nyuuzyou/wikis huggingface dataset:
Ordered Semigroup Isomorphism is Surjective Monomorphism

== Theorem ==
Let $\struct {S, \circ, \preceq}$ and $\struct {T, *, \preccurlyeq}$ be ordered semigroups.
Let $\phi: \struct {S, \circ, \preceq} \to \struct {T, *, \preccurlyeq}$ be a mapping.
Then $\phi$ is an ordered semigroup isomorphism :
:$(1): \quad \phi$ is an ordered semigroup monomorphism
:$(2): \quad \phi$ is a surjection.
== Proof ==
=== Necessary Condition ===
Let $\phi: \struct {S, \circ, \preceq} \to \struct {T, *, \preccurlyeq}$ be an ordered semigroup isomorphism.
Then by definition:
:$\phi$ is a semigroup isomorphism from the semigroup $\struct {S, \circ}$ to the semigroup $\struct {T, *}$
:$\phi$ is an order isomorphism from the ordered set $\struct {S, \preceq}$ to the ordered set $\struct {T, \preccurlyeq}$.
A semigroup isomorphism is by definition:
:A semigroup homomorphism
which is:
:A monomorphism and an epimorphism.
From Order Isomorphism is Surjective Order Embedding, an order isomorphism is an order embedding which is also a surjection.
Putting this all together, we see that an ordered semigroup isomorphism is:
:A monomorphism
:An order embedding
:A surjection.
An ordered semigroup monomorphism is by definition: 
:A monomorphism
which is also
:An order embedding
Hence $\phi$ is:
:An ordered semigroup monomorphism
:A surjection.
=== Sufficient Condition ===
Let $\phi$ be:
:An ordered semigroup monomorphism
:A surjection.
By definition, that means $\phi$ be:
:A monomorphism
:An order embedding
:A surjection.
From Order Isomorphism is Surjective Order Embedding, an order isomorphism is an order embedding which is also a surjection.
A semigroup isomorphism is by definition:
:A semigroup homomorphism
which is:
:A semigroup monomorphism and an semigroup epimorphism.
Thus a semigroup monomorphism which is also a surjection is a semigroup isomorphism.
So $\phi$ is:
:A semigroup isomorphism from the semigroup $\struct {S, \circ}$ to the semigroup $\struct {T, *}$
:An order isomorphism from the ordered set $\struct {S, \preceq}$ to the ordered set $\struct {T, \preccurlyeq}$.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 15$: Ordered Semigroups
Category:Ordered Semigroups
Category:Surjections
Category:Monomorphisms (Abstract Algebra)
Category:Isomorphisms (Abstract Algebra)
Ordered Semigroup Monomorphism into Image is Isomorphism

== Theorem ==
Let $\struct {S, \circ, \preceq}$ and $\struct {T, *, \preccurlyeq}$ be ordered semigroups.
Let $\phi: \struct {S, \circ, \preceq} \to \struct {T, *, \preccurlyeq}$ be an ordered semigroup monomorphism.
Let $S'$ be the image of $\phi$.
Then $\phi$ is an ordered semigroup isomorphism from $\struct {S, \circ, \preceq}$ into $\struct {S', * {\restriction_{S'} }, \preccurlyeq \restriction_{S'} }$.
Here:
:$* {\restriction_{S'}}$ denotes the restriction of $*$ to $S' \times S'$
:$\preccurlyeq \restriction_{S'}$ denotes the restriction of $\preccurlyeq$ to $S' \times S'$.
== Proof ==
Let $\phi: \struct {S, \circ, \preceq} \to \struct {T, *, \preccurlyeq}$ be an ordered semigroup monomorphism.
Then $\phi$ is an injection into $\struct {T, *, \preccurlyeq}$ by definition.
From Restriction of Mapping to Image is Surjection, a mapping from a set to the image of that mapping is a surjection.
Thus the surjective restriction of $\phi$ onto $S'$ is an ordered semigroup monomorphism which is also a surjection.
Hence the result from Ordered Semigroup Isomorphism is Surjective Monomorphism.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 15$: Ordered Semigroups
Category:Semigroup Homomorphisms
Category:Order Isomorphisms
Ordered Set has Lower Bound iff has Smallest Element

== Theorem ==
Let $\struct{S,\preceq}$ be an ordered set.
Then:
:$S$ has a lower bound in $\struct{S,\preceq}$ $\struct{S,\preceq}$ has a smallest element
== Proof ==
This follows immediately from:
:* Definition:Lower Bound
:* Definition:Smallest Element
Category:Smallest Elements
Category:Ordered Sets
Ordered Set has Upper Bound iff has Greatest Element

== Theorem ==
Let $\struct{S,\preceq}$ be an ordered set.
Then:
:$S$ has an upper bound in $\struct{S,\preceq}$ $\struct{S,\preceq}$ has a greatest element
== Proof ==
This follows immediately from:
:* Definition:Upper Bound
:* Definition:Greatest Element
Category:Greatest Elements
Category:Ordered Sets
Ordered Set may not have Maximal Element

== Theorem ==
Let $\struct {S, \preccurlyeq}$ be an ordered set.
It may be the case that $S$ has no maximal elements.
== Proof ==
Consider the set $S$ defined as:
:$S = \N \setminus \set 0$
That is, $S$ is the set of natural numbers with $0$ removed.
Let $\preccurlyeq$ be the ordering on $S$ defined as:
:$\forall a, b \in S: a \preccurlyeq b \iff a \divides b$
where $a \divides b$ denotes that $a$ is a divisor of $b$.
From Divisor Relation on Positive Integers is Partial Ordering, $\struct {S, \preccurlyeq}$ is a partially ordered set.
 $S$ has a maximal element $m$.
Consider the natural number $n = 2 m$.
Then:
:$m \divides n$
but:
:$m \ne n$
and so by definition $m$ is not a maximal element of $S$.
Hence by Proof by Contradiction it follows that there exists no such $m$.
Hence the result.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations
Category:Maximal Elements
Ordered Set may not have Minimal Element

== Theorem ==
Let $\struct {S, \preccurlyeq}$ be an ordered set.
It may be the case that $S$ has no minimal elements.
== Proof ==
Let $\Q_{>0}$ denote the set of (strictly) positive rational numbers.
From Rational Numbers form Totally Ordered Field, the rational numbers $\Q$ are totally ordered by the usual ordering $\le$.
From Subset of Toset is Toset, $\Q_{>0}$ is also totally ordered by $\le$.
Thus $\struct {\Q_{>0}, \le}$ is an ordered set.
 there exists a minimal element $m$ of $\struct {\Q_{>0}, \le}$.
From Minimal Element in Toset is Unique and Smallest, $m$ is also the smallest element of $\struct {\Q_{>0}, \le}$
By Smallest Strictly Positive Rational Number does not Exist, $\struct {\Q_{>0}, \le}$ has no smallest element.
Hence $m$ cannot be a minimal element of $\struct {\Q_{>0}, \le}$.
It follows by Proof by Contradiction that $\struct {\Q_{>0}, \le}$ has no minimal elements.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations
Category:Minimal Elements
Ordered Set of All Mappings is Lattice iff Codomain is Lattice or Domain is Empty/Lemma

== Lemma for Ordered Set of All Mappings is Lattice iff Codomain is Lattice or Domain is Empty ==
Let $S$ be a set.
Let $\struct {T, \preccurlyeq}$ be an ordered set.
Let $\struct {T^S, \preccurlyeq}$ denote the ordered set of all mappings from $S$ to $T$.
Let $S = \O$.
Then $\struct {T^S, \preccurlyeq}$ is a lattice.
== Proof ==
Recall the definition of lattice:
Let $S = \O$.
Then there is one mapping from $S$ to $T$, and that is the empty mapping $e: S \to T$.
Thus we have $T^S = \set e$
From Supremum of Singleton and Infimum of Singleton:
:$\sup \set e = e = \inf \set e$
Hence $T^S$ (trivially, in the degenerate sense) is a lattice.
Category:Ordered Set of All Mappings is Lattice iff Codomain is Lattice or Domain is Empty
Ordered Set of All Mappings is Ordered Set

== Theorem ==
Let $L = \struct {S, \preceq}$ be an ordered set.
Let $X$ be a set.
Then $L^X$ is also an ordered set.
== Proof ==
By definition of ordered set of all mappings:
:$L^X = \struct {S^X, \precsim}$
where
:$\forall f, g \in S^X: f \precsim g \iff f \preceq g$
:$\preceq$ denotes the ordering on mappings,
:$S^X$ denotes the set of all mappings from $X$ into $S$.
=== Reflexivity ===
Let $f \in S^X$.
By definition of reflexivity:
:$\forall x \in X: \map f x \preceq \map f x$
By definition of ordering on mappings:
:$f \preceq f$
Thus by definition of $\precsim$:
:$f \precsim f$
=== Transitivity ===
Let $f, g, h \in S^X$ such that
:$f \precsim g$ and $g \precsim h$
By definition of $\precsim$:
:$f \preceq g$ and $g \preceq h$
By definition of ordering on mappings:
:$\forall x \in X: \map f x \preceq \map g x$
and
:$\forall x \in X: \map g x \preceq \map h x$
By definition of transitivity:
:$\forall x \in X: \map f x \preceq \map h x$
By definition of ordering on mappings:
:$f \preceq h$
Thus by definition of $\precsim$:
:$f \precsim h$
=== Antisymmetry ===
Let $f, g \in S^X$ such that
:$f \precsim g$ and $g \precsim f$
By definition of $\precsim$:
:$f \preceq g$ and $g \preceq f$
By definition of ordering on mappings:
:$\forall x \in X: \map f x \preceq \map g x$
and
:$\forall x \in X: \map g x \preceq \map f x$
By definition of antisymmetry:
:$\forall x \in X: \map f x = \map g x$
Thus by Equality of Mappings:
:$f = g$
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.20$
* 
* 
* 
Category:Ordered Sets
Ordered Set of Auxiliary Relations is Complete Lattice

== Theorem ==
Let $L = \struct {S, \vee, \preceq}$ be a bounded below join semilattice.
Let $\map {\operatorname {Aux} } L$ be the set of all auxiliary relations on $S$.
Let $P = \struct {\map {\operatorname {Aux} } L, \precsim}$ be an ordered set
where $\precsim \mathop = \subseteq \restriction_{\map {\operatorname {Aux} } L \times \map {\operatorname {Aux} } L}$
Then
:$P$ is a complete lattice.
== Proof ==
Let $X \subseteq \map {\operatorname {Aux} } L$
In the case when $X \ne \O$:
By Intersection of Auxiliary Relations is Auxiliary Relation:
:$\bigcap X \in \map {\operatorname {Aux} } L$
By Intersection is Largest Subset, $\bigcap X$ is the infimum of $X$.
In case when $X = \O$:
By proof of Preceding is Top in Ordered Set of Auxiliary Relations:
:$\O$ admits an infimum in $P$
Then:
:$X$ (empty or non-empty) admits an infimum in $P$
By duality of Lattice is Complete iff it Admits All Suprema:
:$P$ is a complete lattice.
== Sources ==
* 
* 
Category:Auxiliary Relations
Ordered Set of Closure Operators and Dual Ordered Set of Closure Systems are Isomorphic

== Theorem ==
Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a complete lattice.
Then $\operatorname{Closure}\left({L}\right)$ and $\operatorname{ClSystems}\left({L}\right)^{-1}$ are order isomorphic
where
:$\operatorname{Closure}\left({L}\right)$ denotes the ordered set of closure operators of $L$,
:$\operatorname{ClSystems}\left({L}\right)$ denotes the ordered set of closure systems oj $L$,
:$\operatorname{ClSystems}\left({L}\right)^{-1}$ denotes the dual to $\operatorname{ClSystems}\left({L}\right)$.
== Proof ==
By definition of ordered set of closure operators:
:$\operatorname{Closure}\left({L}\right) = \left({X, \preceq'}\right)$
where
:$X$ is the set of all closure operators on $L$,
:for all closure operators $f, g$ on $L$: $f \preceq' g \iff f \preceq g$
:$\preceq$ denotes the ordering on mappings.
By definition of ordered set of closure systems:
:$\operatorname{ClSystems}\left({L}\right) = \left({Y, \precsim}\right)$
where
:$Y$ is the set of all closure systems on $L$,
:for all closure systems $s_1 = \left({T_1, \preceq_1}\right)$, $s_2 = \left({T_2, \preceq_2}\right)$ on $L$: $s_1 \precsim s_2 \iff T_1 \subseteq T_2$
Define $F:X \to Y$ by Image of Closure Operator Inherits Infima:
:for all closure operators $f$ on $L$: $F\left({f}\right) = \left({f\left[{S}\right], \mathord\preceq \cap \left({f\left[{S}\right] \times f\left[{S}\right]}\right)}\right)$
By Operator Generated by Closure System is Closure Operator and Image of Operator Generated by Closure System is Set of Closure System:
:$F$ is a surjection.
By Ordering on Closure Operators iff Images are Including:
:$F$ is an order embedding.
Thue by definition:
:$F$ is an order isomorphism.
== Sources ==
* 
* 
Category:Closure Operators
Ordered Set of Closure Systems is Ordered Set

== Theorem ==
Let $L = \left({S, \preceq}\right)$ be an ordered set.
Then $\operatorname{ClSystems}\left({L}\right)$ is an ordered set,
where $\operatorname{ClSystems}\left({L}\right)$ denotes the ordered set of closure systems.
== Proof ==
By definition of ordered set of closure systems:
:$\operatorname{ClSystems}\left({L}\right) = \left({X, \precsim}\right)$
where
:$X$ is the set of all closure systems of $L$,
:dor all closure systems $s_1 = \left({T_1, \preceq_1}\right), s_2 = \left({T_2, \preceq_2}\right)$ of $L$: $s_1 \precsim s_2 \iff T_1 \subseteq T_2$
=== Reflexivity ===
Let $s = \left({T, \preceq'}\right)$ be a closure system.
By Set is Subset of Itself:
:$T \subseteq T$
By definition of $\precsim$:
:$s \precsim s$
Thus by definition
:$\precsim$ is reflexive.
=== Transitivity ===
Let $s_1 = \left({T_1, \preceq_1}\right)$, $s_2 = \left({T_2, \preceq_2}\right)$, $s_3 = \left({T_3, \preceq_3}\right)$ be closure systems such that
:$s_1 \precsim s_2$ and $s_2 \precsim s_3$
By definition of $\precsim$:
:$T_1 \subseteq T_2$ and $T_2 \subseteq T_3$
By Subset Relation is Transitive:
:$T_1 \subseteq T_3$
By definition of $\precsim$:
:$s_1 \precsim s_3$
Thus by definition:
:$\precsim$ is transitive.
=== Antisymmetry ===
Let $s_1 = \left({T_1, \preceq_1}\right)$, $s_2 = \left({T_2, \preceq_2}\right)$ be closure systems such that
:$s_1 \precsim s_2$ and $s_2 \precsim s_1$
By definition of $\precsim$:
:$T_1 \subseteq T_2$ and $T_2 \subseteq T_1$
By definition of set equality:
:$T_1 = T_2$
By definition of ordered subset:
:$\mathord\preceq_1 = \mathord\preceq \cap \left({T_1 \times T_1}\right) = \mathord\preceq_2$
Then
:$s_1 = s_2$
Thus by definition:
:$\precsim$ is antisymmetric
So, by definition:
:$\precsim$ is ordering.
Hence $\operatorname{ClSystems}\left({L}\right)$ is an ordered set,
== Sources ==
* 
* 
Category:Order Theory
Ordered Set with Greatest Element whose Subsets have Infimum is Complete Lattice

== Theorem ==
Let $\struct {S, \preccurlyeq}$ be an ordered set such that:
:$\struct {S, \preccurlyeq}$ has a greatest element $u$
:every non-empty subset of $S$ admits an infimum.
Then $\struct {S, \preccurlyeq}$ is a complete lattice.
== Proof ==
For $\struct {S, \preccurlyeq}$ to be a complete lattice, it has to be such that:
:every non-empty subset of $S$ admits both a supremum and an infimum.
Let $T \subseteq S$ be an arbitrary subset of $S$.
We already have that $T$ admits an infimum.
We are to show that $T$ admits a supremum.
By definition of greatest element:
:$\forall x \in T: x \preccurlyeq u$
Thus by definition $u$ is an upper bound of $T$.
Let $H \subseteq S$ be the set of all upper bounds of $T$.
Because $u$ is an upper bound of $T$, $u \in H$.
Hence $H$ is not empty.
We have that $H$ has an infimum.
That is:
:$\exists v \in H: \forall x \in H: v \preccurlyeq x$
Hence $v$ is by definition a supremum of $T$.
Thus $T$ admits a supremum.
Hence $T$ admits both a supremum and an infimum.
Hence the result.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.11 \ \text{(a)}$
Category:Complete Lattices
Category:Greatest Elements
Ordered Set with Multiple Maximal Elements has no Greatest Element

== Theorem ==
Let $\struct {S, \preccurlyeq}$ be an ordered set.
Let $\struct {S, \preccurlyeq}$ have more than one maximal element.
Then $\struct {S, \preccurlyeq}$ has no greatest element.
== Proof ==
Let $s$ and $t$ both be maximal elements of $\struct {S, \preccurlyeq}$ such that $s \ne t$.
Then by definition:
:$\forall x \in S: s \preccurlyeq x \implies s = x$
and:
:$\forall x \in S: t \preccurlyeq x \implies t = x$
 $S$ has a greatest element $m$.
Then by definition:
:$\forall y \in S: y \preccurlyeq m$
Hence:
:$s \preccurlyeq m$
and also:
:$y \preccurlyeq m$
But because $s$ and $t$ are both maximal elements of $\struct {S, \preccurlyeq}$:
:$s = m$
and:
:$t = m$
This contradicts the fact that $s \ne t$.
Hence by Proof by Contradiction there can be no greatest element of $\struct {S, \preccurlyeq}$.
Category:Maximal Elements
Category:Greatest Elements
Ordered Set with Multiple Minimal Elements has no Smallest Element

== Theorem ==
Let $\struct {S, \preccurlyeq}$ be an ordered set.
Let $\struct {S, \preccurlyeq}$ have more than one minimal element.
Then $\struct {S, \preccurlyeq}$ has no smallest element.
== Proof ==
Let $s$ and $t$ both be minimal elements of $\struct {S, \preccurlyeq}$ such that $s \ne t$.
Then by definition:
:$\forall x \in S: x \preccurlyeq s \implies s = x$
and:
:$\forall x \in S: x \preccurlyeq t \implies t = x$
 $S$ has a smallest element $m$.
Then by definition:
:$\forall y \in S: m \preccurlyeq y$
Hence:
:$m \preccurlyeq s$
and also:
:$m \preccurlyeq t$
But because $s$ and $t$ are both minimal elements of $\struct {S, \preccurlyeq}$:
:$s = m$
and:
:$t = m$
This contradicts the fact that $s \ne t$.
Hence by Proof by Contradiction there can be no smallest element of $\struct {S, \preccurlyeq}$.
Category:Minimal Elements
Category:Smallest Elements
Ordered Set with Order Type of Natural Numbers plus Dual has Minimum Element

== Theorem ==
Let $\struct {S, \preccurlyeq}$ be an ordered structure such that:
:$\map \ot {S, \preccurlyeq} = \omega + \omega^*$
where:
:$\ot$ denotes order type
:$\omega$ denotes the order type of the natural numbers $\N$
:$\omega^*$ denotes the dual of $\omega$
:$+$ denotes addition of order types.
Then $\struct {S, \preccurlyeq}$ has a smallest element.
== Proof ==
By definition of order type addition:
:$\struct {S, \preccurlyeq}$ is isomorphic to $\struct {\N, \le} \oplus \struct {\N, \ge}$
where:
:$\cong$ denotes order isomorphism
:$\oplus$ denotes order sum.
By the Well-Ordering Principle, $\struct {\N, \le}$ has a smallest element.
By definition of order sum, every element of $\struct {\N, \le}$ precedes every element of $\struct {\N, \ge}$.
Hence the smallest element of $\struct {\N, \le}$ is also the smallest element of $\struct {\N, \le} \oplus \struct {\N, \ge}$.
Hence the result.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $28 \ \text {(b)}$
Category:Examples of Order Types
Ordered Subset of Ordered Set is Ordered Set

== Theorem ==
Let $L = \struct {S, \preceq}$ be an ordered set.
Let $\struct {S', \preceq'}$ be an ordered subset of $L$.
Then $\struct {S', \preceq'}$ is an ordered set.
== Proof ==
By definition of ordered subset:
:$S' \subseteq S$
=== Reflexivity ===
Let $x \in S'$.
By definition of subset:
:$x \in S$
By definition of reflexivity:
:$x \preceq x$
Thus by definition of ordered subset:
:$x \preceq' x$
=== Transitivity ===
Let $x, y, z \in S'$ such that:
:$x \preceq' y$ and $y \preceq' z$
By definition of ordered subset:
:$x \preceq y$ and $y \preceq z$
By definition of transitivity:
:$x \preceq z$
Thus by definition of ordered subset:
:$x \preceq' z$
=== Antisymmetry ===
Let $x, y \in S'$ such that:
:$x \preceq' y$ and $y \preceq' x$
By definition of ordered subset:
:$x \preceq y$ and $y \preceq x$
Thus by definition of antisymmetry:
:$x = y$
== Sources ==
* 
* 
* 
Category:Order Theory
Ordered Sum of Tosets is Totally Ordered Set

== Theorem ==
Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be tosets.
Let $S + T = \struct {S \cup T, \preceq}$ be the ordered sum of $S$ and $T$.
Then $\struct {S \cup T, \preceq}$ is itself a toset.
=== General Result ===
== Proof ==
By definition of ordered sum, we have that:
* If $a, b \in S$, then $a \preceq b \iff a \preceq_1 b$.
* Otherwise, if $a, b \in T$, then $a \preceq b \iff a \preceq_2 b$.
* If neither of these is the case, then $a \in S, b \in T \iff a \preceq b$.
First we show that $\preceq$ is connected.
We note that as $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ are both tosets, then $\preceq_1$ and $\preceq_2$ are both connected.
Let $a, b \in S \cup T$.
There are three cases:
: $(1): \quad a, b \in S$, in which case $a \preceq_1 b$ or $b \preceq_1 a$
: $(2): \quad a, b \notin S$, in which case $a \preceq_2 b$ or $b \preceq_2 a$
: $(3): \quad$ Otherwise, in which case either $a \in S, b \notin S$ or $a \notin S, b \in S$.
:::By definition of $\preceq$ it follows that $a \preceq b$ or $b \preceq a$, but clearly not both.
So in all cases $a \preceq b$ or $b \preceq a$, and $\preceq$ is connected.
Now we check in turn each of the criteria for an ordering:
=== Reflexivity ===
We have by definition of set union:
:$\forall a \in S \cup T: a \in S \lor a \in T$
and so either $a \preceq_1 a$ or $a \preceq_2 a$, showing $a \preceq a$.
So $\preceq$ is reflexive.
=== Transitivity ===
Suppose $a \preceq b$ and $b \preceq c$.
We proceed on a case-by-case basis.
There are eight of these, (indirectly) from Cardinality of Power Set of Finite Set.
* Suppose $a, b, c \in S$. Then $a \preceq c$ by transitivity of $\preceq_1$.
* Suppose $a, b \in S, c \notin S$. Then $a \preceq c$ by definition of $\preceq$.
* Suppose $a \in S, b, c \notin S$. Then $a \preceq c$ by definition of $\preceq$.
* Suppose $a, b, c \notin S$. Then $a \preceq c$ by transitivity of $\preceq_2$.
There are four other cases to consider:
* $a, c \in S, b \notin S$
* $b, c \in S, a \notin S$
* $c \in S, a, b \notin S$
* $b \in S, a, c \notin S$.
None of these can happen as otherwise one of $a \npreceq b$ or $b \npreceq c$ would be the case.
So we have shown that $\preceq$ is transitive.
=== Antisymmetry ===
Suppose $a \preceq b$ and $b \preceq a$.
* It can not be the case that $a \in S$ and $b \notin S$ because then $b \npreceq a$.
* It can not be the case that $a \notin S$ and $b \in S$ because then $a \npreceq b$.
So either $a, b \in S$ or $a, b \notin S$.
* If $a, b \in S$ then $a = b$ by antisymmetry of $\preceq_1$.
* If $a, b \notin S$ then $a = b$ by antisymmetry of $\preceq_2$.
Thus in all cases it can be seen that $a \preceq b \preceq a \implies a = b$.
So $\preceq$ is antisymmetric.
So we have shown that:
* $\preceq$ is connected
* $\preceq$ is reflexive, transitive and antisymmetric.
Thus by definition, $\preceq$ is a total ordering and so $\struct {S \cup T, \preceq}$ is a toset.
== Sources ==
* : $\S 3.4$: Ordered sums and products of ordered sets
Category:Total Orderings
Category:Ordered Sums
Ordered Sum of Tosets is Totally Ordered Set/General Result

== Theorem ==
Let $S_1, S_2, \ldots, S_n$ all be tosets.
Let $T_n$ be the ordered sum of $S_1, S_2, \ldots, S_n$:
:$\forall n \in \N_{>0}: T_n = \begin{cases}
S_1 & : n = 1 \\
T_{n - 1} + S_n & : n > 1
\end{cases}$
Then $T_n$ is a toset.
== Proof ==
From Ordered Sum of Tosets is Totally Ordered Set, $S_1 + S_2$ is a toset.
Suppose $T_{n-1}$ is a toset.
Given that $S_n$ is a toset, it follows from Ordered Sum of Tosets is Totally Ordered Set that $T_{n-1} + S_n$ is also a toset.
The result follows by induction.
Category:Total Orderings
Category:Ordered Sums
Ordering/Examples/American Presidency

== Example of Ordering ==
Let $S$ denote the set of .
Let $\PP$ denote the relation on $S$ defined as:
:$a \mathrel \PP b$ $a$ was after or at the same time as $b$.
Because was both before and after , $\PP$ is not an antisymmetric relation.
Thus $\PP$ is not an ordering on $S$.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings
Category:Examples of Orderings