3
$\begingroup$

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.

  • 3
    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.2012-07-23
  • 0
    @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 ;-)2012-07-23
  • 0
    @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.2012-07-23
  • 10
    @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.2012-07-23
  • 1
    If you would like us to make sure we understand certain definitions, it would be nice to provide them.2012-07-23
  • 0
    @Théophile My apologies. I quoted the first part of the exercise verbatim from the book. I have now added the definitions as well for clarification.2012-07-23

1 Answers 1

3

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.