nyuuzyou/wikis huggingface dataset:
Transitivity of Integrality

== Theorem ==
Let $A \subseteq B \subseteq C$ be extensions of commutative rings with unity.
Suppose that $C$ is integral over $B$, and $B$ is integral over $A$.
Then $C$ is integral over $A$.
== Proof ==
First, a lemma:
=== Lemma ===
Now let $x \in C$.
Thus $x$ is supposed integral over $B$.
That is, we can find an expression:
:$(1): \quad x^n + b_{n - 1} x^{n - 1} + \dotsb + b_1 x + b_0 = 0, \quad b_i \in B, \ i = 0, \dotsc, n - 1$
Let $D$ be the subring of $C$ generated by $A \cup \set {b_0, \dotsc, b_{n - 1} }$.
By the lemma, $D$ is finitely generated over $A$.
Moreover, $D \sqbrk x$ is finitely generated over $D$ because of the equation $(1)$.
Therefore by Transitivity of Finite Generation:
:$D \sqbrk x$ is a finitely generated $A$-module.
Finally by Equivalent Definitions of Integral Dependence, $x$ is integral over $A$.
== Linguistic Note ==
== Sources ==
* : Chapter $2$ Ring of Integers: Integral elements: Proposition $2.15$
Category:Algebraic Number Theory
Category:Commutative Algebra
Category:Transitivity of Integrality
Transitivity of Integrality/Lemma

== Lemma for Transitivity of Integrality ==
Let $A \subseteq B$ be a ring extension.
Let $x_1, \dotsc, x_n \in B$ be integral over $A$.
Let $A \sqbrk {x_1, \dotsc, x_n}$ be the subring of $B$ generated by $A \cup \set {x_1, \dotsc, x_n}$ over $A$.
Then $A \sqbrk {x_1, \dotsc, x_n}$ is integral over $A$.
== Proof ==
Let $C$ be the integral closure of $A$ in $B$.
Since the $x_i$ are integral over $A$, they lie in $C$.
So by Integral Closure is Subring, all sums of the form:
:$\ds \sum_{\text {finite} } r x_1^{\alpha_1} \dotsm x_n^{\alpha_n}, \quad r \in A,\ \alpha_j \in \N \cup \set 0$
lie in $C$.
That is, they are integral over $A$. 
But the set of such sums is precisely $A \sqbrk {x_1, \dotsc, x_n}$.
== Linguistic Note ==
Category:Transitivity of Integrality
Transitivity of Separable Field Extensions

== Theorem ==
Let $E / K / F$ be a tower of fields.
Let $E / K$ and $K / F$ be separable.
Then $E / F$ is separable.
== Proof ==
== Also see ==
* Subextension of Separable Field Extension is Separable, the converse
== Source ==
* : Chapter $1$: $\S 4$: Separable and Inseparable Extensions: Proposition $4.21$
* : Chapter $\text V$: $\S4$: Separable Extensions: Theorem $4.3$
Category:Separable Field Extensions
Translation Invariant Measure on Euclidean Space is Multiple of Lebesgue Measure

== Theorem ==
Let $\mu$ be a measure on $\R^n$ equipped with the Borel $\sigma$-algebra $\map \BB {\R^n}$.
Suppose that $\mu$ is translation invariant.
Also, suppose that $\kappa := \map \mu {\hointr 0 1^n} < +\infty$.
Then $\mu = \kappa \lambda^n$, where $\lambda^n$ is the $n$-dimensional Lebesgue measure.
== Proof ==
From Characterization of Euclidean Borel Sigma-Algebra, we have:
:$\map \BB {\R^n} = \map \sigma {\JJ^n_{ho, \text {rat} } }$
where $\JJ^n_{ho, \text {rat} }$ denotes the collection of half-open $n$-rectangles with rational endpoints.
So let $J = \horectr {\mathbf a} {\mathbf b} \in \JJ^n_{ho, \text {rat} }$.
Let $M \in \N$ be a common denominator of the $a_i, b_i$ (which are rational by assumption).
We may then cover $J$ by finitely many pairwise disjoint half-open $n$-rectangles $\hointr 0 {\dfrac 1 M}^n$, in that:
:$\ds J = \bigcup_{i \mathop = 1}^{\map k J} \mathbf x_i + \hointr 0 {\dfrac 1 M}^n$
for some $\map k J \in \N$ and suitable $\mathbf x_i$, where:
:$\mathbf x_i + \hointr 0 {\dfrac 1 M}^n := \horectr {\mathbf x_i} {\mathbf x_i + \dfrac 1 M}$
Using that $\mu$ is translation invariant, this means:
:$\map \mu J = \map k J \, \map \mu {\hointr 0 {\dfrac 1 M}^n}$
Also, by Lebesgue Measure is Translation Invariant:
:$\map {\lambda^n} J = \map k J \, \map {\lambda^n} {\hointr 0 {\dfrac 1 M}^n}$
A moment's thought shows us that the half-open $n$-rectangle $I = \hointr 0 1^n$ may be covered by $M^n$ copies of $\hointr 0 {\dfrac 1 M}^n$, so that:
:$\map k I = M^n$
For brevity, write $I / M$ for $\hointr 0 {\dfrac 1 M}^n$.
Now, using that:
:$\ds \map {\lambda^n} {I / M} = \prod_{i \mathop = 1}^n \frac 1 M = \frac 1 {M^n}$
and $\map \mu I = \kappa$, compute:
Therefore, $\map \mu J = \kappa \, \map {\lambda^n} J$ for all $J \in \JJ^n_{ho, \text {rat} }$.
Let us quickly verify the other conditions for Uniqueness of Measures.
For $(1)$, we have Half-Open Rectangles Closed under Intersection.
For $(2)$, observe the exhausting sequence $\hointr {-k} k^n \mathop \uparrow \R^n$
Finally, for $(4)$, we recall that $\kappa < +\infty$.
Thus, by Uniqueness of Measures, $\mu = \kappa \lambda^n$.
== Sources ==
* : $5.8 \ \text{(ii)}$
Category:Measure Theory
Translation Mapping is Bijection

== Theorem ==
Let $\struct {G, +}$ be an abelian group.
Let $g \in G$.
Let $\tau_g: G \to G$ be the translation by $g$:
:$\forall h \in G: \map {\tau_g} h = h + \paren {-g}$
where $-g$ is the inverse of $g$ with respect to $+$ in $G$.
Then $\tau_g$ is a bijection.
== Proof ==
=== Proof of Injectivity ===
=== Proof of Surjectivity ===
Category:Translation Mappings
Category:Bijections
Translation Mapping is Isometry

== Theorem ==
Let $\Gamma = \R^n$ denote the real Euclidean space of $n$ dimensions.
Let $\tau_\mathbf x$ be a translation on $\Gamma$:
:$\forall \mathbf y \in \R^n: \map {\tau_\mathbf x} {\mathbf y} = \mathbf y - \mathbf x$
where $\mathbf x$ is a vector in $\R^n$.
Then $\tau_\mathbf x$ is an isometry.
== Proof ==
From Translation Mapping is Bijection, $\tau_\mathbf x$ is a bijection.
From Euclidean Metric on Real Number Space is Translation Invariant, $\tau_\mathbf x$ is distance-preserving on $\Gamma$.
The result follows by definition of isometry.
== Sources ==
* 
* 
Category:Translation Mappings
Category:Isometries (Metric Spaces)
Category:Isometries (Euclidean Geometry)
Translation Mapping on Topological Vector Space is Continuous

== Theorem ==
Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$. 
Let $x \in X$.
Let $T_x$ be the translation by $x$ mapping.
Then $T_x$ is continuous.
== Proof ==
From the definition of a topological vector space, the mapping $X \times X \to X$ defined by $\tuple {y, x} \mapsto y + x$ is continuous.
From Horizontal Section of Continuous Function is Continuous, it follows that the $\paren {-x}$-horizontal section $T_x : X \to X$ with $y \mapsto y - x$ is continuous.
Category:Translation Mappings (Vector Spaces)
Category:Topological Vector Spaces
Translation Mapping on Topological Vector Space is Homeomorphism

== Theorem ==
Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$. 
Let $x \in X$. 
Let $T_x$ be the translation by $x$ mapping.
Then $T_x$ is a homeomorphism.
== Proof ==
From Translation Mapping on Topological Vector Space is Continuous, both $T_x$ and $T_{-x}$ are continuous.
It is therefore sufficient to establish that $T_{-x}$ is the inverse mapping of $T_x$.
For all $y \in X$, we have:
and:
So both $T_x \circ T_{-x}$ and $T_{-x} \circ T_x$ are the identity mapping for $X$.
So $T_{-x}$ is the inverse mapping of $T_x$, as required.
== Sources ==
* : $1.6$: Topological vector spaces
Category:Translation Mappings (Vector Spaces)
Category:Topological Vector Spaces
Translation in Euclidean Space is Measurable Mapping

== Theorem ==
Let $\BB$ be the Borel $\sigma$-algebra on $\R^n$.
Let $\mathbf x \in \R^n$, and denote with $\tau_{\mathbf x}: \R^n \to \R^n$ translation by $\mathbf x$.
Then $\tau_{\mathbf x}$ is $\BB \, / \, \BB$-measurable.
== Proof ==
By Characterization of Euclidean Borel Sigma-Algebra, $\BB = \map \sigma {\JJ_{ho}^n}$.
Here, $\JJ_{ho}^n$ is the set of half-open $n$-rectangles, and $\sigma$ denotes generated $\sigma$-algebra.
Now, for any half-open $n$-rectangle $\horectr {\mathbf a} {\mathbf b}$, it is trivial that:
:$\map {\tau_{\mathbf x}^{-1} } {\horectr {\mathbf a} {\mathbf b} } = \horectr {\mathbf a + \mathbf x} {\mathbf b + \mathbf x}$
That is, the preimage of a half-open $n$-rectangle under $\tau_{\mathbf x}$ is again a half-open $n$-rectangle.
In particular, since $\JJ_{ho}^n \subseteq \map \sigma {\JJ_{ho}^n} = \BB$, Mapping Measurable iff Measurable on Generator applies.
Thus it follows that $\tau_{\mathbf x}$ is $\BB \, / \, \BB$-measurable.
== Sources ==
* : $\S 7$, $\S 7$: Problem $1$
Category:Measure Theory
Category:Translation Mappings
Translation of Affine Subspace of Real Euclidean Space is Vector Space

== Theorem ==
Let $\R^3$ denote the (real) Euclidean $3$-space.
Let $\SS$ be an affine subspace of $\R^3$.
Let $\SS'$ be the image of a translation of $\SS$.
Then $\SS'$ is a vector space.
== Proof ==
== Sources ==
* 
Category:Affine Subspaces
Category:Euclidean Space
Category:Vector Spaces
Translation of Axes/Examples/Cartesian Plane

== Example of Translation of Axes ==
Let $\CC$ be a Cartesian plane.
Let $\CC'$ be another Cartesian plane whose axes are parallel to the corresponding axes of $\CC$.
Let $\tuple {h, k}$ be the coordinates of the origin of $\CC'$ $\CC$.
Let $\tuple {x, y}$ be the coordinates of a point $P$ in $\CC$.
Let $\tuple {x', y'}$ be the coordinates of $P$ in $\CC'$.
Then:
== Sources ==
* 
* 
Category:Examples of Translation of Axes
Translation of Axes/Examples/Equation of Circle

== Example of Translation of Axes ==
Let $\CC$ be a Cartesian plane.
Let $\KK$ be a circle embedded in $\CC$ whose equation is given as:
:$\paren {x - 3}^2 + \paren {y - 5}^2 = 7$
Then by translation of axes to new axes whose origin is at $\tuple {3, 5}$, $\KK$ will have the equation:
:$x'^2 + y'^2 = 7$
== Sources ==
* 
* 
Category:Examples of Translation of Axes
Translation of Closed Set in Topological Vector Space is Closed Set

== Theorem ==
Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$. 
Let $F$ be a closed set in $X$. 
Let $x \in X$.
Then $F + x$ is a closed set in $X$.
== Proof 1 ==
== Proof 2 ==
Category:Translation of Subsets of Vector Spaces
Category:Topological Vector Spaces
Category:Translation of Closed Set in Topological Vector Space is Closed Set
Translation of Closed Set in Topological Vector Space is Closed Set/Proof 1

== Theorem ==
== Proof ==
We aim to show that $X \setminus \paren {F + x}$ is open.
We are given $F$ is closed in $X$.
So, $X \setminus F$ is open.
It follows from Translation of Open Set in Topological Vector Space is Open that $\paren {X \setminus F} + x$ is open.
From Translation of Complement of Set in Vector Space, we have:
:$X \setminus \paren {F + x} = \paren {X \setminus F} + x$
Since we have established that $\paren {X \setminus F} + x$ is open, we have that $X \setminus \paren {F + x}$ is open.
Hence by definition, $F + x$ is closed in $X$.
Category:Translation of Closed Set in Topological Vector Space is Closed Set
Translation of Closed Set in Topological Vector Space is Closed Set/Proof 2

== Theorem ==
== Proof ==
Define a mapping $T_{-x} : X \to X$ by:
:$\map {c_\lambda} y = y + x$
for each $y \in X$.
From Translation Mapping on Topological Vector Space is Homeomorphism, $T_{-x}$ is a homeomorphism.
From Definition 4 of a homeomorphism, $T_{-x}$ is therefore a closed mapping.
Hence $T_{-x} \sqbrk F = F + x$ is closed.
Category:Translation of Closed Set in Topological Vector Space is Closed Set
Translation of Complement of Set in Vector Space

== Theorem ==
Let $K$ be a field.
Let $X$ be a vector space over $K$.
Let $F \subseteq X$ be a non-empty set.
Let $x \in X$.
Then:
:$X \setminus \paren {F + x} = \paren {X \setminus F} + x$
== Proof ==
It is immediate that if:
:$y \in F$
we have:
:$y + x \in F + x$.
Conversely, if:
:$y + x \in F + x$
then:
:$y + x = u + x$
for some $u \in F$.
That is:
:$y = u$
and so:
:$y \in F$.
Hence we have:
:$y \in F$ $y + x \in F + x$
Hence for $y \in X$ we have:
:$y + x \not \in F + x$ $y \not \in F$
That is:
:$y + x \in X \setminus \paren {F + x}$ $y \in X \setminus F$ $y + x \in \paren {X \setminus F} + x$.
So:
:$y + x \in X \setminus \paren {F + x}$ $y + x \in \paren {X \setminus F} + x$.
We can therefore conclude:
:$X \setminus \paren {F + x} = \paren {X \setminus F} + x$
Category:Translation of Subsets of Vector Spaces
Translation of Convex Set in Vector Space is Convex

== Theorem ==
Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a vector space over $\Bbb F$.
Let $C \subseteq X$ be convex.
Let $x \in X$.
Then $C + x$, the translation of $C$ by $x$, is convex.
== Proof ==
Let $t \in \closedint 0 1$ and $u, v \in C + x$.
Then there exists $u', v' \in C$ such that: 
:$u = u' + x$
and:
:$v = v' + x$ 
Then, we have:
Since $C$ is convex, we have:
:$t u' + \paren {1 - t} v' \in C$
so:
:$t u + \paren {1 - t} v \in C + x$
So we have that $C + x$ is convex.
Category:Translation of Subsets of Vector Spaces
Category:Convex Sets (Vector Spaces)
Translation of Index Variable of Product

== Theorem ==
:$\ds \prod_{\map R j} a_j = \prod_{\map R {c \mathop + j} } a_{c \mathop + j} = \prod_{\map R {c \mathop - j} } a_{c \mathop - j}$
where:
:$\ds \prod_{\map R j} a_j$ denotes the product over $a_j$ for all $j$ that satisfy the propositional function $\map R j$
:$c$ is an integer constant which is not dependent upon $j$.
== Proof ==
== Also see ==
* Translation of Index Variable of Summation
== Sources ==
* : $\S 1.2.3$: Sums and Products
Category:Continued Products
Translation of Index Variable of Summation

== Theorem ==
Let $R: \Z \to \set {\T, \F}$ be a propositional function on the set of integers.
Let $\ds \sum_{\map R j} a_j$ denote a summation over $R$.
Then:
:$\ds \sum_{\map R j} a_j = \sum_{\map R {c \mathop + j} } a_{c \mathop + j} = \sum_{\map R {c \mathop - j} } a_{c \mathop - j}$
where $c$ is an integer constant which is not dependent upon $j$.
=== Corollary ===
== Proof ==
First we consider the case where the fiber of truth of $R$ is finite.
=== Infinite case ===
Now consider the case where the fiber of truth of $R$ is infinite.
== Examples ==
== Also known as ==
This operation is often referred to as Adjustment of Indices.
Some expositions, when invoking this result, refer to it as an instance of Permutation of Indices of Summation, otherwise known as permutation of the range.
== Also see ==
* Translation of Index Variable of Product
== Sources ==
* : $\S 1.2.3$: Sums and Products
Category:Summations
Translation of Index Variable of Summation/Examples

== Examples of Translation of Index Variable of Summation ==
:$\ds \sum_{1 \mathop \le j \mathop \le n} a_j = \sum_{1 \mathop \le j - 1 \mathop \le n} a_{j - 1} = \sum_{2 \mathop \le j \mathop \le n + 1} a_{j - 1}$
== Proof ==
A direct instance of Translation of Index Variable of Summation.
== Sources ==
* : $\S 1.2.3$: Sums and Products: $(6)$
Category:Summations
Translation of Index Variable of Summation/Infinite Series

== Theorem ==
Let $R: \Z \to \set {\T, \F}$ be a propositional function on the set of integers.
Let $\ds \sum_{\map R j} a_j$ denote a summation over $R$.
Let the fiber of truth of $R$ be infinite.
Then:
:$\ds \sum_{\map R j} a_j = \sum_{\map R {c \mathop + j} } a_{c \mathop + j} = \sum_{\map R {c \mathop - j} } a_{c \mathop - j}$
where $c$ is an integer constant which is not dependent upon $j$.
== Proof ==
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $7$
Category:Summations
Translation of Integer Interval is Bijection

== Theorem ==
Let $a, b, c \in \Z$ be integers.
Let $\closedint a b$ denote the integer interval between $a$ and $b$.
Then the mapping $T: \closedint a b \to \closedint {a + c} {b + c}$ defined as:
:$\map T k = k + c$
is a bijection.
== Proof ==
Note that if $k \in \closedint a b$, then indeed $k + c \in \closedint {a + c} {b + c}$.
=== Injectivity ===
Let $k, l \in \closedint a b$ with $k + c = l + c$.
By Integer Addition is Cancellable, $k = l$.
Thus $T$ is injective.
=== Surjectivity ===
Let $m \in \closedint {a + c} {b + c}$.
Then $m - c \in \closedint a b$.
Then $\map T {m - c} = m - c + c = m$.
Thus $T$ is surjective.
== Also see ==
* Cardinality of Integer Interval
Category:Set Theory
Translation of Intersection of Subsets of Vector Space

== Theorem ==
Let $K$ be a field. 
Let $X$ be a vector space over $K$.
Let $\family {E_\alpha}_{\alpha \mathop \in I}$ be an indexed family of subsets of $X$.
Let $x \in X$.
Then:
:$\ds \paren {\bigcap_{\alpha \mathop \in I} E_\alpha} + x = \bigcap_{\alpha \mathop \in I} \paren {E_\alpha + x}$
where $E_\alpha + x$ denotes the translation of $E_\alpha$ by $x$.
== Proof ==
Let $v \in X$.
We have:
:$\ds v \in \paren {\bigcap_{\alpha \mathop \in I} E_\alpha} + x$
:
:$\ds v - x \in \bigcap_{\alpha \mathop \in I} E_\alpha$
:
:$v - x \in E_\alpha$ for each $\alpha \in I$
:
:$v \in E_\alpha + x$ for each $\alpha \in I$
:
:$\ds v \in \bigcap_{\alpha \mathop \in I} \paren {E_\alpha + x}$
That is:
:$\ds \paren {\bigcap_{\alpha \mathop \in I} E_\alpha} + x = \bigcap_{\alpha \mathop \in I} \paren {E_\alpha + x}$
Category:Translation of Subsets of Vector Spaces
Category:Set Intersection
Translation of Local Basis in Topological Vector Space

== Theorem ==
Let $K$ be a topological field.
Let $\struct {X, \tau}$ be a topological vector space over $K$.
Let $\sequence {U_\alpha}_{\alpha \mathop \in A}$ be a local basis at ${\mathbf 0}_X$.
Let $x \in X$.
Then $\sequence {U_\alpha + x}_{\alpha \in A}$ is an local basis at $x$.
== Proof ==
From Translation of Open Set in Topological Vector Space is Open, $U_\alpha + x$ is an open neighborhood of $x$ in $\struct {X, \tau}$ for each $\alpha \in A$. 
Let $U$ be an open neighborhood of $x$ in $\struct {X, \tau}$.
From Translation of Open Set in Topological Vector Space is Open, $U - x$ is an open neighborhood of ${\mathbf 0}_X$ in $\struct {X, \tau}$.
Since $\sequence {U_\alpha}_{\alpha \mathop \in A}$ is local basis at ${\mathbf 0}_X$, there exists $\beta \in A$ such that:
:$U_\beta \subseteq U - x$ 
Then:
:$U_\beta + x \subseteq U$
Since $U$ was arbitrary, we obtain that $\sequence {U_\alpha + x}_{\alpha \in A}$ is an local basis at ${\mathbf 0}_X$.
Category:Local Bases
Category:Topological Vector Spaces
Category:Translation of Subsets of Vector Spaces
Translation of Open Ball in Invariant Pseudometric on Vector Space

== Theorem ==
Let $K$ be a field.
Let $X$ be a vector space over $K$.
Let $d$ be an invariant pseudometric on $X$.
For $x \in X$ and $\epsilon > 0$, let $\map {B_\epsilon} x$ be the open ball centered at $x$ with radius $\epsilon$. 
Let $y \in X$.
Then:
:$\map {B_\epsilon} x + y = \map {B_\epsilon} {x + y}$
== Proof ==
Let $z \in X$.
Then we have $z \in \map {B_\epsilon} x + y$ $z = u + y$ for $u \in \map {B_\epsilon} x$. 
That is, $z - y \in \map {B_\epsilon} x$. 
This is equivalent to:
:$\map d {z - y, x} < \epsilon$
Since $d$ is invariant, this is equivalent to:
:$\map d {z, x + y} < \epsilon$ 
So, we have $z \in \map {B_\epsilon} x + y$ $z \in \map {B_\epsilon} {x + y}$.
Hence $\map {B_\epsilon} x + y = \map {B_\epsilon} {x + y}$.
Category:Invariant Pseudometrics on Vector Spaces
Translation of Open Set in Normed Vector Space is Open

== Theorem ==
Let $\struct {X, \norm \cdot}$ be a normed vector space.
Let $U \subseteq X$ be an open set.
Let $x \in X$.
Then:
:$U + x$ is open.
== Proof ==
Let: 
:$v \in U + x$
then:
:$v = u + x$
for some $u \in U$.
So:
:$v - x \in U$ 
Since $U$ is open, there exists $\epsilon > 0$ such that whenever $v' \in X$ and: 
:$\norm {\paren {v - x} - \paren {v' - x} } < \epsilon$
we have $v' - x \in U$.
That is: 
:$v' \in U + x$
Note that: 
:$\norm {\paren {v - x} - \paren {v' - x} } = \norm {v - v'}$
So, whenever $v \in U + x$ and $v' \in X$ is such that: 
:$\norm {v - v'} < \epsilon$
we have: 
:$v' \in U + x$
Since $v$ was arbitrary: 
:$U + x$ is open.
Category:Open Sets (Normed Vector Spaces)
Category:Translation of Subsets of Vector Spaces
Translation of Open Set in Topological Vector Space is Open

== Theorem ==
Let $\struct {X, \tau}$ be a topological vector space.
Let $U \subseteq X$ be an open set.
Let $x \in X$.
Then:
:$U + x$ is open.
== Proof ==
Let $T_{-x}$ be the translation by $x$ mapping. 
From Translation Mapping on Topological Vector Space is Homeomorphism, we have: 
:$T_{-x}$ is a homeomorphism.
So, since $U$ is open, we have: 
:$T_{-x} \sqbrk U$ is open.
That is: 
:$U - \paren {-x} = U + x$ is open.
== Sources ==
* : $1.6$: Topological vector spaces
Category:Topological Vector Spaces
Category:Translation of Subsets of Vector Spaces
Translation of Union of Subsets of Vector Space

== Theorem ==
Let $K$ be a field. 
Let $X$ be a vector space over $K$.
Let $\family {E_\alpha}_{\alpha \mathop \in I}$ be an indexed family of subsets of $X$.
Let $x \in X$.
Then:
:$\ds \paren {\bigcup_{\alpha \mathop \in I} E_\alpha} + x = \bigcup_{\alpha \mathop \in I} \paren {E_\alpha + x}$
where $E_\alpha + x$ denotes the translation of $E_\alpha$ by $x$.
== Proof ==
We have:
:$\ds v \in \paren {\bigcup_{\alpha \mathop \in I} E_\alpha} + x$
:
:$v = u + x$ for some $\ds u \in \bigcup_{\alpha \mathop \in I} E_\alpha$.
This is equivalent to: 
:there exists $\alpha \in I$ and $u \in E_\alpha$ such that $v = u + x$.
That is: 
:there exists $\alpha \in I$ such that $v \in E_\alpha + x$.
So by the definition of set equality, we have: 
:$\ds \paren {\bigcup_{\alpha \mathop \in I} E_\alpha} + x = \bigcup_{\alpha \mathop \in I} \paren {E_\alpha + x}$
Category:Translation of Subsets of Vector Spaces
Category:Set Union
Transplant (Abstract Algebra)/Examples

== Examples of Transplants ==
=== Multiplication on $\Z$ under Doubling ===
=== Multiplication on $\R$ under $10^x$ ===
=== Multiplication on $\R$ under $1 - x$ ===
=== Addition on $\R_{>0}$ under $x^2$ ===
=== Addition on $\R_{>0}$ under $\log_{10} x$ ===
Category:Transplants
Transplant (Abstract Algebra)/Examples/Addition on Positive Reals under Log Base 10

== Example of Transplant ==
Let $\struct {\R_{>0}, +}$ be the set of strictly positive real numbers under addition.
Let $f: \R_{>0} \to \R$ be the permutation defined as:
:$\forall x \in \R_{>0}: \map f x = \log_{10} x$
The transplant $\oplus$ of $+$ under $f$ is given by:
:$x \oplus y = \map {\log_{10} } {10^x + 10^y}$
== Proof ==
From Logarithm on Positive Real Numbers is Group Isomorphism, $f$ is a bijection.
The inverse of $f$ is given as:
:$\forall x \in \R: \map {f^{-1} } x = 10^x$
Hence from the Transplanting Theorem:
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Exercise $6.8 \ \text {(c)}$
Category:Transplants
Transplant (Abstract Algebra)/Examples/Addition on Positive Reals under Squaring

== Example of Transplant ==
Let $\struct {\R_{>0}, +}$ be the set of strictly positive real numbers under addition.
Let $f: \R_{>0} \to \R_{>0}$ be the permutation defined as:
:$\forall x \in \R_{>0}: \map f x = x^2$
The transplant $\oplus$ of $+$ under $f$ is given by:
:$x \oplus y = x + y + 2 \sqrt {x y}$
== Proof ==
From Restriction of Real Square Mapping to Positive Reals is Bijection, $f$ is a bijection.
The inverse of $f$ is given as:
:$\forall x \in \R_{>0}: \map {f^{-1} } x = \sqrt x$
Hence from the Transplanting Theorem:
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Exercise $6.8 \ \text {(b)}$
Category:Transplants
Transplant (Abstract Algebra)/Examples/Multiplication on Integers under Doubling

== Example of Transplant ==
Let $\struct {\Z, \times}$ be the set of integers under multiplication.
Let $E$ be the set of even integers.
Let $f: \Z \to E$ be the mapping from $\Z$ to $E$ defined as:
:$\forall n \in \Z: \map f n = 2 n$
The transplant $\otimes$ of $\times$ on $\Z$ under $f$ is given by:
:$\forall n, m \in E: n \otimes m = \dfrac {n m} 2$
== Proof ==
From Bijection between Integers and Even Integers, $f$ is a bijection.
The inverse of $f$ is given as:
:$\forall n \in E: \map {f^{-1} } n = \dfrac n 2$
Hence from the Transplanting Theorem:
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures
Category:Transplants