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from the set $\{a, b, c, d\}$?

Of the one's I have tried, it at best is two of the three, but never all.

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Doesn't $\{(a,b)\}$ work? – Kevin Carlson Mar 8 '14 at 19:03
@KevinCarlson No, $\{(a,b)\}$ would be transitive. – dtldarek Mar 8 '14 at 19:04
Ah, yes, that's right. One needs a case in which $R(a,b)$ and $R(b,c)$. – Kevin Carlson Mar 8 '14 at 19:05
up vote 9 down vote accepted

Consider the ordered pair, such that $\{(x,y)\,|\,x,y\in\{a,b,c,d\}\}$. The following relation satisfies those conditions:

$$\{(a,b) ,(b,c), (c,d), (d,a)\}$$

Clearly, this relation is not reflexive since there is no ordered pair with same members i.e. $(x,x)$. This relation is anti-symmetric since for instance, there is no ordered pair $(b,a)$. This relation is also not transitive (which is left for you to work out).

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I think even $\{(a,b),(b,c)\}$ is enough. – DKal Mar 8 '14 at 19:08
I believe you took "anti-symmetric" to mean asymmetric, but this is also anti-symmetric trivially. – Hayden Mar 8 '14 at 19:08
Thanks for the note. ;) – NasuSama Mar 8 '14 at 19:09

If non-transitive means "never transitive for any triple", that is impossible for a complete tournament (irreflexive, never-symmetric relation) with $4$ or more players. The question sounds like an exercise of rediscovering that fact.

If partial tournaments are allowed then it can be done with any number of players, just partition them into 3 categories A,B,C, and have every A player always beat the B players who always beat C who always beat A, and no matches within any category.

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