Zorn's lemma states

Suppose a partially ordered set $P$ has the property that every chain (i.e. totally ordered subset) has an upper bound in $P$. Then the set $P$ contains at least one maximal element.

Now, of course, the rationals $\mathbb{Q}$ satisfy the condition that every chain has an upper bound in $\mathbb{Q}$. But $\mathbb{Q}$ does not contain a maximal element, a contradiction.

Where is my line of reasoning going wrong?

  • $\begingroup$ $\mathbb Q$ is particularly badly chosen, because even bounded chains may not have an upper bound in $\mathbb Q$, like a chain of rationals converging towards $\sqrt 2$ from below. $\endgroup$ – Denis Sep 5 '13 at 14:22
  • $\begingroup$ dkuper: Bounded chains in $\mathbb Q$ do have an upper bound but may lack a least upper bound. $\endgroup$ – nonpop Sep 5 '13 at 14:35

What is an upper bound of the chain $\mathbb{Z}$ in $\mathbb{Q}$? (or $\mathbb{Q}$ itself, for that matter.)

  • $\begingroup$ Suppose $x, y \in \Bbb{Q}$.Take any rational y in 0 < x < 1 such that x < y. $\endgroup$ – Don Larynx Sep 5 '13 at 12:39
  • $\begingroup$ @Josie: I am having trouble making heads or tails of your comment above. But ask yourself this question: is there a rational number $q$ such that $n \leq q$ holds for every integer $n$? $\endgroup$ – user642796 Sep 5 '13 at 12:42
  • $\begingroup$ That wasn't readily obvious, thanks! $\endgroup$ – Don Larynx Sep 5 '13 at 13:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.