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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!

John Toh
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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.

Ittay Weiss
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  • Thanks! I've rephrased (fine-tuned) my question. I guess I was not clear about the term 'complete suborder'. – John Toh Nov 19 '12 at 08:08
  • 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.

     – John Toh Nov 19 '12 at 08:17
  • I suggest you collect your thoughts and post a new question, giving definitions for the terms you use. – Ittay Weiss Nov 19 '12 at 08:20