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I have this set A = {-4,4} and the relation is R={(a,b)|$a^5=b^5$}. So this is partial order, I'm guessing Hasse diagram for this relation is just two vertices with no edge at all or I can't draw Hasse diagram if there's no edge?

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After looking at the problem again, you are probably right. For some reason, I'm under impression that a graph with vertices with self loop only is partial and equivalence . – Andy Apr 10 '13 at 14:39

I don't see a problem with having a Hasse diagram with no edges.

In any case, it would be a matter of definition as to whether or not we allow a Hasse diagram with no edges. If a definition is a problem, particularly in these technical border cases, we simply change the definition. In this case, in order to ensure that a Hasse diagram is defined for all partial orders, we need to define it for the partial order defined by equality.

You've probably never seen an example of one since (a) they're not particularly illustrative and (b) usually Hasse diagrams are drawn for lattices where the only Hasse diagrams without edges would have $0$ or $1$ vertices.

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