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I'm trying to show that if $(B, i)$ is the (BA) completion of any partial order $P$ and $A$ is a complete subalgebra of $B$, then $i^{-1}[A]$ is a complete suborder of $P$.

Pure hunch says it's true, but i'm stuck at whether a complete subalgebra $A$ of a complete boolean $B$ algebra always intersect all dense subsets of $B$.

Thanks in advance!

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It is not generally true that if $i:P\to B$ is the injection of the completion of $P$ and $A\subseteq B$ is a complete subalgebra of $B$ then $i^{-1}(A)$ is complete. As an example take $A=B$, which is clearly a subalgebra of $B$ but $i^{-1}(A)=P$ which clearly need not be complete.

The second assertion is also not generally true. Let $B$ be a complete boolean algebra with the property that $B$ with the top and bottom elements removed is dense. Let $A$ be the two-element set consisting of just the top and bottom elements in $B$. Clearly, $A$ is complete yet does not intersect $B-\{\top , \bot \}$, which is assumed dense.

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    Thanks! I've rephrased (fine-tuned) my question. I guess I was not clear about the term 'complete suborder'.2012-11-19
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    I think my reduction of the problem is too strong. What if I change the statement to: if _(B, i)_ is the completion of _P_ and _A_ is a complete subalgebra of _B_ such that _A^{+}_ is a complete suborder of _B^{+}_, then _i^{-1}[A]_ is a complete suborder of _P_. I am not sure of the nomenclature I used is conventionally correct. By complete suborder I mean, _A_ is a complete suborder of _B_ if the inclusion map from _A_ to _B_ is a complete embedding.2012-11-19
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    I suggest you collect your thoughts and post a new question, giving definitions for the terms you use.2012-11-19