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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.

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    $\begingroup$ 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. $\endgroup$ – PtF Jun 26 '18 at 17:49
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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).

enter image description here

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