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While trying to show that two completions of a Boolean algebra must be isomorphic following a hint in Jech, I stumbled upon this question. After reading it several times, I can't seem to understand the following step:

Now, if $b\leq_C c$ then $u\wedge b\leq_Cu\wedge c\leq_C c$. Therefore $u\leq_C c$.

From the first sentence, we infer that $c$ is an upper bound for the set $X=\{u\wedge b:b\in B , b\le_C c\}\subset B$; so it would be less than the supremum of $X$ in C. However, we know that $u$ equals the supremum of $X$ in D, so why can we say $u\le_C c$? The set over which we are taking the supremum lies in $B$, but it seems were mixing the two orders.

I also considered a proof by contradiction; if $u\not\le_C c$, then $u-c\neq 0$, and by density we can find $b_1\in B$ such that $0

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    I'm not sure that you understand the point of the proof from the linked question. Do you know how to prove that any two Dedekind completions of the rational numbers are isomorphic? – 2017-02-12
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    @AsafKaragila as I understand it, the $\iff$ shown in that question is used to show that the composition of the two morphisms is the identity. – 2017-02-12
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    Forget that question for a second. Do you know how to prove that any Dedekind completion of the rational numbers is isomorphic to the real numbers? – 2017-02-12
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    @AsafKaragila I don't have a proof at hand, but I think I read (probably from Spivak) it at some point, yes. – 2017-02-12
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    Try to think how you would go about proving something like that. If $D$ is another completion of the rationals, and $d\in D$, what would be the real number to which you'd map $d$? – 2017-02-12
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    @AsafKaragila as a cut, the set of rationals strictly below $d$ seems like a reasonable choice. – 2017-02-12
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    That is correct. And the same thing happens here. We have a Boolean algebra, to complete it we add the needed suprema. So the uniqueness follows simply by mapping the new suprema to the corresponding suprema in "the other completion". – 2017-02-12
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    @AsafKaragila By adding the needed suprema, you mean the Dedekind cut construction? So you'd be mapping the cut to its suprema in the other completion? – 2017-02-12
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    Yes, precisely. When we complete a Boolean algebra, we add missing 'cuts' (as much as you can call them cuts anyway). The uniqueness just follows from the construction. – 2017-02-12
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    @AsafKaragila Ok, I think I see what you're saying, although I'll think about it some more. Thank you very much for your help! – 2017-02-12

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