1. Suppose we can define a relation $R$ over the sets $X_1, …, X_k$ for any natural number $k$, note not specified for a particular $k$. I was wondering if there is some definition or conditions concerning the following situation:

    For any natural number $k$, and any elements $\{ x_1 \in X_1, …, x_k \in X_k \}$, existence of the relation for any two of the elements and existence of the relation for these $k$ elements imply each other? In other words, existence of pairwise relation and existence of mutual relation are equivalent?

  2. For example,

    In probability theory, for a (finite, countably infinite, uncountably infinite) set of events, mutual independence implies pairwise independence, but pairwise independence does not imply mutual independence. I was wondering why? Specifically what kind of property does measure space lack to make the two equivalent?

Thanks and regards!


I think that with a second part you provide a counterexample for a first part of your question. I will say, that the case with independence (when pairwise do not imply mutual) is a "usual" (general) case while implication is a special case. I think it's natural to find properties which leads to the implication $pairwise\to mutual$ than vice-versa like you are trying.

A nice example also is a relation of intersection. If any two sets in the class intersects it doesn't mean that there exists a common intersection.

On the other hand for the relations $=$ and $\neq$ admit this implication, so transitivity is not necessary.

  • $\begingroup$ Thanks! In the last sentence, how does transitivity not necessary come from = and ≠? $\endgroup$ – Tim Apr 13 '11 at 16:56
  • $\begingroup$ $\neq$ is not transitive. But for this relation an implication does hold. $\endgroup$ – Ilya Apr 13 '11 at 19:56

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