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In Kunen's new Set-Theory book, the hint for exercise V.5.13 requires proving the following:

Given an atomless countably closed poset, there is an antichain $\{p_n | n\in \omega\}$ such that the boolean supremum $\bigvee_n p_n$ is not realized.

But if we consider $2^{<\omega_1}$ with $p\leq q$ iff $p$ end-extends $q$, then it is atomless and countably-closed (since the union of a countable chain is still there) and every subset has a supremum (there is a greatest common initial segment*). So, what am I missing? Is my example wrong? Am I misunderstanding the notion of "boolean supremum"? Or something else?

*Proof: If $A\subset 2^{<\omega_1}$, set for each distinct $p,q\in A$ - $\alpha(p,q)=\min\{\alpha|p(\alpha)\ne q(\alpha)\}$, take $\alpha^*=\min\{\alpha(p,q)|p,q\in A\}$, so for every $p,q\in A$: $p\upharpoonright\alpha^*=q\upharpoonright\alpha^*$, and there are $p,q\in A$ such that $p(\alpha^*)\ne q(\alpha^*)$. So taking any $p\in A$, $p\upharpoonright\alpha^*$ is the supremum of $A$.

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    I'm guessing that he means the supremum in the Boolean completion. Because a Boolean algebra (without $0$, that is) is never countably closed, at most it has a countably closed dense subset. So the idea is that the antichain adds a new element in the completion, as opposed to an element from the original forcing. – 2017-02-13
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    @AsafKaragila so you mean that although there is a supremum in the original poset, in the completion there will be a new (lower) supremum? – 2017-02-13
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    I don't know what I mean, I just guessed what Kunen might have meant. – 2017-02-13

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