# Maximal and Minimal Elements

In my textbook, the give an example for finding maximal and minimal elements on a set. The set is $(\{2,4,5,10,12,20,25\},|)$. To find the maximal and minimal elements of the set, the draw a Hasse diagram to find them, saying that the elements on the "top" of the diagram are the maxima, and the ones on the bottom are minima. From the diagram, the author tries to intimate that there can be more than one maximum and minimum. My question is, if we didn't draw this graph, how would you know there are more than one? The definition they provide for maximum and minimums seem to suggest that there can only be one of each.

Here is the author's discussion on this topic, "That is, $a$ is maximal in the poset $(S,\preceq)$ if there is no $b∈S$ such that $a≺b$. Similarly, an element of a poset is called minimal if it is not greater than any element of the poset. That is, $a$ is minimal if there is no element $b∈S$ such that $b≺a$.

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Beware -- it is not necessarily true that "maximum" and "maximal element" means the same here. Certainly "maximal element" and "largest element" means different things in a partial order, and I think I've seen "maximum" used to mean either of these two. –  Henning Makholm Nov 12 '12 at 14:48
One simple example: in the set $\{16,35\}$ ordered by divisibility, each element is both maximal and minimal -- because neither divides the other. –  Henning Makholm Nov 12 '12 at 14:50

Consider the collection of sets $\{\{1, 2, 3\}, \{4\}, \{5, 6\}, \{1, 2, 3, 5,6\}\}$. The minimal members of this collection are $\{1, 2, 3\}, \{4\}, \{5, 6\}$, i.e., These do not contain proper subsets which are members of this collection. The maximal members of this collection are {4}, {1, 2, 3, 5, 6}, i.e., these are not proper subsets of other sets which are members of this collection.
To clarify, the order you're using for your collection is $\subseteq$. And you can know without drawing the diagram by just thinking "which sets contain no proper subsets from the collection", etc. –  Mark S. Nov 13 '12 at 1:40