Discrete Mathematics - POSETs

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My task is to find out what is the lowest # of elements a partial ordered set can have with the following characteristics. If such a set exists I should show it and if it doesn't I must prove it.

1) has supremum of all its subsets, but there is a subset with no infimum
2) has two maximal and two minimaln elements
3) has two greatest elements
4) has one minimal but no least element

2) should be easy. We can just take Hasse diagram for divides relation of the set {${3,5}$}and we get two maximal and two minimal elements.

3) should be impossible since greatest/least element can only be one.

4) seems like it should be impossible (at least in fininte sets) as well even though I am not sure on this one and 1) I have no idea.

I am pretty new to this stuff and I am trying to understand the basics so I appreciate all the help I can get.

Thanks.

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Your solutions to $(2)$ and $(3)$ are correct.

$(1)$ is a bit tricky. You might think that the Hasse diagram below would do the trick, with $\{b,c\}$ as a subset with no infimum:

                      a  
                     / \  
                    b   c

However, the supremum of the empty set is the infimum of the whole partial order, so if every subset has a supremum, the partial order must have a least element, which the one shown clearly does not. In fact the situation described in $(1)$ is impossible. Suppose that $\langle P,\le\rangle$ is a partial order in which every subset has a supremum, and let $S\subseteq P$. Let $L$ be the set of all lower bounds for $S$. By hypothesis $L$ has a supremum $x$. If $s\in S$, then $s$ is an upper bound for $L$, so $x\le s$. Thus, $x$ is actually the greatest element of $L$ and as such is clearly the infimum of $S$.

You are correct that $(4)$ is impossible for a finite partial order. If $\langle P,\le\rangle$ is a finite partial order, and $x$ is minimal in $P$ but not the least element of $P$, let $Q=\{p\in P:p\not\le x\}$. Then $Q$ is finite and non-empty, so it has a minimal element, say $q$, and I’ll leave it to you to verify that $q$ is a minimal element of $P$ distinct from $x$.

It is possible, however, if $P$ is infinite, as you can see from the following Hasse diagram:

                 *  
                / \  
               *   x  
               |  
               *  
               |  
               *  
               |  
               ·  
               ·  
               ·

Here $x$ is minimal, but the partial order clearly has no least element.