# Hasse diagram for the relation $\{(a,a),(a,b),(a,c),(b,a),(b,b),(b,b),(b,c),(c,a),(c,b),(c,c)\}$ on set $M=\{a,b,c\}$

Hasse diagram for the the relation $$\{(a,a),(a,b),(a,c),(b,a),(b,b),(b,b),(b,c),(c,a),(c,b),(c,c)\}$$ on set $$M=\{a,b,c\}$$

I tried to make the Hasse-Diagram for the relation. But as every element is in relation to all other elements, there can be no edges between the elements in the Hasse-diagram. Is this right or am I wrong? Would the Hasse-diagram then simply be $$c, b, a$$ without edges?

• Hasse diagrams are ways of picturing Posets, partially ordered sets, a set and a relation where the relation is a partial order, that is a relation which is reflexive, transitive and antisymmetric. Your relation is an equivalence relation and not a partial order. It does not qualify for a Hasse diagram. You may draw a graph for it anyways, for instance as a $K_3$ with loops on every vertex, but that is not a Hasse diagram. – JMoravitz Oct 16 at 16:02
• "Why is it antisymmetric?" What "it" are you referring to here? Your relation? Your relation is symmetric. A partial order? Why is a partial order antisymmetric? Because that is a part of the definition of what it means to be a partial order in the first place... – JMoravitz Oct 16 at 16:11
• ... there are only a few partial orders on three elements. You have the ones where they all appear in a chain like $a<b<c$ and you have others where you have some but not all are comparable like $a<c, b<c$ but $a$ and $b$ are not comparable... and then you have where none are comparable at all. If you insist on talking about "removing some elements to make it a partial order"... you need to make it antisymmetric, so removing at least one of $(x,y)$ or $(y,x)$ for each $x\neq y$ in your set, and then ensure that it was still transitive afterwards. $a<b<c<a$ for instance doesn't work. – JMoravitz Oct 16 at 16:41
• for each $x\neq y$. No. That is still not a partial order... you have both $(a,c)$ and $(c,a)$. At least one of those needs to go away. You have $(b,c)$ and $(c,b)$, at least one of those needs to go away... – JMoravitz Oct 16 at 16:51
• That is one example of many, yes, and corresponds to $a<b<c$ whose Hasse diagram is just a vertex labeled $c$ at the top with a line straight down connecting it to a vertex labeled $b$ with a line straight down from that connecting it to a vertex labeled $a$. – JMoravitz Oct 16 at 16:56

The only Hasse diagrams I'm familiar with are the diagrams of an order $$<$$. In other words, the diagram is of an antisymmetric relation.