nyuuzyou/wikis huggingface dataset:
Open and Closed Sets in Multiple Pointed Topology

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $A$ be a finite set whose cardinality is greater than $1$.
Let $D = \struct {A, \set {\O, A} }$ be the indiscrete space on $A$.
Let $T \times D$ be a multiple pointed topological space generated from $T$ and $D$.
Let $H \subseteq T$.
Then:
:$H \times A$ is open in $T \times D$ $H$ is open in $T$
:$H \times A$ is closed in $T \times D$ $H$ is closed in $T$
:$H \times \O$ is both open and closed in $T \times D$
Let $\O \subsetneqq H_A \subsetneqq A$, that is, let $H_A$ be a non-empty proper subset of $A$.
Then $H \times H_A$ is neither open nor closed in $T \times D$.
== Proof ==
By definition of multiple pointed topology, $T \times D$ is the product space of $T$ with $D$:
:$T \times D = \struct {S \times A, \tau}$
where $\tau$ is the product topology on $S \times A$.
Consider the set $H \times A$.
We have that:
:$H \times A = \pr_1^{-1} \sqbrk H$
where $\pr_1: T \times D \to T$ is the first projection from $T \times D$ to $T$.
By definition of the product topology, $\pr_1: T \times D \to T$ is a continuous mapping.
=== Open Sets ===
By definition of product topology, if $H$ is open in $T$ then $H \times A$ is open in $T \times D$.
As the Product Topology is Coarsest Topology such that Projections are Continuous, it follows that if $H$ is not open in $T$ then $H \times A$ is not open in $T \times D$.
=== Closed Sets ===
Let $H \subseteq S$ be closed in $T$.
We have that $\pr_1: T \times D \to T$ is a continuous mapping.
By Continuity Defined from Closed Sets, it follows that $H \times A$ is closed in $T \times D$.
As the Product Topology is Coarsest Topology such that Projections are Continuous, it follows that if $H$ is not closed in $T$ then $H \times A$ is not closed in $T \times D$.
=== Cartesian Product with Empty Set ===
From Cartesian Product is Empty iff Factor is Empty, $H \times \O = \O$.
From Open and Closed Sets in Topological Space, $\O$ is both open and closed in $T \times D$.
=== Neither Open Nor Closed ===
Let $\O \subsetneqq H_A \subsetneqq A$.
That is, $H_A \ne \O$ and $H_A \ne A$.
So $H_A$ is neither open nor closed in the indiscrete topology on $A$
We have that:
:$S \times H_A = \pr_2^{-1} \sqbrk {H_A}$
where $\pr_2: T \times D \to D$ is the second projection from $T \times D$ to $D$.
By definition of the product topology, $\pr_2: T \times D \to D$ is a continuous mapping.
We have that the Product Topology is Coarsest Topology such that Projections are Continuous.
So if $H_A$ is neither open nor closed in $D$ then $S \times H_A$ is neither open nor closed in $T \times D$.
Hence, by definition of the product topology, nor is $H \times H_A$.
Category:Multiple Pointed Topology
Category:Open Sets
Category:Closed Sets
Category:Clopen Sets
Open and Closed Sets in Topological Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Then $S$ and $\O$ are both both open and closed in $T$.
== Proof ==
From the definition of closed set, $U$ is open in $T$ $S \setminus U$ is closed in $T$.
From Underlying Set of Topological Space is Clopen, $S$ is both open and closed in $T$.
From Empty Set is Element of Topology, $\O$ is open in $T$.
From Empty Set is Closed in Topological Space, we have that $\O$ is closed in $T$.
== Sources ==
* : Chapter $\text {I}$: Topological Spaces: $1$. Open Sets and Closed Sets
Category:Clopen Sets
Open iff Upper and with Property (S) in Scott Topological Lattice

== Theorem ==
Let $T = \struct {S, \preceq, \tau}$ be an up-complete topological lattice.
Let $A$ be a subset of $S$.
Then $A$ is open $A$ is upper and with property (S).
== Proof ==
=== Sufficient Condition ===
Let $A$ be open.
Thus by definition of Scott topology:
:$A$ is an upper section.
Let $D$ be a directed subset of $S$ such that
:$\sup D \in A$
By definition of Scott topology:
:$A$ is inaccessible by directed suprema.
By definition of inaccessible by directed suprema:
:$A \cap D \ne \O$
By definition of non-empty:
:$\exists y: y \in A \cap D$
By definition of intersection:
:$y \in A$ and $y \in D$
Thus $y \in D$.
Thus by definition of upper section:
:$\forall x \in D: y \preceq x \implies x \in A$
=== Necessary Condition ===
Assume that
:$A$ is upper and with property (S).
According to the definition of Scott topology it should be proved that
:$A$ is upper and inaccessible by directed suprema.
Thus by assumption:
:$A$ is upper.
Let $D$ be a directed subset of $S$ such that
:$\sup D \in A$
By definition of property (S):
:$\exists y \in D: \forall x \in D: y \preceq x \implies x \in A$
By definition of reflexivity:
:$y \preceq y$
Then
:$y \in A$
By definition of intersection:
:$y \in A \cap D$
Thus by definition of non-empty:
:$A \cap D \ne \O$
== Sources ==
* 
* 
Category:Topological Order Theory
Category:Open Sets
Open implies There Exists Way Below Element

== Theorem ==
Let $L = \struct {S, \preceq, \tau}$ be a continuous topological lattice with Scott topology.
Let $p \in S, A \subseteq S$ such that:
:$A$ is open and $p \in A$.
Then:
:$\exists q \in A: q \ll p$
where $q \ll p$ denotes $q$ is way below $p$.
== Proof ==
By definition of continuous ordered set:
:$p^\ll$ is directed
and
:$L$ satisfies the axiom of approximation.
By the axiom of approximation:
:$p = \map \sup {p^\ll}$
By definition of Scott topology:
:$A$ is inaccessible by directed suprema.
By definition of inaccessible by directed suprema:
:$A \cap p^\ll \ne \O$
By definition of non-empty set:
:$\exists q: q \in A \cap p^\ll$
Thus by definition of intersection:
:$q \in A$
By definition of intersection:
:$q \in p^\ll$
By definition of way below closure:
:$q \ll p$
== Sources ==
* 
* 
Category:Topological Order Theory
Category:Way Below Relation
Category:Continuous Lattices
Category:Scott Topology
Openness Relation on Topological Spaces is Transitive

== Theorem ==
Let $T_1 = \struct {S_1, \tau_1}$ be a topological space
Let $S_2 \subseteq S_1$ be a subset of $S_1$.
Let $S_3 \subseteq S_2$ be a subset of $S_2$.
Let $T_2 = \struct {S_2, \tau_2}$ be the topological subspace of $T_1$ such that $\tau_2$ is the subspace topology induced by $\tau_1$.
Let $T_3 = \struct {S_3, \tau_3}$ be the topological subspace of $T_2$ such that $\tau_3$ is the subspace topology induced by $\tau_2$.
Let:
:$S_2$ be an open set of $T_1$
:$S_3$ be an open set of $T_2$.
Then:
:$S_3$ is an open set of $T_1$.
== Proof ==
We have by definition of subspace topology that:
:$\tau_2 = \set {U \cap S_2: U \in \tau_1}$
Then we have that:
:$S_3 \in \tau_2$
and so:
:$S_3 \in \set {U \cap S_2: U \in \tau_1}$
That is, $S_3$ is the intersection of $U$ and $S_2$, both of which are open sets of $T_1$.
Hence by , $S_3$ is an open set of $T_1$.
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 10$
Category:Topological Subspaces
Operand is Upper Bound of Way Below Closure

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $x \in S$.
Then
:$x$ is upper bound for $x^\ll$
where $x^\ll$ denotes the way below closure of $x$.
== Proof ==
Let $y \in x^\ll$
By definition of way below closure:
:$y \ll x$
where $\ll$ denotes the way below relation.
Thus by Way Below implies Preceding:
:$y \preceq x$
Thus by definition:
:$x$ is upper bound for $x^\ll$
== Sources ==
* 
* 
Category:Way Below Relation
Operating on Ordered Group Inequalities

== Theorem ==
Let $\struct {G, \circ, \preccurlyeq}$ be an ordered group.
Let $\prec$ be the reflexive reduction of $\preceq$.
Let $x, y, z, w \in \struct {G, \circ, \preceq}$.
Then the following implications hold:
If $x \prec y$ and $z \prec w$, then $x \circ z \prec y \circ w$.
If $x \prec y$ and $z \preccurlyeq w$, then $x \circ z \prec y \circ w$.
If $x \preccurlyeq y$ and $z \prec w$, then $x \circ z \prec y \circ w$.
If $x \preccurlyeq y$ and $z \preccurlyeq w$, then $x \circ z \preccurlyeq y \circ w$.
== Proof ==
Because:
:$\struct {G, \circ, \preccurlyeq}$ is a group
:$\preccurlyeq$ is compatible with $\circ$
it follows from Reflexive Reduction of Relation Compatible with Group Operation is Compatible that $\prec$ is compatible with $\circ$ .
By the definition of an ordering, $\preccurlyeq$ is transitive and antisymmetric.
Therefore by Reflexive Reduction of Transitive Antisymmetric Relation is Strict Ordering, $\prec$ is transitive.
The result follows by Operating on Transitive Relationships Compatible with Operation.
== Also see ==
* Properties of Ordered Group
Category:Ordered Groups
Operating on Transitive Relationships Compatible with Operation

== Theorem ==
Let $\struct {S, \circ}$ be a magma.
Let $\RR$ be a transitive relation on $S$ which is compatible with $\circ$.
Let $\RR^=$ be the reflexive closure of $\RR$.
Let $x, y, z, w \in S$.
Then the following implications hold:
$(1): \quad$ If $x \mathrel \RR y$ and $z \mathrel \RR w$, then $x \circ z \mathrel \RR y \circ w$.
$(2): \quad$ If $x \mathrel \RR y$ and $z \mathrel {\RR^=} w$, then $x \circ z \mathrel \RR y \circ w$.
$(3): \quad$ If $x \mathrel {\RR^=} y$ and $z \mathrel \RR w$, then $x \circ z \mathrel \RR y \circ w$.
$(4): \quad$ If $x \mathrel {\RR^=} y$ and $z \mathrel {\RR^=} w$, then $x \circ z \mathrel {\RR^=} y \circ w$.
== Proof ==
We will prove $(1)$ and $(2)$. $(3)$ can be proven using precisely the same argument as $(2)$ and $(4)$ follows from $(1)$.
Suppose that $x \mathrel \RR y$ and $z \mathrel \RR w$.
By the definition of compatibility:
:$x \mathrel \RR y \implies x \circ z \mathrel \RR y \circ z$
:$z \mathrel \RR w \implies y \circ z \mathrel \RR y \circ w$
By transitivity:
:$(1): \quad \paren {x \mathrel \RR y} \land \paren {z \mathrel \RR w} \implies x \circ z \mathrel \RR y \circ w$
Suppose $x \mathrel \RR y$ and $z \mathrel {\RR^=} w$.
By Reflexive Closure of Relation Compatible with Operation is Compatible, $\RR^=$ is compatible with $\circ$.
By the definition of compatibility:
:$x \mathrel \RR y \implies x \circ z \mathrel \RR y \circ z$
:$z \mathrel {\RR^=} w \implies y \circ z \mathrel {\RR^=} y \circ w$
By Extended Transitivity:
:$(2): \quad \paren {x \mathrel \RR y} \land \paren {z \mathrel {\RR^=} w} \implies x \circ z \mathrel \RR y \circ w$
$(3)$ can be proven using precisely the same argument as $(2)$.
By Reflexive Closure of Relation Compatible with Operation is Compatible, $\RR^=$ is compatible with $\circ$.
By Reflexive Closure of Transitive Relation is Transitive, $\RR^=$ is transitive.
Thus $(4)$ follows from $(1)$.
Category:Compatible Relations
Category:Transitive Relations
Operation/Examples/Addition

== Example of Operation ==
An example of an operation, from conventional arithmetic, is "$+$", as in, for example, $2 + 3 = 5$.
The operands (in this particular instance) are $2$ and $3$.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions
* 
* 
Category:Examples of Operations
Operation Induced by Permutation on Magma is Closed

== Theorem ==
Let $\struct {S, \circ}$ be a magma.
Let $\sigma: S \to S$ be a permutation on $S$.
Let $\circ_\sigma$ be the operation on $S$ induced by $\sigma$:
:$\forall x, y \in S: x \circ_\sigma y := \map \sigma {x \circ y}$
Then $\circ_\sigma$ is closed on $S$
== Proof ==
Suppose $S$ is the empty set.
Let $\sigma: S \to S$ be a permutation on $S$.
Since $S$ is empty, by definition, $\sigma$ is the empty map. 
Since $\circ_\sigma$ is the operation on $S$ induced by $\sigma$, it follows that $\circ_\sigma$ is the empty map.
It is vacuously true that $\circ_\sigma$ is closed on $S$, as required.
Suppose $S$ is non-empty.
Let $a, b \in S$.
By definition of magma, $\circ$ is closed on $S$.
Thus:
:$a \circ b \in S$
Hence:
:$a \circ b \in \Dom \sigma$
As $\sigma$ is a permutation on $S$, it follows directly that:
:$\map \sigma {a \circ b} \in S$
The operation on $S$ induced by $\sigma$ is well-defined.
In other words, $\circ_\sigma$ is well-defined.
Hence:
:$a \circ_\sigma b \in S$
Category:Operations Induced by Permutations
Category:Magmas
Operation Induced by Permutation on Semigroup is not necessarily Associative

== Theorem ==
Let $\struct {S, \circ}$ be a semigroup.
Let $\sigma: S \to S$ be a permutation on $S$.
Let $\circ_\sigma$ be the operation on $S$ induced by $\sigma$:
:$\forall x, y \in S: x \circ_\sigma y := \map \sigma {x \circ y}$
Then $\circ_\sigma$ is not necessarily associative on $S$.
== Proof ==
;Proof by Counterexample
Let $S = \set {a, b, c}$.
Let $\circ$ denote the right operation on $S$:
:$\forall x, y \in S: x \to y = y$
From Structure under Right Operation is Semigroup, $\struct {S, \circ}$ is a semigroup.
Hence we have:
:$a \circ \paren {b \circ c} = c = \paren {a \circ b} \circ c$
Let $\sigma$ denote the permutation on $S$ defined as:
We have:
Then:
So $a \circ_\sigma \paren {b \circ_\sigma c} \ne \paren {a \circ_\sigma b} \circ_\sigma c$ and the result follows.
Category:Operations Induced by Permutations
Category:Semigroups
Category:Associativity
Operation is Isomorphic to Power Structure Operation on Set of Singleton Subsets

== Theorem ==
Let $\struct {S, \circ}$ be a magma.
Let $\struct {\powerset S, \circ_\PP}$ be the power structure of $\struct {S, \circ}$.
Let $S'$ denote the set of singleton elements of $\powerset S$.
Then $\struct {S, \circ}$ is isomorphic to $\struct {S', \circ_\PP}$.
== Proof ==
Let $\phi: S \to S'$ be the mapping defined as:
:$\forall x \in S: \map \phi x = \set x$
We have that:
demonstrating that $\phi$ is an injection.
Then we have:
demonstrating that $\phi$ is a surjection.
Hence by definition $\phi$ is a bijection.
By the definition of a magma, $S$ is closed under $\circ$.
That is:
:$\forall a, b \in S, a \circ b \in S$
Hence:
:$a \circ b \in \Dom \phi$
Also, by Power Structure of Magma is Magma, $S'$ is closed under $\circ_\PP$.
Hence:
:$\set a \circ_\PP \set b \in S'$
Now:
That is, $\phi$ is a homomorphism.
So $\phi$ is a bijective homomorphism.
Hence the result by definition of isomorphism.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 9$: Compositions Induced on the Set of All Subsets: Exercise $9.5$
Category:Power Structures
Category:Singletons
Category:Examples of Isomorphisms (Abstract Algebra)
Operation is Left Operation iff Anticommutative with Right Cancellable Element

== Theorem ==
Let $\struct {S, \circ}$ be an semigroup.
Then:
:$\circ$ is the left operation
:
:$\circ$ is anticommutative and has a right cancellable element.
== Proof ==
=== Sufficient Condition ===
Let $\circ$ be the left operation.
Then from Left Operation is Anticommutative we have that $\circ$ is anticommutative.
Let $x \in S$ be arbitrary.
Let $y, z \in S$ such that:
:$z \circ x = y \circ x$
Then:
That is, $x$ is a right cancellable element for all $x \in S$.
Thus:
:$\circ$ is anticommutative and has a right cancellable element.
=== Necessary Condition ===
Let $\circ$ be anticommutative and have a right cancellable element $z$.
As $\struct {S, \circ}$ is a semigroup it follows from that $\circ$ is associative.
Hence from Associative and Anticommutative:
:$\forall x, y, z \in S: x \circ y \circ z = x \circ z$
As $z$ is right cancellable:
:$\forall x, y \in S: x \circ y = x$
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.17$
Category:Left Operation
Category:Anticommutativity
Category:Cancellability
Operation is Right Operation iff Anticommutative with Left Cancellable Element

== Theorem ==
Let $\struct {S, \circ}$ be a semigroup.
Then:
:$\circ$ is the right operation
:
:$\circ$ is anticommutative and has a left cancellable element.
== Proof ==
=== Sufficient Condition ===
Let $\circ$ be the right operation.
Then from Right Operation is Anticommutative we have that $\circ$ is anticommutative.
Let $x \in S$ be arbitrary.
Let $y, z \in S$ such that:
:$x \circ z = x \circ y$
Then:
That is, $x$ is a left cancellable element for all $x \in S$.
Thus:
:$\circ$ is anticommutative and has a left cancellable element.
=== Necessary Condition ===
Let $\circ$ be anticommutative and have a left cancellable element $z$.
As $\struct {S, \circ}$ is a semigroup it follows from that $\circ$ is associative.
Hence from Associative and Anticommutative:
:$\forall x, y, z \in S: z \circ x \circ y = z \circ y$
As $z$ is left cancellable:
:$\forall x, y \in S: x \circ y = y$
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.17$
Category:Right Operation
Category:Anticommutativity
Category:Cancellability
Operation on Set for which Every Equivalence Relation is Congruence

== Theorem ==
Let $S$ be a set with at least $3$ elements.
Let $\circ$ be an operation on $S$ such that every equivalence relation on $S$ is a congruence relation for $\circ$.
Then $\circ$ is one of the following:
:the right operation $\to$
:the left operation $\gets$
:the constant operation $\sqbrk c$ for some $c \in S$.
== Proof ==
First we note that from:
:Equivalence Relation is Congruence for Constant Operation
:Equivalence Relation is Congruence for Left Operation
:Equivalence Relation is Congruence for Right Operation
every equivalence relation on $S$ is a congruence relation for the right operation, the left operation and the constant operation.
Let $\circ$ be an operation on $S$ which is not one of those three.
Then there exists $x_1, x_2, y_1, y_2 \in S$ such that:
:$x_1 \circ y_1 = z_1$
:$x_2 \circ y_2 = z_2$
where:
:$z_1 \ne z_2$
:$\map \lnot {z_1 = x_1 \land z_2 = x_2}$
:$\map \lnot {z_1 = y_1 \land z_2 = y_2}$
The above are possible only if there are more than $3$ elements in $S$.
Let $\RR$ be an equivalence relation on $S$ such that:
:$x_1 \mathrel \RR x_2$
:$y_1 \mathrel \RR y_2$
but that:
:$\map \lnot {z_1 \mathrel \RR z_2}$
Then $\RR$ is not a congruence relation.
The result follows from the Rule of Transposition.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Exercise $11.14$
Category:Congruence Relations
Operation on Structure of Cardinality 4+ where Every Permutation is Automorphism is Left or Right Operation

== Theorem ==
Let $S$ be a set whose cardinality is at least $4$.
Let $\struct {S, \circ}$ be an algebraic structure on $S$ such that every permutation on $S$ is an automorphism on $\struct {S, \circ}$.
Then $\circ$ is either the left operation or the right operation.
== Proof ==
From Structure of Cardinality 3+ where Every Permutation is Automorphism is Idempotent, we have that $\circ$ is idempotent:
:$\forall a \in S: a \circ a = a$
 $\circ$ is such that:
:$x \circ y = z$
for some distinct $x, y, z \in S$.
As $S$ has cardinality of at least $4$, there exists $w \in S$ such that $w \ne x$, $w \ne y$ and $w \ne z$.
Let $f$ be a permutation on $S$ such that:
:$\map f x = x$
:$\map f y = y$
:$\map f z = w$
Then:
This contradicts our assertion that $w$ and $z$ are distinct.
Hence we have shown that:
:$\forall x, y \in S: x \circ y = x \lor x \circ y = y$
 $\circ$ is neither the left operation nor the right operation.
From the above, that means there exist $w, x, y, z \in S$ such that:
Let $f$ be a permutation on $S$ such that:
Then we have:
This contradicts our assertion that $w$ and $z$ are distinct.
Hence $\circ$ is either the left operation or the right operation.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Exercise $6.11 \ \text {(d)}$
Category:Left and Right Operations
Operation over which Every Commutative Associative Operation is Distributive is either Left or Right Operation

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
Let $\circ$ be such that every operation on $S$ which is both commutative and associative is distributive over $\circ$.
Then $\circ$ is either the left operation $\gets$ or the right operation $\to$.
== Proof ==
Recall from:
:Every Operation is Distributive over Right Operation
and:
:Every Operation is Distributive over Left Operation
that if $\circ$ is either $\gets$ or $\to$, then every operation is distributive over it, whether commutative or associative.
=== Lemma ===
Hence $\odot$ must be distributive over $\circ$.
First note that we have:
But we have that $\odot$ is distributive over $\circ$.
That is:
:$\forall x, y, z \in S: x \odot \paren {y \circ z} = \paren {x \odot z} \circ \paren {y \odot z}$
and so as $a$ and $b$ are arbitrary:
:$\forall a, b \in S: \set {a \circ a, a \circ b, b \circ a, b \circ b} = \set {a, b}$
This of course also applies to:
:$y \circ z \in \set {y, z}$
and:
:$x \circ y \in \set {x, y}$
As $a$ and $b$ are arbitrary, the result applies for all $a, b \in S$.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.26 \ \text {(c)}$
Category:Left Operation
Category:Right Operation
Category:Commutativity
Category:Associativity
Category:Distributive Operations
Category:Operation over which Every Commutative Associative Operation is Distributive is either Left or Right Operation
Operation over which Every Commutative Associative Operation is Distributive is either Left or Right Operation/Lemma

== Theorem ==
Let $S$ be a set.
Let $A \subsetneqq S$ be a proper subset of $S$.
Let $a \in A$ and $b \in \relcomp S A$.
Let $\odot$ be the operation on $S$ defined as:
:$\forall x, y \in S: x \odot y = \begin {cases} a & : \set {x, y} \subseteq A \\ b & : \set {x, y} \not \subseteq A \end {cases}$
Then $\odot$ is commutative and associative.
== Proof ==
=== Commutativity ===
Trivially:
=== Associativity ===
There are $2$ cases to attend to:
;Case $1$: $\set {x, y, z} \subseteq A$
That is:
:$x \odot \paren {y \odot z} = \paren {x \odot y} \odot z = a$
;Case $2$: $\set {x, y, z} \not \subseteq A$
Then at least one of $\set {x, y, z}$ is in $\relcomp S A$.
, suppose $x \in \relcomp S A$.
Then:
:$\forall y \in S: \set {x, y} \not \subseteq A$
and so:
:$\forall y \in S: x \odot y = b$
Then we note that similarly:
:$\forall y \in S: \set {b, y} \not \subseteq A$
and so:
:$\forall y \in S: y \odot b = b$
Hence:
:$x \odot \paren {y \odot z} = \paren {x \odot y} \odot z = b$
Both cases have been examined, and we see that:
:$\forall x, y, z \in S: x \odot \paren {y \odot z} = \paren {x \odot y} \odot z$
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.26 \ \text {(b)}$
Category:Operation over which Every Commutative Associative Operation is Distributive is either Left or Right Operation
Operation which is Left Distributive over Every Commutative Associative Operation is Right Operation

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
Let $\circ$ have the property that:
:for every arbitrary operation $*$ on $S$ which is both commutative and associative, $\circ$ is left distributive over $*$.
Then $\circ$ is the right operation $\to$:
:$\forall a, b \in S: a \to b = b$
== Proof ==
First recall from Right Operation is Left Distributive over All Operations that the right operation is indeed left distributive over all operations, whether commutative or associative.
Let $*$ be an arbitrary operation on $S$ which is both commutative and associative.
As asserted, let $\circ$ be left distributive over $*$.
Let $c \in S$ be arbitrary.
Consider the constant operation $\sqbrk c$:
:$\forall a, b \in S: a \sqbrk c b := c$
We have from Constant Operation is Commutative and Constant Operation is Associative that $\sqbrk c$ is both commutative and associative.
Hence $\circ$ must be left distributive over $\sqbrk c$.
But then from Condition for Operation to be Left Distributive over Constant Operation:
:$x \circ c = c$
As $c$ is arbitrary:
:$\forall a, b \in S: a \circ b = b$
so $\circ$ has to be the right operation.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.25$
Category:Right Operation
Category:Distributive Operations
Category:Commutativity
Category:Associativity