.
By Mapping whose Domain is Small Class is Small, it follows that $F {\restriction_A}$ is a small class.
== Sources ==
* : $\S 6.16$
Category:Zermelo-Fraenkel Class Theory
Restriction of Measurable Function is Measurable on Trace Sigma-Algebra

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space. 
Let $f : X \to \overline \R$ be a $\Sigma$-measurable functions. 
Let $E \in \Sigma$. 
Let $\Sigma_E$ be the trace $\sigma$-algebra of $E$ in $\Sigma$. 
Then the restriction $f \restriction_E$ is $\Sigma_E$-measurable. 
== Proof ==
From the definition of a $\Sigma_E$-measurable function, we aim to show that: 
:$\set {x \in E : \map f x \le \alpha} \in \Sigma_E$
for each $\alpha \in \R$.
Let $\alpha \in \R$. 
We have: 
:$\set {x \in E : \map f x \le \alpha} = \set {x \in X : \map f x \le \alpha} \cap E$
Since $f$ is $\Sigma$-measurable, we have: 
:$\set {x \in X : \map f x \le \alpha} \in E$
So, from the definition of trace $\sigma$-algebra, we have: 
:$\set {x \in X : \map f x \le \alpha} \cap E \in \Sigma_E$
So we have: 
:$\set {x \in E : \map f x \le \alpha} \in \Sigma_E$
for each $\alpha \in \R$.
So:
:$f \restriction_E$ is $\Sigma_E$-measurable.
Category:Measurable Functions
Category:Trace Sigma-Algebras
Restriction of Measure to Trace Sigma-Algebra of Measurable Set is Measure

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a measure on $\struct {X, \Sigma}$. 
Let $A \in \Sigma$. 
Let $\Sigma_A$ be the trace $\sigma$-algebra of $A$ in $\Sigma$.
Let $\mu \restriction_{\Sigma_A}$ be the restriction of $\mu$ to $\Sigma_A$.
Then $\mu \restriction_{\Sigma_A}$ is a measure on $\struct {A, \Sigma_A}$. 
== Proof ==
We verify the three conditions required of a measure for $\mu \restriction_{\Sigma_A}$.
Note that from Trace Sigma-Algebra of Measurable Set, we have $\Sigma_A \subseteq \Sigma$.
=== Condition $(1)$ ===
Let $E \in \Sigma_A$. 
Then, we have $E \in \Sigma$ and: 
=== Condition $(2)$ ===
Let $\sequence {E_n}_{n \mathop \in \N}$ be a sequence of pairwise disjoint $\Sigma_A$-measurable sets.
Then $\sequence {E_n}_{n \mathop \in \N}$ be a sequence of pairwise disjoint $\Sigma$-measurable sets, and we have: 
=== Condition $(3)$ ===
From Empty Set is Null Set, we have that $\O$ is $\mu$-null.
From Sigma-Algebra Contains Empty Set, we have $\O \in \Sigma_A$.
So, we have: 
Category:Measures
Category:Trace Sigma-Algebras
Restriction of Non-Connected Relation is Not Necessarily Non-Connected

== Theorem ==
Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a non-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 not necessarily a non-connected relation on $T$.
== Proof ==
Proof by Counterexample:
Let $S = \set {a, b}$.
Let $\RR = \set {\tuple {a, a}, \tuple {b, b} }$.
$\RR$ is a non-connected relation, as can be seen by definition: neither $a \mathrel \RR b$ nor $b \mathrel \RR a$.
Now let $T = \set a$.
Then $\RR {\restriction_T} = \set {\tuple {a, a} }$.
Then $\RR {\restriction_T}$ is trivially connected on $T$.
== Also see ==
* Properties of Relation Not Preserved by Restriction‎ for other similar results.
Category:Connected Relations
Category:Restrictions
Restriction of Non-Continuous Mapping on Metric Space to Subspace may be Continuous

== Theorem ==
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be a metric spaces.
Let $f: A_1 \to A_2$ be a mapping.
Let $Y \subseteq A_1$.
Let $f {\restriction_Y}: Y \to A_2$ be the restriction of $f$ to $Y$.
Let $f {\restriction_Y}$ be $\tuple {d_Y, d_2}$-continuous.
Then it is not necessarily the case that $f$ is also $\tuple {d_1, d_2}$-continuous.
== Proof ==
Proof by Counterexample:
Let $f: \R \to \R$ be given by:
:$\map f x = \begin {cases}
0 & : x \in \Q \\
1 & : x \in \R \setminus \Q
\end {cases}$
where $\Q$ is the set of rational numbers.
Then $f {\restriction_\Q}: \Q \to \R$ is the constant function $f_0$ with value $0$.
By Constant Mapping is Continuous, $f {\restriction_\Q}$ is continuous at every point.
However, $f$ is not continuous at any point on $\R$.
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.2$: Examples: Example $2.2.6$
Category:Continuity
Category:Metric Spaces
Restriction of Non-Reflexive Relation is Not Necessarily Non-Reflexive

== Theorem ==
Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a non-reflexive 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 not necessarily a non-reflexive relation on $T$.
== Proof ==
Proof by Counterexample:
Let $S = \set {a, b}$.
Let $\RR = \set {\tuple {b, b} }$.
$\RR$ is a non-reflexive relation, as can be seen by definition:
:$\tuple {a, a} \notin \RR$
:$\tuple {b, b} \in \RR$
Now let $T = \set a$.
Then $\RR {\restriction_T} = \O$.
So:
:$\forall x \in T: \tuple {x, x} \notin \RR {\restriction_T}$
That is, $\RR {\restriction_T}$ is an antireflexive relation on $T$.
That is, specifically not a non-reflexive relation.
== Also see ==
* Properties of Relation Not Preserved by Restriction‎ for other similar results.
Category:Non-Reflexive Relations
Category:Restrictions
Restriction of Non-Symmetric Relation is Not Necessarily Non-Symmetric

== Theorem ==
Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a non-symmetric 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 not necessarily a non-symmetric relation on $T$.
== Proof ==
Proof by Counterexample:
Let $S = \set {a, b}$.
Let $\RR = \set {\tuple {a, b}, \tuple {b, b} }$.
$\RR$ is a non-symmetric relation, as can be seen by definition.
Now let $T = \set b$.
Then $\RR {\restriction_T} \ = \set {\tuple {b, b} }$.
So:
:$\forall x, y \in T: \tuple {x, y} \in \RR {\restriction_T} \implies \tuple {y, x} \in \RR {\restriction_T}$
as can be seen by setting $x = y = b$.
So $\RR {\restriction_T}$ is a symmetric relation on $T$.
That is, $\RR {\restriction_T}$ is not a non-symmetric relation on $T$.
== Also see ==
* Properties of Relation Not Preserved by Restriction‎ for other similar results.
Category:Non-Symmetric Relations
Category:Restrictions
Restriction of Non-Transitive Relation is Not Necessarily Non-Transitive

== Theorem ==
Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a non-transitive 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 not necessarily a non-transitive relation on $T$.
== Proof ==
Proof by Counterexample:
Let $S = \set {a, b}$.
Let $\RR = \set {\tuple {a, b}, \tuple {b, a}, \tuple {b, b} }$.
$\RR$ is a non-transitive relation, as can be seen by definition.
Now let $T = \set b$.
Then:
:$\RR {\restriction_T} = \set {\tuple {b, b} }$
So:
:$\forall x, y \in T: \tuple {x, y} \in \RR {\restriction_T} \land \tuple {y, z} \in \RR {\restriction_T} \implies \tuple {y, z} \in \RR {\restriction_T}$
as can be seen by setting $x = y = z = b$.
So $\RR {\restriction_T}$ is a transitive relation on $T$.
That is, $\RR {\restriction_T}$ is not a non-transitive relation on $T$.
== Also see ==
* Properties of Relation Not Preserved by Restriction‎ for other similar results.
Category:Non-Transitive Relations
Category:Restrictions
Restriction of Norm on Vector Space to Subspace is Norm

== Theorem ==
Let $\Bbb F$ be a subfield of $\C$. 
Let $X$ be a vector space over $\Bbb F$.
Let $\norm \cdot_X : X \to \hointr 0 \infty$ be a norm on $X$.
Let $Y$ be a vector subspace of $X$.
Let $\norm \cdot_Y$ denote the restriction of $\norm \cdot_X$ to $Y$.
Then $\norm \cdot_Y$ is a norm on $Y$. 
== Proof ==
Since $\norm x_X \ge 0$ for any $x \in X$, we have $\norm y_Y \ge 0$ for any $y \in Y$.
We verify each of the axioms for a norm on a vector space.
=== ===
Note that for $y \in Y$ we have $\norm y_Y = 0$ $\norm y_X = 0$.
Then by $(\text N 1)$ for $\norm \cdot_X$, we have $y = 0$. 
=== ===
Let $\lambda \in \Bbb F$ and $y \in Y$. 
Then, we have:
=== ===
Let $x, y \in Y$.
Then, we have: 
Category:Normed Vector Spaces
Restriction of Operation Distributivity

== Theorem ==
Let $\struct {S, *, \circ}$ be an algebraic structure.
Let $T \subseteq S$.
If the operation $\circ$ is distributive over $*$ in $\struct {S, *, \circ}$, then it is also distributive over $*$ on a restriction $\struct {T, * \restriction_T, \circ \restriction_T}$.
== Proof ==
Similarly for $\paren {a *_T b} \circ_T c = \paren {a \circ_T c} *_T \paren {b \circ_T c}$.
Category:Abstract Algebra
Category:Distributive Operations
Restriction of Ordering is Ordering

== Theorem ==
Let $S$ be a set.
Let $\preceq$ be an ordering on $S$.
Let $T \subseteq S$ be a subset of $S$.
Let $\preceq \restriction_T$ be the restriction of $\preceq$ to $T$.
Then $\preceq \restriction_T$ is an ordering on $T$.
== Proof ==
Let $\preceq$ be an ordering on $S$.
Then, by definition:
:$\preceq$ is a reflexive relation on $S$
:$\preceq$ is an antisymmetric relation on $S$
:$\preceq$ is a transitive relation on $S$.
Then:
:from Restriction of Reflexive Relation is Reflexive, $\preceq \restriction_T$ is a reflexive relation on $T$
:from Restriction of Antisymmetric Relation is Antisymmetric, $\preceq \restriction_T$ is an antisymmetric relation on $T$
:from Restriction of Transitive Relation is Transitive, $\preceq \restriction_T$ is a transitive relation on $T$
and so it follows by definition that $\preceq \restriction_T$ is an ordering on $T$.
== Also see ==
* Restriction of Equivalence Relation is Equivalence
* 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
* : $\S \text I.1$: Theorem $1$
Category:Orderings
Category:Restrictions
Restriction of Real Square Mapping to Positive Reals is Bijection

== Theorem ==
Let $f: \R \to \R$ be the real square function:
:$\forall x \in \R: \map f x = x^2$
Let $g: \R_{\ge 0} \to R_{\ge 0} := f {\restriction_{\R_{\ge 0} \times R_{\ge 0} } }$ be the restriction of $f$ to the positive real numbers $\R_{\ge 0}$.
Then $g$ is a bijective restriction of $f$.
== Proof ==
From Order is Preserved on Positive Reals by Squaring, $f$ is strictly increasing on $\R_{\ge 0}$.
By definition, a strictly increasing real function is strictly monotone.
The result follows from Strictly Monotone Real Function is Bijective.
== Sources ==
* : Chapter $3$. Mappings: Exercise $3$
Category:Square Function
Category:Examples of Bijections