nyuuzyou/wikis huggingface dataset:
Relation/Examples/Ordering on Arbitrary Sets of Integers

== Example of Relation ==
Let $A = \set {1, 2, 3, 4}$ and $B = \set {1, 2, 3}$ be sets of integers.
Consider the following diagram, where:
:$A$ runs along the top
:$B$ runs down the 
:a relation $\RR$ between $A$ and $B$ is indicated by marking with $\bullet$ every ordered pair $\tuple {a, b} \in A \times B$ which is in the truth set of $\RR$
:$\begin {array} {r|rrrr}
A \times B & 1 & 2 & 3 & 4 \\
\hline
1 & \bullet & \bullet & \bullet & \circ \\
2 & \bullet & \bullet & \circ & \circ \\
3 & \bullet & \circ & \circ & \circ \\
\end {array}$
This relation $\RR$ can be described as:
:$\RR = \set {\tuple {x, y} \in A \times B: x + y \le 4}$
=== Not Many-to-One ===
== Sources ==
* : $\S 4$. Relations; functional relations; mappings: Example $4.1$
Category:Examples of Relations
Relation/Examples/Ordering on Arbitrary Sets of Integers/Not Many-to-One

== Example of Relation which is not Many-to-One ==
$\RR$ is not a many-to-one relation.
== Proof ==
For example we have:
:$\tuple {1, 1} \in \RR$
and:
:$\tuple {1, 2} \in \RR$
Hence $\RR$ is not many-to-one by definition.
== Sources ==
* : $\S 4$. Relations; functional relations; mappings: Example $4.4$
Category:Examples of Relations
Relation/Examples/Sisterhood

== Example of Relation ==
Let $S$ be the set of all human females.
Let $T$ be the set of all human beings.
Let $\RR$ be the set defined as:
:$\RR = \set {\tuple {a, b}: a \in S, b \in T, \text {$a$ is a sister of $b$} }$
Then $\RR$ is a relation in $S$ to $T$.
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Relations
Category:Examples of Relations
Relation/Examples/Subsets of Initial Segment of Natural Numbers

== Example of Relation ==
Let $S$ be the set of all the subsets of the initial segment of the natural numbers $\set {1, 2, 3, \ldots, n}$.
Let $\RR$ be the set defined as:
:$\RR = \set {\tuple {S_1, S_2}: S_1 \subseteq S_2, S_1 \in S, S_2 \in S}$
Then $\RR$ is a relation on $S$.
== Sources ==
* : Chapter $2$: 'And you do addition?': $\S 2.3$: Relations: Example $2.3.2$
Category:Examples of Relations
Relation Between First and Second Form of Binet Form

== Theorem ==
Let $m \in \R$.
Define:
=== First Form ===
=== Second Form ===
For any given value of $m$:
:$U_{n - 1} + U_{n + 1} = V_n$
== Proof ==
Proof by induction:
Let $\map P n$ be the proposition:
:$U_{n - 1} + U_{n + 1} = V_n$
=== Basis for the Induction ===
We have:
Therefore $\map P 1$ and $\map P 2$ are true.
This is the basis for the induction.
=== Induction Hypothesis ===
This is our induction hypothesis:
:For some $k \in \N_{> 0}$, both $\map P k$ and $\map P {k + 1}$ are true.
That is:
:$U_{k - 1} + U_{k + 1} = V_k$
:$U_k + U_{k + 2} = V_{k + 1}$
Now we need to show true for $n = k + 2$:
:$\map P {k + 2}$ is true.
That is:
:$U_{k + 1} + U_{k + 3} = V_{k + 2}$
=== Induction Step ===
This is our induction step:
This show that $\map P {k + 2}$ is true.
By Principle of Mathematical Induction, $\map P n$ is true for all $n \in \N _{>0}$.
== Sources ==
* 
Category:Binet Form
Relation Between Rank and Exponent of Convergence

== Theorem ==
Let $f: \C \to \C$ be an entire function.
Let $k$ be its rank and $\tau$ be its exponent of convergence.
Then:
* $k=\tau=0$ if $f$ has finitely many zeroes.
* $k<\tau\leq k+1$ otherwise.
== Proof ==
Category:Entire Functions
Relation Compatibility in Totally Ordered Semigroup

== Theorem ==
Let $\left({S, \circ, \preceq}\right)$ be an ordered semigroup such that:
:$(1): \quad$ All the elements of $\left({S, \circ, \preceq}\right)$ are cancellable for $\circ$
:$(2): \quad \preceq$ is a total ordering.
Then:
:$\forall x, y, z \in S: x \circ z \preceq y \circ z \iff x \preceq y$
== Proof ==
From Strict Ordering Preserved under Cancellability in Totally Ordered Semigroup:
: $x \circ z \prec y \circ z \implies x \prec y$
From the definition of cancellable element:
: $x \circ z = y \circ z \implies x = y$
Category:Ordered Semigroups
Relation Compatible with Group Operation is Reflexive or Antireflexive

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $\RR$ be a relation on $G$ that is compatible with $\circ$.
Then $\RR$ is reflexive or antireflexive.
== Proof ==
Suppose that $\RR$ is not antireflexive.
Then there is some $x \in G$ such that $x \mathrel \RR x$.
Let $y \in G$.
Then by the definition of compatibility:
:$\paren {x \circ \paren {x^{-1} \circ y} } \mathrel \RR \paren {x \circ \paren {x^{-1} \circ y} }$
By and :
:$y \mathrel \RR y$
Since this holds for arbitrary $y$, $\RR$ is reflexive.
Thus $\RR$ is either reflexive or antireflexive.
Category:Relations Compatible with Group Operation
Category:Reflexive Relations
Category:Antireflexive Relations
Relation Compatible with Group Operation is Strongly Compatible

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $\RR$ be a relation on $G$ compatible with $\circ$.
$\RR$ is strongly compatible with $\circ$:
:$\forall x, y, z \in G:$
::$x \mathrel \RR y \iff x \circ z \mathrel \RR y \circ z$
::$x \mathrel \RR y \iff z \circ x \mathrel \RR z \circ y$
=== Corollary ===
== Proof ==
Since $\RR$ is compatible with $\circ$:
:$\forall a, b, c \in G: a \mathrel \RR b \implies a \circ c \mathrel \RR b \circ c$
In particular, letting $a = x$, $b = y$, and $c = z$, we see that:
:$x \mathrel \RR y \implies x \circ z \mathrel \RR y \circ z$
On the other hand, letting $a = x \circ z$, $b = y \circ z$, and $c = z^{-1}$, we see that:
:$x \circ z \mathrel \RR y \circ z \implies \paren {x \circ z} \circ z^{-1} \mathrel \RR \paren {y \circ z} \circ z^{-1}$
By the associativity of $\circ$ and the definition of inverse, this reduces to:
:$x \circ z \mathrel \RR y \circ z \implies x \mathrel \RR y$
We have thus shown that:
:$x \mathrel \RR y \iff x \circ z \mathrel \RR y \circ z$
A similar argument shows that:
:$x \mathrel \RR y \iff z \circ x \mathrel \RR z \circ y$
so $\RR$ is strongly compatible with $\circ$.
Category:Relations Compatible with Group Operation
Relation Compatible with Group Operation is Strongly Compatible/Corollary

== Theorem ==
Let $\struct {G, \circ}$ be a group with identity $e$.
Let $\RR$ be a relation compatible with $\circ$.
The following equivalences hold:
:$\forall x, y \in G:$
::$(1): \quad x \mathrel \RR y \iff e \mathrel \RR y \circ x^{-1}$
::$(2): \quad x \mathrel \RR y \iff e \mathrel \RR x^{-1} \circ y$
::$(3): \quad x \mathrel \RR y \iff x \circ y^{-1} \mathrel \RR e$
::$(4): \quad x \mathrel \RR y \iff y^{-1} \circ x \mathrel \RR e$
== Proof ==
Applying Relation Compatible with Group Operation is Strongly Compatible to $x$, $y$, and $x^{-1}$ we obtain:
:$x \mathrel \RR y \iff x \circ x^{-1} \mathrel \RR y \circ x^{-1}$
:$x \mathrel \RR y \iff x^{-1} \circ x \mathrel \RR x^{-1} \circ y$
Applying Relation Compatible with Group Operation is Strongly Compatible to $x$, $y$, and $y^{-1}$, on the other hand, yields:
:$x \mathrel \RR y \iff x \circ y^{-1} \mathrel \RR y \circ y^{-1}$
:$x \mathrel \RR y \iff y^{-1} \circ x \mathrel \RR y^{-1} \circ y$
By the definition of inverse element:
:$x \circ x^{-1} = x^{-1} \circ x = y \circ y^{-1} = y^{-1} \circ y = e$
Making these substitutions proves the theorem.
Category:Relations Compatible with Group Operation
Relation Contains Mapping is Equivalent to AoC

== Theorem ==
Let $S$ and $T$ be sets.
Let $\RR \subseteq S \times T$ be a relation on $S \times T$.
Then:
:there exists a mapping $f \subseteq \RR$ whose domain is the same as the preimage of $\RR$
:the axiom of choice holds.
== Proof ==
== Sources ==
* : $\S 15$: The Axiom of Choice
Relation Induced by Mapping is Equivalence Relation

== Theorem ==
Let $f: S \to T$ be a mapping.
Let $\RR_f \subseteq S \times S$ be the relation induced by $f$:
:$\tuple {s_1, s_2} \in \RR_f \iff \map f {s_1} = \map f {s_2}$
Then $\RR_f$ is an equivalence relation.
== Proof ==
We need to show that $\RR_f$ is an equivalence relation.
Checking in turn each of the criteria for equivalence:
=== Reflexive ===
$\RR_f$ is reflexive:
:$\forall x \in S: \map f x = \map f x \implies x \mathrel {\RR_f} x$
=== Symmetric ===
$\RR_f$ is symmetric:
=== Transitive ===
$\RR_f$ is transitive:
Thus $\RR_f$ is reflexive, symmetric and transitive, and is therefore an equivalence relation.
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Factoring Functions: Theorem $10$
* : $\S 6$. Indexed families; partitions; equivalence relations: Example $6.6$
* : Chapter $1$: Theory of Sets: $\S 7$: Relations: Exercise $2$
* : Chapter $4$: Mappings: Exercise $10$
* : Chapter $1$: Sets and mappings: $\S 1.4$: Equivalence relations
* : $1$: Set Theory and Logic: $\S 3$: Relations: Exercise $3.4 \ \text{(a)}$
Category:Mapping Theory
Category:Equivalence Relations
Category:Examples of Equivalence Relations
Relation Induced by Partition is Equivalence

== Theorem ==
Let $\mathbb S$ be a partition of a set $S$.
Let $\RR$ be the relation induced by $\mathbb S$.
Then:
:$(1): \quad \RR$ is unique
:$(2): \quad \RR$ is an equivalence relation on $S$.
Hence $\mathbb S$ is the quotient set of $S$ by $\RR$, that is:
:$\mathbb S = S / \RR$
== Proof ==
From the definition of Relation Induced by Partition, we define the relation $\RR$ on $S$ by:
:$\RR = \set {\tuple {x, y}: \paren {\exists T \in \mathbb S: x \in T \land y \in T} }$
=== Test for Equivalence ===
We are to show that $\RR$ is an equivalence relation.
Checking in turn each of the criteria for equivalence:
==== Reflexive ====
$\RR$ is reflexive:
==== Symmetric ====
$\RR$ is symmetric:
==== Transitive ====
$\RR$ is transitive:
So $\RR$ is reflexive, symmetric and transitive.
Hence, by definition, $\RR$ is an equivalence relation.
=== Test for Uniqueness ===
Now by definition of a partition, we have that:
:$\mathbb S$ partitions $S \implies \forall x \in S: \exists T \in \mathbb S: x \in T$
Also:
Thus $\mathbb S$ is the set of $\RR$-classes constructed above, and no other relation can be constructed in this way.
== Also see ==
* Fundamental Theorem on Equivalence Relations for the converse.
== Sources ==
* : Introduction $\S 3$: Equivalence relations
* : $\S 7$: Relations
* : $\S 1.2$
* : $\S 2.4$. Equivalence classes: Example $37$
* : Chapter $2$. Equivalence Relations: Exercise $4$
* : Chapter $\text {II}$: New Structures from Old: $\S 10$: Equivalence Relations: Theorem $10.2$
* : $\S 0.3$: Theorem $5$ Part $\text {II}$
* : $\S 1.4$: Decomposition of a set into classes. Equivalence relations: Theorem $4$
* : Appendix: Elementary set and number theory
* : Chapter $1$: Equivalence Relations: $\S 17 \alpha$
* : $\S 6$. Indexed families; partitions; equivalence relations: Theorem $6.4$
* : $3$: Continuity generalized: topological spaces: $3.8$: Quotient spaces
* : Appendix $\text{A}.3$: Equivalence Relations: Problem Set $\text{A}.3$: $16$
* : Chapter $1$: Sets and mappings: $\S 1.4$: Equivalence relations
* : Part $1$: Not Entirely Naive Set Theory: Chapter $1$: Pairs, Relations, and Functions: Exercise $10 \ \text {(b)}$
* : $1$: Set Theory and Logic: $\S 3$: Relations
* : Chapter $2$: 'And you do addition?': $\S 2.3$: Relations: Remark $2.3.1$
Category:Quotient Sets
Category:Equivalence Relations
Category:Examples of Equivalence Relations
Relation Induced by Quotient Set is Equivalence

== Theorem ==
Let $S$ be a set.
Let $\RR$ be an equivalence relation on $S$.
Let $S / \RR$ be the quotient set of $S$ determined by $\RR$.
Let $\RR'$ be the equivalence relation induced by $S / \RR$ on $S$.
Then $\RR' = \RR$.
== Proof ==
Let $\RR$ be an equivalence relation on $S$.
Let $\tuple {x, y} \in \RR$.
By definition of equivalence class, $y \in \eqclass x \RR$ and $x \in \eqclass x \RR$.
By definition of quotient set, $x$ and $y$ both belong to the same element of $S / \RR$.
So, by definition of $\RR'$, it follows that $\tuple {x, y} \in \RR'$.
That is:
:$\tuple {x, y} \in \RR \implies \tuple {x, y} \in \RR'$
and so by definition of subset:
:$\RR \subseteq \RR'$
Now let $\tuple {x, y} \in \RR'$.
Then $y$ belongs to the same element of $S / \RR$ that $x$ does.
That is:
:$y \in \eqclass x \RR$
and so $\tuple {x, y} \in \RR$.
That is:
:$\tuple {x, y} \in \RR' \implies \tuple {x, y} \in \RR$
and so by definition of subset:
:$\RR' \subseteq \RR$
The result follows by definition of set equality.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 10$: Equivalence Relations: Theorem $10.3$
Category:Quotient Sets
Category:Equivalence Relations
Category:Examples of Equivalence Relations
Relation Induced by Strict Positivity Property is Asymmetric and Antireflexive

== Theorem ==
Let $\struct {D, +, \times}$ be an ordered integral domain where $P$ is the (strict) positivity property.
Let the relation $<$ be defined on $D$ as:
:$\forall a, b \in D: a < b \iff \map P {-a + b}$
Then $<$ is asymmetric and antireflexive.
== Proof ==
From the trichotomy law of ordered integral domains, for all $x \in D$ exactly one of the following applies:
:$(1): \quad \map P x$
:$(2): \quad \map P {-x}$
:$(3): \quad x = 0$
Let $a, b \in D: a < b$.
Then by definition:
:$\map P {-a + b}$
Suppose also that $b < a$.
Then by definition:
:$\map P {-b + a}$
and so:
:$\map P {-\paren {-a + b} }$
But then that contradicts the trichotomy law of ordered integral domains:
Either:
:$(1): \quad \map P {-a + b}$
or:
:$(2): \quad \map P {-\paren {-a + b} }$
So $(1)$ and $(2)$ can not both apply, and so if $a < b$ it is not possible that $b < a$.
Thus $<$ is asymmetric.
Now suppose $\exists a \in D: a < a$.
Then that would mean $\map P {-a + a}$, that is, $\map P {0_D}$.
But this also contradicts the conditions of the trichotomy law of ordered integral domains.
Hence:
:$\forall a \in D: a \not < a$
and so $<$ is antireflexive.
Category:Ordered Integral Domains
Relation Induced by Strict Positivity Property is Compatible with Addition

== Theorem ==
Let $\struct {D, +, \times}$ be an ordered integral domain where $P$ is the (strict) positivity property.
Let the relation $<$ be defined on $D$ as:
:$\forall a, b \in D: a < b \iff \map P {-a + b}$
Then $<$ is compatible with $+$, that is:
:$\forall x, y, z \in D: x < y \implies \paren {x + z} < \paren {y + z}$
:$\forall x, y, z \in D: x < y \implies \paren {z + x} < \paren {z + y}$
=== Corollary ===
== Proof ==
Let $a < b$:
And so $<$ is seen to be compatible with $+$.
== Sources ==
* : Chapter $2$: Ordered and Well-Ordered Integral Domains: $\S 7$. Order: Theorem $9: \ \text{O} 1$
Category:Ordered Integral Domains
Relation Induced by Strict Positivity Property is Compatible with Addition/Corollary

== Corollary to Relation Induced by Strict Positivity Property is Compatible with Addition ==
Let $\struct {D, +, \times}$ be an ordered integral domain where $P$ is the (strict) positivity property.
Let $\le$ be the relation defined on $D$ as:
:$\le \ := \ < \cup \Delta_D$
where $\Delta_D$ is the diagonal relation.
Then $\le$ is compatible with $+$.
== Proof ==
Let $a \le b$.
If $a \ne b$ then:
:$a < b$
and Relation Induced by Strict Positivity Property is Compatible with Addition applies.
Otherwise $a = b$.
But $\struct {D, +}$ is the additive group of $\struct {D, +, \times}$ and the Cancellation Laws apply:
:$a + c = b + c \iff a = b \iff c + a = c + b$
So $\le$ is seen to be compatible with $+$.
Category:Ordered Integral Domains
Relation Induced by Strict Positivity Property is Compatible with Multiplication

== Theorem ==
Let $\struct {D, +, \times}$ be an ordered integral domain where $P$ is the (strict) positivity property.
Let the relation $<$ be defined on $D$ as:
:$\forall a, b \in D: a < b \iff \map P {-a + b}$
Then $<$ is compatible with $\times$ in the following sense:
:$\forall x, y, z \in D: x < y, \map P z \implies \paren {z \times x} < \paren {z \times y}$
:$\forall x, y, z \in D: x < y, \map P z \implies \paren {x \times z} < \paren {y \times z}$
== Proof ==
If $x < y$ then $\map P {-x + y}$.
Hence:
The other result follows from the fact that $\times$ is commutative in an integral domain.
== Sources ==
* : Chapter $2$: Ordered and Well-Ordered Integral Domains: $\S 7$. Order: Theorem $9: \ \text{O} 2$
Category:Ordered Integral Domains
Relation Induced by Strict Positivity Property is Transitive

== Theorem ==
Let $\struct {D, +, \times}$ be an ordered integral domain where $P$ is the (strict) positivity property.
Let the relation $<$ be defined on $D$ as:
:$\forall a, b \in D: a < b \iff \map P {-a + b}$
Then $<$ is a transitive relation.
== Proof ==
Let $a < b$ and $b < c$.
Thus:
And so $<$ is seen to be transitive.
== Sources ==
* : Chapter $2$: Ordered and Well-Ordered Integral Domains: $\S 7$. Order: Theorem $8$
Category:Ordered Integral Domains
Category:Transitive Relations
Relation Induced by Strict Positivity Property is Trichotomy

== Theorem ==
Let $\struct {D, +, \times}$ be an ordered integral domain where $P$ is the (strict) positivity property.
Let the relation $<$ be defined on $D$ as:
:$\forall a, b \in D: a < b \iff \map P {-a + b}$
Then $\forall a, b \in D:$ exactly one of the following conditions applies:
:$a < b$
:$a = b$
:$a > b$
That is, $<$ is a trichotomy.
== Proof ==
Take any $a, b \in D$ and consider $-a + b$.
From the trichotomy law of ordered integral domains, exactly one of the following applies:
:$(1): \quad \map P {-a + b}$
:$(2): \quad \map P {-\paren {-a + b} }$
:$(3): \quad -a + b = 0$
Taking each of these in turn and taking into account that $\struct {D, +}$ is the additive group of $\struct {D, +, \times}$:
The result has been proved.
== Sources ==
* : Chapter $2$: Ordered and Well-Ordered Integral Domains: $\S 7$. Order: Theorem $9: \ \text{O} 3$
Category:Ordered Integral Domains
Relation Intersection Inverse is Greatest Symmetric Subset of Relation

== Theorem ==
Let $\RR$ be a relation on a set $S$.
Let $\powerset \RR$ be the power set of $\RR$.
By definition, $\powerset \RR$ is the set of all relations on $S$ that are subsets of $\RR$.
Then the greatest element of $\powerset \RR$ that is symmetric is:
:$\RR \cap \RR^{-1}$
== Proof ==
By Intersection of Relation with Inverse is Symmetric Relation:
:$\RR \cap \RR^{-1}$ is a symmetric relation.
Suppose for some $\SS \in \powerset \RR$ that $S$ is symmetric and not equal to $\RR \cap \RR^{-1}$.
We will show that it is a proper subset of $\RR \cap \RR^{-1}$.
Suppose $\tuple {x, y} \in \SS$.
Then as $\SS \subseteq \RR$:
:$\tuple {x, y} \in \RR$
As $\SS$ is symmetric:
:$\tuple {x, y} \in \SS$
So as $\SS \subseteq \RR$:
:$\tuple {x, y} \in \RR$
By definition of inverse relation:
:$\tuple {x, y} \in \RR^{-1}$
By definition of intersection:
:$\tuple {x, y} \in \RR \cap \RR^{-1}$
Thus:
:$\tuple {x, y} \in \SS \implies \tuple {x, y} \in \RR \cap \RR^{-1}$
By definition of subset:
:$\SS \subseteq \RR \cap \RR^{-1}$
Finally, as we assumed $\SS \ne \RR \cap \RR^{-1}$:
:$\SS \subset \RR \cap \RR^{-1}$
Hence the result.
Category:Relation Theory
Category:Symmetric Relations
Relation Isomorphism Preserves Antisymmetry

== Theorem ==
Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be relational structures.
Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be (relationally) isomorphic.
Then $\RR_1$ is an antisymmetric relation $\RR_2$ is also an antisymmetric relation.
== Proof ==
Let $\phi: S \to T$ be a relation isomorphism.
By Inverse of Relation Isomorphism is Relation Isomorphism it follows that $\phi^{-1}: T \to S$ is also a relation isomorphism.
, it therefore suffices to prove only that if $\RR_1$ is antisymmetric, then also $\RR_2$ is antisymmetric.
So, suppose $\RR_1$ is an antisymmetric relation.
Let $y_1, y_2 \in T$ such that both $y_1 \mathrel {\RR_2}, y_2$ and $y_2 \mathrel {\RR_2} y_1$.
Let $x_1 = \map {\phi^{-1} } {y_1}$ and $x_2 = \map {\phi^{-1} } {y_2}$.
As $\phi$ is a bijection it follows from Inverse Element of Bijection that $y_1 = \map \phi {x_1}$ and $y_2 = \map \phi {x_2}$.
As $\phi^{-1}$ is a relation isomorphism it follows that:
:$x_1 = \map {\phi^{-1} } {y_1} \mathrel {\RR_1} \map {\phi^{-1} } {y_2} = x_2$
:$x_2 = \map {\phi^{-1} } {y_2} \mathrel {\RR_1} \map {\phi^{-1} } {y_1} = x_1$
As $\RR_1$ is an antisymmetric relation it follows that $x_1 = x_2$.
As $\phi$ is a bijection it follows that:
:$y_1 = \map \phi {x_1} = \map \phi {x_2} = y_2$
Hence $y_1 = y_2$ and so by definition, $\RR_2$ is an antisymmetric relation.
Hence the result.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.9 \ \text{(c)}$
Category:Relation Isomorphisms
Category:Antisymmetric Relations
Relation Isomorphism Preserves Equivalence Relations

== Theorem ==
Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be relational structures.
Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be (relationally) isomorphic.
Then $\RR_1$ is an equivalence relation $\RR_2$ is also an equivalence relation.
== Proof ==
Let $\phi: S \to T$ be a relation isomorphism.
By Inverse of Relation Isomorphism is Relation Isomorphism it follows that $\phi^{-1}: T \to S$ is also a relation isomorphism.
, it thus is necessary to prove only that if $\RR_1$ is an equivalence relation then $\RR_2$ is an equivalence relation.
So, suppose $\RR_1$ is an equivalence relation.
By definition:
:$(1): \quad \RR_1$ is reflexive
:$(2): \quad \RR_1$ is symmetric
:$(3): \quad \RR_1$ is transitive.
It follows that:
:From Relation Isomorphism Preserves Reflexivity, $\RR_2$ is reflexive.
:From Relation Isomorphism Preserves Symmetry, $\RR_2$ is symmetric.
:From Relation Isomorphism Preserves Transitivity, $\RR_2$ is transitive.
So by definition $\RR_2$ is an equivalence relation.
Hence the result.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.9 \ \text{(c)}$
Category:Relation Isomorphisms
Category:Equivalence Relations
Relation Isomorphism Preserves Reflexivity

== Theorem ==
Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be relational structures.
Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be (relationally) isomorphic.
Then $\RR_1$ is a reflexive relation $\RR_2$ is also a reflexive relation.
== Proof ==
 it is necessary to prove only that if $\RR_1$ is reflexive then $\RR_2$ is reflexive.
Let $\phi: S \to T$ be a relation isomorphism.
Let $y \in T$.
Let $x = \map {\phi^{-1} } y$.
As $\phi$ is a bijection it follows from Inverse Element of Bijection that:
:$y = \map \phi x$
As $\RR_1$ is a reflexive relation it follows that:
:$x \mathrel {\RR_1} x$
As $\phi$ is a relation isomorphism it follows that:
:$\map \phi x \mathrel {\RR_2} \map \phi x$
Hence the result.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.9 \ \text{(c)}$
Category:Relation Isomorphisms
Category:Reflexive Relations
Relation Isomorphism Preserves Symmetry

== Theorem ==
Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be relational structures.
Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be (relationally) isomorphic.
Then $\RR_1$ is a symmetric relation $\RR_2$ is also a symmetric relation.
== Proof ==
Let $\phi: S \to T$ be a relation isomorphism.
By Inverse of Relation Isomorphism is Relation Isomorphism it follows that $\phi^{-1}: T \to S$ is also a relation isomorphism.
, it suffices to prove only that if $\RR_1$ is symmetric, then also $\RR_2$ is symmetric.
So, suppose $\RR_1$ is a symmetric relation.
Let $y_1, y_2 \in T$ such that $y_1 \mathrel {\RR_2} y_2$.
Let $x_1 = \map {\phi^{-1} } {y_1}$ and $x_2 = \map {\phi^{-1} } {y_2}$.
As $\phi$ is a bijection it follows from Inverse Element of Bijection that $y_1 = \map \phi {x_1}$ and $y_2 = \map \phi {x_2}$.
As $\phi^{-1}$ is a relation isomorphism it follows that:
:$x_1 = \map {\phi^{-1} } {y_1} \mathrel {\RR_1} \map {\phi^{-1} } {y_2} = x_2$
As $\RR_1$ is a symmetric relation it follows that:
:$x_2 \mathrel {\RR_1} x_1$
As $\phi$ is a relation isomorphism it follows that:
:$y_2 = \map \phi {x_2} \mathrel {\RR_2} \map \phi {x_1} = y_1$
So:
:$y_1 \mathrel {\RR_2} y_2 \implies y_2 \mathrel {\RR_2} y_1$
and so by definition $\RR_2$ is a symmetric relation.
Hence the result.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.9 \ \text{(c)}$
Category:Relation Isomorphisms
Category:Symmetric Relations