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If we have some poset $\langle$$\lbrace$1, 2, 3, 4, 6, 24, 36, 72$\rbrace$, |$\rangle$, with $|$ being the divides relation, how should I go about finding the least upper bound and greatest lower bound of the subset $\lbrace$4, 6$\rbrace$?

Since the subset $\lbrace$4, 6$\rbrace$ has only two elements, would the greatest lower bound be 4 and the least upper bound be 6?

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up vote 3 down vote accepted

$4$ is not a lower bound of $\{4,6\}$, because $4\nmid 6$. The lower bounds of $\{4,6\}$ are by definition the numbers in your set that divide both $4$ and $6$; those numbers are $1$ and $2$, and $1\mid 2$, so the $2$ is the greatest lower bound of $\{4,6\}$ in this order.

Similarly, $6$ is not an upper of $\{4,6\}$ at all, again because $4\nmid 6$. The upper bounds of $\{4,6\}$ are by definition the members of the set that are multiples of both $4$ and $6$, i.e., $24,36$, and $72$. Now, however, you have to be careful. $24$ is numerically the smallest of these upper bounds, but it is not the least upper bound of $\{4,6\}$ in this partial order.

Recall that if $\langle P,\le\rangle$ is a partial order, and $A\subseteq P$, then an element $u\in P$ is the least upper bound of $A$ if and only if (1) $u$ is an upper bound of $A$, meaning that $a\le u$ for all $a\in A$, and (2) if $v$ is any upper bound of $A$, then $u\le v$.

In your partial order $24$ meets the first criterion with respect to the set $\{4,6\}$: it is an upper bound of that set. It does not, however, meet the second criterion, because $36$ is also an upper bound of $\{4,6\}$, and $24\nmid 36$. In fact the set $\{4,6\}$ has no least upper bound in this partial order.

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