Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
add comment

1 Answer

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.

share|improve this answer
    
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 8:08
    
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 8:17
    
I suggest you collect your thoughts and post a new question, giving definitions for the terms you use. –  Ittay Weiss Nov 19 '12 at 8:20
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.