Talk:Euclidean relation

Need b and c be distinct?

Is there any requirement that b and c be distinct? The article doesn't mention it. If we do indeed dispense with that requirement, then "sibling of" is not a Euclidean relation. We don't consider anyone to be their own sibling. — Kevin Nelson (talk) 04:09, 30 January 2009 (UTC)[reply]

There is no such requirement. You're right, "sibling of" is not a Euclidean relation. — Emil J. 10:59, 30 January 2009 (UTC)[reply]

Equivalence?

I don't see equivalence here, only implication in one direction. It is true that if a relation is Euclidean and reflexive that implies it is symmetric, and if it is Euclidean and symmetric that implies it is transitive, but the opposite is not true. For example, the relation of divisibility in the set of natural numbers is both transitive and reflexive but is neither symmetric nor Euclidean. So it is only necessary and quite sufficient to say that "if a relation is symmetric, then it is Euclidean if and only if it is transitive". 91.148.102.143 (talk) 01:38, 24 April 2009 (UTC)[reply]

Could you clarify what are you talking about? The article does not say that a reflexive transitive relation is symmetric or Euclidean. — Emil J. 10:19, 24 April 2009 (UTC)[reply]

Modified form of transitivity

The articles states that the relation "satisfy a modified form of transitivity", but it's unclear to me what this means? Is it alluding simply to the syntactic similarities with transitivity or is there some meaning behind this? The sentence seems opaque nonetheless. I removed it and believe the article reads much clearer now. — Preceding unsigned comment added by 94.254.51.243 (talk) 10:38, 15 July 2017 (UTC)[reply]

Incomplete image?

If (for all a, b and c), (aRb and aRc) implies that bRc, then cRb should also hold. The same should then also apply to left Euclidean (i.e. bRc and also cRb). -- (1) I guess this somewhat hidden property is what is behind the post from April 2009. Maybe add an explicit hint in the "Properties" listing? -- (2) If I didn't misread the definitions, then the image to the right of the definitions seems incomplete as it only displays the following three arrows {(a,b), (a,c), (b,c)}. Shouldn't the image also contain a dashed (c,b) arrow? - Rehierl (talk) 08:54, 3 August 2018 (UTC)[reply]

I agree, and implemented your suggestion (1). As for (2), one could argue that the picture just reflects the bare definition, rather than its consequences. That is, it shows what is to be done in order to prove that some given relation is right Euclidean. Another reason is of course that changing the image would require some work... - Jochen Burghardt (talk) 12:39, 3 August 2018 (UTC)[reply]

I see what you mean. If only Wikipedia would directly support simple straight forward SVG diagrams. - Rehierl (talk) 16:11, 3 August 2018 (UTC)[reply]

Topic Closed - Counter examples for properties 5, 7 and 8

After the clarification made in replies, I realize counter examples below are invalid, topic closed. Thank you.

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The relations R and Q serve as counter examples to properties 5 and 7, and the relation S below as a counter example for property 8.

Given the set X = {x,y,z}, the relation R = {(x,y),(y,z)} is vacuously right and left Euclidean.

Given the set Y = {x,y,z,a,b}, the relation Q = {(x,y),(y,z),(a,b),(a,c),(b,c),(c,b)} is right Euclidean.

Property 5 says:

The range of a right Euclidean relation is always a subset of its domain.

Similarly, the domain of a left Euclidean relation is a subset of its range.

R's range isn't a subset of its domain since it includes z. Similarly, Q's range isn't a subset of its domain either.

Property 7 says:

A right Euclidean relation is always quasitransitive, as is a left Euclidean relation.

Neither R nor Q are quasi-transitive since they don't include (x,z).

Property 8 says:

A connected right Euclidean relation is always transitive; and so is a connected left Euclidean relation.

Given X = {x,y,z}, the relation S = {(x,y),(y,z),(z,x)} is connected and vacuously right and left Euclidean but it's not transitive.

In the references, these properties are justified by noting the equivalence relation on the equivalence closures but that means they don't hold for all Euclidean relations. Are these counter examples invalid? Dipidydip (talk) 10:24, 7 November 2024 (UTC)[reply]

R is not right Euclidean, since xRy and xRy, but not yRy. I guess (but didn't check) that your other counter-examples lack some "reflexive elements" in a similar way. - Jochen Burghardt (talk) 22:18, 8 November 2024 (UTC)[reply]

Thanks, it all makes sense now. All the counter examples I listed above are invalid because I overlooked that xRy gives yRy for any right Euclidean relation. Dipidydip (talk) 04:47, 9 November 2024 (UTC)[reply]