# Is there partial order over $\{1,2,3,4\}$ such that there're two maximal elements and one least element?

Let set $A=\{1,2,3,4\}$. Is there such a partial order relation over $A$ such that there're two maximal elements and one least element?

I think we could define a partial order as follows: $$xRy \iff x>y \quad \land \quad\text{x and y are prime}$$ Thus for $A$ we have $\{(3,1), (4,1)\}$. There're no other elements less than $1$ so it's the least element. And this way $3$ and $4$ are maximal elements.

Not sure if this is correct.

• I suggest you to study how to draw a Hasse diagram out of an order relation. Those diagrams make it easy to determine maximal/minimal, least/gratest elements, etc. Based upon a Hasse diagram the most natural choice would be $$\leq =\{(1, 1), (1, 2), (1, 4), (1, 3), (2, 2), (2, 4), (2, 3), (3, 3), (4, 4)\},$$ which provides you the $Y$ picture mentioned in the answer below. As to your example, there are no least element. You would not have lower bounds. – PtF Jun 26 '18 at 17:49

Y

(the partially ordered set whose poset diagram looks like the alphabet Y, where we assume the order is increasing from bottom to top. see below image for more details).