Recently I found the following problem.
Suppose $R$ is a partial order on $A$, $B\subseteq A$ and $b\in B$. Prove that if $b$ is the smallest element of $B$, then it is also the greatest lower bound of $B$.
But how that is possible?
I mean even if we have set $L$ of elements which are lower bounds of $B$, partial order does not guarantee that $L$ will have the greatest element $z$ such that $\forall x\in L(xRz)$.
Or maybe I miss something?
Consider the set $L$ of lower bounds of $B$. By the definition of a lower bound, for any $l\in L$ and any $b\in B$, we have $l\le b$. But that also implies that any element in $B$ is an upper bound of $L$. That especially is true of the minimum of $B$, if it exists.
So we have: If $b$ is the minimum of $B$, then $b\in L$ and $b$ is an upper bound of $L$. But that's exactly the definition of the maximum.
Thus if $B$ has a minimum, the set of lower bounds of $B$ has a maximum, and that maximum is the minimum of $B$.