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While it is true that $\mathcal P(\kappa)$ is a complete Boolean algebra, it is not necessarily true that $\mathcal P(\kappa)/I$ is complete for an ideal $I$. In particular if we consider $I=J_{bd}$ the ideal of bounded subsets of $\kappa$.

For example, Hausdorff showed that there is an $(\omega_1,\omega_1)$ gap in $\mathcal P(\omega)/\textrm{Fin}$. That is, there is a sequence of order type $\omega_1+\omega_1^*$ which is strictly increasing, but there is no $A$ which realizes the cut between the $\omega_1$ and $\omega_1^*$ parts. Consider now the lower $\omega_1$ sets, if there was a supremum to this collection, it would have to be an element realizing the gap.

Assuming $V=L$. How complete is $\mathcal P(\kappa)/J_{bd}$? If the answer is strictly less than $\operatorname{cf}(\kappa)$, what if we limit ourselves to strictly increasing sequences? Can we get $\operatorname{cf}(\kappa)$-completeness?

That is, assuming that $\gamma<\kappa^+$ and $\langle A_\alpha\subseteq\kappa\mid\alpha<\gamma\rangle$ is strictly increasing modulo $J_{bd}$. Is there a least upper bound to the sequence?

I'd be interested to know (at least some partial) answer outside $V=L$, although that is less important.

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    $\begingroup$ You don't need a gap to show that suprema don't exist in $\mathcal{P}(\omega)/\mathrm{Fin}$; take any sequence $A_n$ ($n < \omega$) of pairwise-disjoint, infinite subsets of $\omega$. If $A$ is a $\subseteq_*$-upper bound for this sequence, then one can find $B\subseteq A$ such that $A\setminus B$ is infinite and $B$ is still an upper bound, by removing one element from each intersection $A\cap A_n$ . Hence the sequence $A_n$ ($n < \omega$) has no least upper bound. $\endgroup$ – Paul McKenney Aug 29 '13 at 18:03
  • $\begingroup$ Thanks Paul, that's quite helpful. This seems to be easily generalized to large cardinalities. Moreover if we take $A'_n=A_1\cup\ldots\cup A_n$ then we get the same result for an increasing sequence, so both my questions have a negative answer. (Nice seeing you here, by the way!) $\endgroup$ – Asaf Karagila Aug 29 '13 at 18:42
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If $\kappa$ is regular, then $\mathcal{P}(\kappa)/J_{bd}$ is a $\kappa$-complete boolean algebra. If $\langle A_\alpha : \alpha < \delta < \kappa \rangle$ is a sequence of subsets of $\kappa$, then the union of these is a least upper bound. This uses the $\kappa$-completeness of $J_{bd}$. It is not $\kappa^+$-complete for the same reason Paul mentioned for $\omega$. Take any partition of $\kappa$ into $\kappa$ many unbounded sets, $\langle A_\alpha : \alpha < \kappa \rangle$, let $B$ be an upper bound, and let $C \subseteq B$ have one point from each $A_\alpha \cap B$ removed. So $C$ is an upper bound and strictly below $B$. For singular $\kappa$, the algebra is $cf(\kappa)$-complete but not $cf(\kappa)^+$-complete: Take any unbounded $A$ set of size $cf(\kappa)$ and consider the algebra below $A$. It is isomorphic to $\mathcal{P}(cf(\kappa))/J_{bd}$.

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  • $\begingroup$ So basically $\mathcal P(\kappa)/J$ inherits the completeness of $J$? $\endgroup$ – Asaf Karagila Sep 1 '13 at 12:30
  • $\begingroup$ Yes, but the completeness of the algebra can exceed that of the ideal. $\endgroup$ – Monroe Eskew Sep 1 '13 at 21:13
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This answer expands on Paul McKenney's comment, and concerns only the gap structure in $\mathcal{P}(\omega)$/fin. I don't know much about what happens for larger cardinals.

The fact that $\mathcal{P}(\omega)$/fin is highly incomplete actually has interesting consequences, and in particular is useful in showing that $\mathcal{P}(\omega)$/fin is universal for Boolean algebras of size at most $\aleph_1$. To give a better sense of the story here, suppose $\langle A_i: i < \omega \rangle$ is a strictly $\subseteq_*$-increasing sequence of subsets of $\omega$. Then if we iterate Paul's construction, we can find a strictly decreasing sequence $\langle B_j: j <\omega \rangle$ of upper bounds for $\langle A_i: i < \omega \rangle$. A natural question is whether the resulting pair of sequences $A_0 \subseteq_* A_1 \subseteq_* \ldots \subseteq_* B_1 \subseteq_* B_0$ is ever a gap. It turns out the answer is no: one can always find a $C$ such that for all $i, j$ we have $A_i \subseteq_* C \subseteq_* B_j$.

This says that there are no $(\omega, \omega)$-gaps in $\mathcal{P}(\omega)$/fin. Rothberger proved that the minimum cardinal $\kappa$ for which there exists an $(\omega, \kappa)$-gap is the bounding number $\mathfrak{b}$. If the CH holds we have $\mathfrak{b} = \omega_1$, and so in this case we can find $(\omega, \omega_1)$-gaps as well as the more celebrated $(\omega_1, \omega_1)$-gaps of Hausdorff.

Even without CH, we have that $\mathcal{P}(\omega)$/fin has the following four properties:

  • it is atomless (as a Boolean algebra)
  • it has no countable limit points
  • it has no countable cofinal sequences
  • it has no $(\omega, \omega)$-gaps

One can prove from these that $\mathcal{P}(\omega)$/fin embeds every Boolean algebra of size at most $\aleph_1$. In the case of CH, any other Boolean algebra of size $\aleph_1$ with these four properties is isomorphic to $\mathcal{P}(\omega)$/fin. The proof for both facts goes by extending countable partial embeddings one element at a time. If there existed countable limit points in $\mathcal{P}(\omega)$/fin, as you asked about in your question, such a proof would not be possible: such points would block certain partial embeddings from being extended.

An encyclopedic reference for these results and the set theory of $\mathcal{P}(\omega)$/fin more generally is the book Hausdorff Gaps and Limits by Frankiewicz and Zbierski.

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