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Suppose $ \mathbb{P} := (P, {\leq_P}, 1_P) $ is a separative partial order. Let $ \mathbb{B} := \operatorname{RO}(\mathbb{P}) $ denote the Boolean completion.

Fix some dense embedding $ i \colon P \to B^+ $. Then $ i $ is an order-isomorphism (one-to-one, order-preserving) to its image.

Assume that $ \mathbb{P} $ is isomorphic to each of its cones $ \mathbb{P} \mathord{\upharpoonright} p := \{ q \in P : q \leq_P p \} $ (where $ p \in P $ is some condition).

Question. Is $ \mathbb{B} $ isomorphic to each of its cones? If not, is $ \mathbb{B} $ isomorphic to each cone of the form $ \mathbb{B} \mathord{\upharpoonright} i(p) $ (where $ p \in P $) at least?

Attempt to proof the second statement. Fix $ p \in P $. Let $ f \colon \mathbb{P} \to \mathbb{P} \mathord{\upharpoonright} p $ denote an isomorphism. Then $ i \circ f \circ i^{-1} \colon \mathbb{B} \to \mathbb{B} \mathord{\upharpoonright} i(p) $ is a canonical candidate for the isomorphism since $ i \mathord{\upharpoonright} (\mathbb{P} \mathord{\upharpoonright} p) $ should be a dense embedding of $ \mathbb{P} \mathord{\upharpoonright} p $ into $ \mathbb{B} \mathord{\upharpoonright} i(p) $, right?

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Your second statement is correct, simply because boolean completions are unique up to isomorphism.

For the stronger statement, let $b \in \mathbb{B}^+$. Let $\{ p_\alpha : \alpha < \kappa \}$ be a maximal antichain of elements of $i[\mathbb{P}]$ below $b$. Pick some maximal antichain $\{ q_\alpha : \alpha < \delta \}$ in $\mathbb{P}$ where $\delta \geq \kappa$. If $\delta > \kappa$, then using the fact that $\mathbb{P} \cong \mathbb{P} \restriction p_0$, enlarge the first antichain to one of size $\delta$; denote it $\{ p'_\alpha : \alpha < \delta \}$. So $\mathbb{B} \restriction b$ is isomorphic to the complete boolean sum of $\delta$ copies of $\mathbb{B}$, since $\mathbb{B} \cong \mathbb{B} \restriction p'_\alpha$ for each $\alpha$. But so is $\mathbb{B}$ itself.

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  • $\begingroup$ Thanks for you answer! I have a couple of questions: (1) For the first part, you use $ \operatorname{RO}(\mathbb{P} \restriction p) \cong \operatorname{RO}(\mathbb{P}) \restriction i(p) $, right? Is this clear? What is about my attempt (as stated in the question)? (2) There's a typo: It should be $ \mathbb{P} \cong \mathbb{P} \restriction i^{-1}(p_0) $, right? So we find that antichain $ \{ p'_\alpha : \alpha < \delta \} $ of size $ \delta $ in $ (\mathbb{B} \restriction b) \cap i[\mathbb{P}] $. That's ok for me. $\endgroup$
    – Justus87
    Jul 23, 2014 at 10:43
  • $\begingroup$ (3) Is $ \{ p'_\alpha : \alpha < \delta \} $ a maximal antichain in $ \mathbb{B} \restriction b $? (4) Now, it seems like you use a theorem (which is unknown to me) that states: Given a maximal antichain in a complete Boolean algebra, this complete Boolean algebra is isomorphic to the complete Boolean sum of all the cones below the members of that antichain. Can you give a reference or is this clear? $\endgroup$
    – Justus87
    Jul 23, 2014 at 10:44
  • $\begingroup$ (5) If I understand your argument correctly, you apply the mentioned theorem twice: (A) to $ \mathbb{B} \restriction b $ and $ \{ p'_\alpha : \alpha < \delta \} $; (B) to $ \mathbb{B} $ and $ \{ i(q_\alpha) : \alpha < \delta \} $. Is this correct? Thanks in advance! $\endgroup$
    – Justus87
    Jul 23, 2014 at 10:45
  • $\begingroup$ (1) Yes, $i : \mathbb{P} \restriction p \to RO(\mathbb{P}) \restriction i(p)$ is dense, so the isomorphism follows. (2) Sure but both are correct. (3) Yes. (4) Yes it's clear, define the isomorphism in the obvious way. (5) Sure I guess. You're being extremely formal! $\endgroup$
    – Monroe Eskew
    Jul 23, 2014 at 15:24

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