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Exercise 1.2.8 (Part 1), 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: Make sure you understand the definition of meet and join in a preorder $X$. Think of some simple finite preordered sets in which meets and joins do not exist.

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You should know that whilst it's perfectly okay to ask and answer your own questions here, you should do it in a way which improves the overall quality of the site. It's not completely clear that posting and answering textbook exercises does that. By all means, carry on - but perhaps try to keep the frequency down. If you post a lot of questions, they take up room on the front page, and other questions will get pushed down. – Chris Taylor Jul 23 '12 at 17:34
@ChrisTaylor Thanks for the feedback. I want to have a record of my work. So far I have only worked out one and a half exercises from this text, so it shouldn't overwhelm the site ;-) – Code-Guru Jul 23 '12 at 17:38
@ChrisTaylor Also, I would appreciate feed back on my work as well as, in this case, other examples that can help clarify these definitions for me. I probably should have added this as a comment as soon as I posted my question and answer. – Code-Guru Jul 23 '12 at 17:40
@Code-Guru That purpose of wanting a record of your work is not in agreement with the purpose of this site. This is not a note-taking site. – Thomas Andrews Jul 23 '12 at 17:42
If you would like us to make sure we understand certain definitions, it would be nice to provide them. – Théophile Jul 23 '12 at 17:45
up vote 3 down vote accepted

Let $X = {a, b}$. Define a preorder on $X$ as $a \le a$ and $b \le b$. Now suppose that $\vee X$ is a join of $X$. Then $\vee X$ is an upperbound of $X$. So $a \le \vee X$ and $b \le \vee X$. So $\vee X = a$ and $\vee X = b$, which is a contradiction. Therefor, $\vee X$ does not exist. A similar proof will show that $X$ does not have a meet, either.

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