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

Question

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.

Answer 1

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).