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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?

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

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  • $\begingroup$ 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

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