Euclidean relation

In mathematics, Euclidean relations are a class of binary relations that formalize "Axiom 1" in Euclid's Elements: "Magnitudes which are equal to the same are equal to each other."

Definition

Right Euclidean property: solid and dashed arrows indicate antecedents and consequents, respectively.

A binary relation R on a set X is Euclidean (sometimes called right Euclidean) if it satisfies the following: for every a, b, c in X, if a is related to b and c, then b is related to c.^[1] To write this in predicate logic:

\forall a,b,c\in X\,(a\,R\,b\land a\,R\,c\to b\,R\,c).

Dually, a relation R on X is left Euclidean if for every a, b, c in X, if b is related to a and c is related to a, then b is related to c:

\forall a,b,c\in X\,(b\,R\,a\land c\,R\,a\to b\,R\,c).

Properties

Due to the commutativity of ∧ in the definition's antecedent, aRb ∧ aRc even implies bRc ∧ cRb when R is right Euclidean. Similarly, bRa ∧ cRa implies bRc ∧ cRb when R is left Euclidean.
The property of being Euclidean is different from transitivity. For example, ≤ is transitive, but not right Euclidean,^[2] while xRy defined by 0 ≤ x ≤ y + 1 ≤ 2 is not transitive,^[3] but right Euclidean on natural numbers.
For symmetric relations, transitivity, right Euclideanness, and left Euclideanness all coincide. However, also a non-symmetric relation can be both transitive and right Euclidean, for example, xRy defined by y=0.
A relation that is both right Euclidean and reflexive is also symmetric and therefore an equivalence relation.^[1]^[4] Similarly, each left Euclidean and reflexive relation is an equivalence.
The range of a right Euclidean relation is always a subset^[5] of its domain. The restriction of a right Euclidean relation to its range is always reflexive,^[6] and therefore an equivalence. Similarly, the domain of a left Euclidean relation is a subset of its range, and the restriction of a left Euclidean relation to its domain is an equivalence.
A relation R is both left and right Euclidean, if, and only if, the domain and the range set of R agree, and R is an equivalence relation on that set.^[7]
A right Euclidean relation is always quasitransitive,^[8] and so is a left Euclidean relation.^[9]
A semi-connex right Euclidean relation is always transitive;^[10] and so is a semi-connex left Euclidean relation.^[11]
If X has at least 3 elements, a semi-connex right Euclidean relation R on X cannot be antisymmetric,^[12] and neither can a semi-connex left Euclidean relation on X.^[13] On the 2-element set X = { 0, 1 }, e.g. the relation xRy defined by y=1 is semi-connex, right Euclidean, and antisymmetric, and xRy defined by x=1 is semi-connex, left Euclidean, and antisymmetric.
A relation R on a set X is right Euclidean if, and only if, the restriction R’ := R|_ran(R) is an equivalence and for each x in X\ran(R), all elements to which x is related under R are equivalent under R’.^[14] Similarly, R on X is left Euclidean if, and only if, R’ := R|_dom(R) is an equivalence and for each x in X\dom(R), all elements that are related to x under R are equivalent under R’.
A left Euclidean relation is left-unique if, and only if, it is antisymmetric. Similarly, a right Euclidean relation is right unique if, and only if, it is anti-symmetric.
A left Euclidean and left unique relation is vacuously transitive, and so is a right Euclidean and right unique relation.
A left Euclidean relation is left quasi-reflexive. For left-unique relations, the converse also holds. Dually, each right Euclidean relation is right quasi-reflexive, and each right unique and right quasi-reflexive relation is right Euclidean.^[15]

References

^ ^a ^b Fagin, Ronald (2003), Reasoning About Knowledge, MIT Press, p. 60, ISBN 978-0-262-56200-3.
^ e.g. 0 ≤ 2 and 0 ≤ 1, but not 2 ≤ 1
^ e.g. 2R1 and 1R0, but not 2R0
^ xRy and xRx implies yRx.
^ Equality of domain and range isn't necessary: the relation xRy defined by y=min{x,2} is right Euclidean on the natural numbers, and its range, {0,1,2}, is a proper subset of its domain, ℕ.
^ If y is in the range of R, then xRy ∧ xRy implies yRy, for some suitable x. This also proves that y is in the domain of R.
^ The only if direction follows from the previous paragraph. — For the if direction, assume aRb and aRc, then a,b,c are members of the domain and range of R, hence bRc by symmetry and transitivity; left Euclideanness of R follows similarly.
^ If xRy ∧ ¬yRx ∧ yRz ∧ ¬zRy holds, then both y and z are in the range of R. Since R is an equivalence on that set, yRz implies zRy. Hence the antecedent of the quasi-transitivity definion formula cannot be satisfied.
^ A similar argument applies, observing that x,y are in the domain of R.
^ If xRy ∧ yRz holds, then y and z are in the range of R. Since R is semi-connex, xRz or zRx or x=z holds. In case 1, nothing remains to be shown. In cases 2 and 3, also x is in the range. Hence, xRz follows from the symmetry and reflexivity of R on its range, respectively.
^ Similar, using that x, y are in the domain of R.
^ Since R is semi-connex, at least two distinct elements x,y are in its range, and xRy ∨ yRx holds. Since R is symmetric on its range, even xRy ∧ yRx holds. This contradicts the antisymmetry property.
^ By a similar argument, using the domain of R.
^ Only if: R’ is an equivalence as shown above. If x∈X\ran(R) and xR’y₁ and xR’y₂, then y₁Ry₂ by right Euclideaness, hence y₁R’y₂. — If: if xRy ∧ xRz holds, then y,z∈ran(R). In case also x∈ran(R), even xR’y ∧ xR’z holds, hence yR’z by symmetry and transitivity of R’, hence yRz. In case x∈X\ran(R), the elements y and z must be equivalent under R’ by assumption, hence also yRz.
^ Jochen Burghardt (Nov 2018). Simple Laws about Nonprominent Properties of Binary Relations (Technical Report). arXiv:1806.05036v2. Lemma 44-46.

[fagin-1] Fagin, Ronald (2003), Reasoning About Knowledge, MIT Press, p. 60, ISBN 978-0-262-56200-3.

[2] .g. 0 ≤ 2 and 0 ≤ 1, but not 2 ≤ 1

[3] .g. 2R1 and 1R0, but not 2R0

[4] xRy and xRx implies yRx.

[5] Equality of domain and range isn't necessary: the relation xRy defined by y=min{x,2} is right Euclidean on the natural numbers, and its range, {0,1,2}, is a proper subset of its domain, ℕ.

[6] If y is in the range of R, then xRy ∧ xRy implies yRy, for some suitable x. This also proves that y is in the domain of R.

[7] The only if direction follows from the previous paragraph. — For the if direction, assume aRb and aRc, then a,b,c are members of the domain and range of R, hence bRc by symmetry and transitivity; left Euclideanness of R follows similarly.

[8] If xRy ∧ ¬yRx ∧ yRz ∧ ¬zRy holds, then both y and z are in the range of R. Since R is an equivalence on that set, yRz implies zRy. Hence the antecedent of the quasi-transitivity definion formula cannot be satisfied.

[9] A similar argument applies, observing that x,y are in the domain of R.

[10] If xRy ∧ yRz holds, then y and z are in the range of R. Since R is semi-connex, xRz or zRx or x=z holds. In case 1, nothing remains to be shown. In cases 2 and 3, also x is in the range. Hence, xRz follows from the symmetry and reflexivity of R on its range, respectively.

[11] Similar, using that x, y are in the domain of R.

[12] Since R is semi-connex, at least two distinct elements x,y are in its range, and xRy ∨ yRx holds. Since R is symmetric on its range, even xRy ∧ yRx holds. This contradicts the antisymmetry property.

[13] By a similar argument, using the domain of R.

[14] Only if: R’ is an equivalence as shown above. If x∈X\ran(R) and xR’y₁ and xR’y₂, then y₁Ry₂ by right Euclideaness, hence y₁R’y₂. — If: if xRy ∧ xRz holds, then y,z∈ran(R). In case also x∈ran(R), even xR’y ∧ xR’z holds, hence yR’z by symmetry and transitivity of R’, hence yRz. In case x∈X\ran(R), the elements y and z must be equivalent under R’ by assumption, hence also yRz.

[15] Jochen Burghardt (Nov 2018). Simple Laws about Nonprominent Properties of Binary Relations (Technical Report). arXiv:1806.05036v2. Lemma 44-46.

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