# Uniqueness of meets and joins in posets

Exercise 1.2.8 (Part 2), p.8, from Categories for Types by Roy L. Crole.

Definition: Let $X$ be a preordered set and $A \subseteq X$. A join of $A$, if such exists, is a least element in the set of upper bounds for $A$. A meet of $A$, if such exists, is a greatest element in the set of lower bounds for $A$.

Exercise: Suppose that $X$ is a poset (and thus also a preorder). Show that meets and joins in a poset are unique if they exist.

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I have posted my solution as an answer and would appreciate any feedback. Thanks. –  Code-Guru Jul 23 '12 at 17:41
Please choose titles that describe the problem, not it's location. –  Bill Dubuque Jul 23 '12 at 18:37
@BillDubuque Is that better? –  Code-Guru Jul 23 '12 at 18:39
Much better, thanks. –  Bill Dubuque Jul 23 '12 at 19:03
Perhaps you should have linked to your question with the first part of this exercise. (Definitions given there are relevant for this part, too.) –  Martin Sleziak Jul 24 '12 at 10:41

Let $A \subseteq X$ and suppose $x$ and $y$ are both joins of $A$. Then $x$ and $y$ are both upper bounds of $A$. So $A \le x$ and $A \le y$. But since $x$ and $y$ are joins of $A$ and $x, y \in X$, $y \le x$ and $x \le y$. Therefore, $x = y$ because because $\le$ is anti-symmetric. Meets are unique by a similar argument.
Your solution seems fine. Just a remark about notation - I am not sure whether $A\le x$ is frequently used as a shortcut for $(\forall a\in A) a\le x$. (But if it used this way in your course/book, then it is certainly ok for you to use this notation.) –  Martin Sleziak Jul 24 '12 at 10:44