Prikry forcing is a well-known example of a forcing which adds an $\omega$-sequence to some cardinal $\kappa$, but does not add bounded subsets to $\kappa$.

There are other examples of forcings which add $\omega$-sequences, but do not add reals (for example Namba forcing in the presence of $\sf CH$).

In these situations the forcing is not $\sigma$-distributive, namely there is a countable sequence of dense open sets whose intersection is empty; but the added $\omega$-sequences live relatively "far up stairs" in terms of the von Neumann hierarchy.

Question. Is there a well-known characterizations of "$\Bbb P$ does not add new sets of rank $\alpha$" in terms of dense open sets and the intersection thereof?

  • $\begingroup$ I guess that using properties of the Boolean completion of the forcing poset is not useful for this question, right? $\endgroup$ – Yair Hayut Nov 8 '18 at 13:55
  • $\begingroup$ What do you mean? $\endgroup$ – Asaf Karagila Nov 8 '18 at 14:37
  • $\begingroup$ Are you sure you mean "rank"? Because Namba adds new set of rank $\omega+8$. $\endgroup$ – Not Mike Dec 9 '18 at 8:59
  • $\begingroup$ @NotMike: Yes, I'm pretty sure I mean rank. Also, since reals have rank $\omega$, I don't see how $\omega+8$ contradicts my statement. $\endgroup$ – Asaf Karagila Dec 9 '18 at 9:02
  • 1
    $\begingroup$ @NotMike: For crying out loud, don't use $\Bbb N$ to denote stuff. It's enough we have $\Bbb Q$ and $\Bbb R$ outside their standard mathematical notation (and even $\Bbb C$ sometimes). But $\Bbb N$ is a step too far, mate! $\endgroup$ – Asaf Karagila Dec 9 '18 at 17:52

This is just a comment on the question (but too long for a comment box), not an answer.

Let $\kappa$ and $\lambda$ be cardinals. A complete Boolean algebra $\mathbb B$ is called $(\kappa,\lambda)$-distributive if $$\prod_{\alpha < \kappa} \ \sum_{\beta < \lambda} u_{\alpha,\,\beta} \,=\, \sum_{f: \kappa \rightarrow \lambda} \ \prod_{\alpha < \kappa} u_{\alpha,\,f(\alpha)}$$ for any $u_{\alpha,\,\beta}$ in $\mathbb B$. In Jech's set theory book (third edition, Theorem 15.38, p. 246) it is proved that

$\mathbb B$ is $(\kappa,\lambda)$-distributive if and only if every $f: \kappa \rightarrow \lambda$ in the generic extension by $\mathbb B$ is already in the ground model.

So for example, $\mathbb B$ does not add new reals if and only if it is $(\aleph_0,2)$-distributive, and (the Boolean completion of) the Prikry forcing is a good example of an algebra that is $(\aleph_0,2)$-distributive but not $\aleph_0$-distributive. (It adds new $\omega$-sequences, but not new reals.)

Using the highlighted theorem above, we can deduce that

$\mathbb B$ adds a set of rank $\leq\!\alpha$ if and only if it fails to be $(|V_\alpha|,2)$-distributive.

The reason is that a set of rank $\leq\!\alpha$ is just a subset of $V_\alpha$, and can be identified with a characteristic function $V_\alpha \rightarrow 2$. So getting new sets of rank $\leq\!\alpha$ is equivalent to getting new functions of this kind.

Of course, I'm not sure this directly answers your question (because $(\kappa,\lambda)$-distributivity is not defined in terms of dense open sets). But here is something a little closer: $\mathbb B$ is $(\kappa,\lambda)$-distributive if and only if every collection of $\kappa$ partitions of $\mathbb B$, each containing at most $\lambda$ members of $\mathbb B$, has a common refinement. (This is Lemma 15.37 from Jech.)

  • 1
    $\begingroup$ Right. I am aware of these things that you mention. But they are not given in terms of open sets, and ultimately I want to apply them (also) in ZF, where chain conditions are meaningless (so the common refinement argument is somewhat irrelevant, or at least much harder to appeal to). $\endgroup$ – Asaf Karagila Nov 8 '18 at 14:37
  • 1
    $\begingroup$ OK, that's fair enough -- I'll leave this here in case anyone finds it helpful, but you're right that it's not really an answer. $\endgroup$ – Will Brian Nov 8 '18 at 14:38
  • 3
    $\begingroup$ I wonder if you can get a version that makes sense with $\neg\text{AC}$ by using pre-dense sets, instead of antichains. Something like: the intersection of $\kappa$ many dense sets, each having a pre-dense subset of size $\lambda$, is dense. $\endgroup$ – Joel David Hamkins Jan 10 '19 at 14:11

This is just a translation of @WillBrian's answer into a statement about dense sets.

For each $\delta \in \mathsf{On}$, fix $\dot{R}_\delta \in V^{\mathbb{P}}$ of minimal rank, such that $1 \Vdash_\mathbb{P} \dot{R}_\delta = \{ x : \mathsf{rank}(x) < \check{\delta}\}$. Then, because of how the rank function works, we always have that the least $\delta \in \mathsf{On}$, such that $1 \Vdash_{\mathbb{P}} \dot{R}_\delta \neq \check{V}_\delta$ is a successor; so the thing to do is characterize when $1 \Vdash_{\mathbb{P}} \dot{R}_{\gamma + 1} = \check{V}_{\gamma + 1}$. To this end,

Theorem: For every $\delta \in \mathsf{On}$ such that $1 \Vdash_{\mathbb{P}} \dot{R}_\delta = \check{V}_\delta$, the following are equivalent,

  1. $1 \Vdash_{\mathbb{P}} \dot{R}_{\delta+1} = \check{V}_{\delta+1}$,

  2. For every $p \in \mathbb{P}$ and every $\{ \mathcal{B}_x: x \in V_{\delta} \} \subset \mathcal{P}(\mathbb{P})$, there is some $p_0 \le p$, such that, for any $x \in V_{\delta}$, either $$\{q_0 \in \mathbb{P}: (\exists s \in \mathcal{B}_x)( q_0 \le s)\}\text{, or }\{q_0 \in \mathbb{P}: (\forall s \in \mathcal{B}_x)( q_0 \perp s)\},$$ is dense below $p_0$.

Proof: Fix $\delta \in \mathsf{On}$, and assume that $1 \Vdash_{\mathbb{P}} \dot{R}_{\delta+1} = \check{V}_{\delta+1}$. Then, for every $p\in \mathbb{P}$ and $\dot{g} \in V^{\mathbb{P}}$ such that $p \Vdash_{\mathbb{P}} \dot{g}:\check{V}_\delta \rightarrow \check{2}$, the set $\{ q \le p : (\exists f: V_\delta\rightarrow 2)(q \Vdash_{\mathbb{P}} \dot{g} = \check{f} ) \}$ is dense. So let $\{ \mathcal{B}_x : x \in V_\delta\} \subset \mathcal{P}(\mathbb{P})$ be arbitrary, and define $\dot{g},\sigma_x, \tau_x \in V^{\mathbb{P}}$ ($x\in V_\delta$) as follows $\dot{g} = \{ \langle\sigma_x, 1\rangle: x\in V_\delta \}$, $\sigma_x = \mathsf{pair}_\mathbb{P}(\check{x}, \tau_x)$, and $\tau_x = \{ \langle \emptyset, p \rangle: p \in \mathcal{B}_x \}$. Then, $1 \Vdash_{\mathbb{P}} \dot{g}: \check{V}_\delta \rightarrow \check{2}$ and so by assumption the set $\{ q \in \mathbb{P}: (\exists f: V_\delta\rightarrow 2)(q \Vdash_{\mathbb{P}} \dot{g} = \check{f} )\}$ is dense open, hence given $p \in \mathbb{P}$ we can find $p_0 \le p$ and $f:V_\delta \rightarrow 2$ such that, for every $x \in V_\delta$, $p_0 \Vdash \dot{g}(\check{x}) = \check{f}(\check{x})$ or equivalently, $ p_0 \Vdash_{\mathbb{P}} "\tau_x = \check{f}(\check{x})" \iff$ $f(x) = 1$ and $\{q_0 \in \mathbb{P}: (\exists s \in \mathcal{B}_x)( q_0 \le s)\}$ is dense below $p_0$, or $f(x)=0$ and $\{q_0 \in \mathbb{P}: (\forall s \in \mathcal{B}_x)( q_0 \perp s)\}$ is dense below $p_0$.

For the converse, assume (2), then fix $p\in\mathbb{P}$ and $\dot{a}\in V^{\mathbb{P}}$ such that $p \Vdash_\mathbb{P} \dot{a} \in \dot{R}_{\delta+1}$. Then, for $x\in V_\delta$ defining $\mathcal{B}_x = \{ q \in \mathbb{P}: q\Vdash_{\mathbb{P}} \check{x} \in \dot{a} \}$, we can find $p_0 \le p$ (by 2) such that for every $x$, either $$\{q_0 \in \mathbb{P}: (\exists s \in \mathcal{B}_x)( q_0 \le s)\}\text{, or }\{q_0 \in \mathbb{P}: (\forall s \in \mathcal{B}_x)( q_0 \perp s)\},$$ is dense below $p_0$. As such, letting $$A = \{ x \in V_\delta : \{q_0 \in \mathbb{P}: (\exists s \in \mathcal{B}_x)( q_0 \le s)\} \text{ is dense below } p_0 \}$$ we have $p_0 \Vdash \dot{a} = \check{A} \in \check{V}_{\delta+1}$. $\square$

Remark: If in (2), the requirement "either $A^x_1=\{q_0 \in \mathbb{P}: (\exists s \in \mathcal{B}_x)( q_0 \le s)\}$, or $A^x_0=\{q_0 \in \mathbb{P}: (\forall s \in \mathcal{B}_x)( q_0 \perp s)\}$ is dense below $p_0$", is changed to "$p_0$ is an element of $A^x_0$ or $A^x_1$", then the new statement is equivalent to $\mathbb{P}$ being $\vert V_\delta \vert$-baire.


The distributivity properties such as $(\kappa,\lambda)$-distributivity that characterizes when a forcing extension does not add any new functions from $\kappa$ to $\lambda$ are actually instances of the property “the intersection of at most $\kappa$ many open sets is open ” and to a lesser extent the $\kappa$-Baire property when we consider the complete Boolean algebra as a topological space whose points are the generic ultrafilters. Since complete Boolean algebras are frames and not topological spaces, and since the points in complete Boolean algebras only exist in forcing extensions, we shall formulate this answer in terms of point-free topology. For a background in point-free topology, please see the textbook Frames and Locales-Topology without points by Jorge Picado and Ales Pultr. And be sure to be aware of the result in Will Brian's answer before reading this one.

To see how distributivity relates to the intersection of open sets, let us look at some point-free topology.

For this post, we shall always assume that $\kappa$ is an uncountabe regular cardinal. A regular space is said to be a $P$-space if the intersection of countably many open sets is open. It is not too hard to show that every regular $P$-space is zero-dimensional. More generally, if $\kappa$ is a regular cardinal, then a regular space $X$ is a $P_{<\kappa}$-space if whenever $U_{i}$ is open for $i\in I$ and $|I|<\kappa$, the intersection $\bigcap_{i\in I}U_{i}$ is open. We say that a space $X$ is a $<\lambda$-Baire space if the intersection of less than $\lambda$ many dense open sets is still dense.

For example, if $X$ is any $P_{<\kappa}$-space and $C_{n}$ is a subspace of $X$ for $n\in\omega$ where each $C_{n}$ is nowhere dense in $C_{n+1}$, then the union $\bigcup_{n\in\omega}C_{n}$ is a $P_{<\kappa}$-space which is not a Baire space. Therefore, the notions of being a $<\kappa$-Baire space is independent of the notion of being a $P_{<\kappa}$-space. See this answer for more details about the relations between the $<\kappa$-Baire spaces and the $P_{<\kappa}$-spaces.

There are several different ways to generalize the notion of a $P_{<\kappa}$-space to point-free topology, and some of these generalizations are simply versions of distributivity.

A frame is a complete lattice that satisfies the infinite distributivity law $$x\wedge\bigvee_{i\in I}y_{i}= \bigvee_{i\in I}(x\wedge y_{i}).$$ Frames are the objects that people study in the field of point-free topology; if $(X,\mathcal{T})$ is a topological space, then $\mathcal{T}$ is a frame.

We say that a frame $L$ is $(<\kappa,\lambda)$-distributive if whenever $J$ is a set with $|J|<\kappa$ and $C_{j}\subseteq L,|C_{j}|\leq\lambda$ for each $j\in J$, then $$\bigwedge_{j\in J}\bigvee C_{j}=\bigvee\{\bigwedge_{j\in J}x_{j}\mid x_{j}\in C_{j}\,\text{for}\,j\in J\}.$$ We say $L$ is $<\kappa$-distributive if it is $(<\kappa,\lambda)$-distributive for all $\lambda$.

We say that a frame $L$ is weakly $<\kappa$-distributive if whenever $|I|<\kappa$, we have $$x\vee\bigwedge_{i\in I}y_{i}=\bigwedge_{i\in I}(x\vee y_{i}).$$

Every complete Boolean algebra is weakly $<\kappa$-distributive for each regular cardinal $\kappa$.

Theorem: Let $(X,\mathcal{T})$ be a regular space. Then the following are equivalent:

  1. $(X,\mathcal{T})$ is a $P_{<\kappa}$-space.

  2. $\mathcal{T}$ is $<\kappa$-distributive.

  3. $\mathcal{T}$ is weakly $<\kappa$-distributive.

  4. $\mathcal{T}$ is $(<\kappa,2)$-distributive.

  5. $\mathcal{T}$ is $(<\kappa,\lambda)$-distributive.

Observation: Suppose that $L$ is a regular frame and $\lambda_{2}\leq \lambda_{1}$. Each of the following conditions implies the next one.

  1. $L$ is $<\kappa$-distributive.

  2. $L$ is $(<\kappa,\lambda_{1})$-distributive.

  3. $L$ is $(<\kappa,\lambda_{2})$-distributive.

  4. $L$ is $(<\kappa,2)$-distributive.

  5. $L$ is weakly $<\kappa$-distributive.

Theorem: A regular frame $L$ is weakly $<\kappa$-distributive if and only if whenever $|I|<\kappa$ and $C_{i}$ is a closed sublocale of $L$ for $i\in I$, the least upper bound $\bigvee_{i\in I}C_{i}$ in the lattice of sublocales of $L$ is closed.

Weak $<\kappa$-distributivity is too weak of a generalization of the notion of a $P_{<\kappa}$-space to distinguish between complete Boolean algebras.

Theorem: A regular frame $L$ is a complete Boolean algebra if and only if it is weakly $<\kappa$-distributive for all cardinals $\kappa.$

By the above results and observations, we realize that the notions of $(<\kappa,2),(<\kappa,\lambda),<\kappa$-distributivity are actually a generalizations of the notion of a $P_{<\kappa}$-space to point-free topology.

The notion of a $P_{<\kappa}$-space is quite unusual; most well-known notions of general topology extend without a problem to point-free topology to only one point-free notion. However, there are many ways of generalizing the notion of a $P_{<\kappa}$-space to point-free topology, and several of these notions (such as that of a $P$-frame and a $P_{0}$-frame) do not even refer to distributivity at all. Furthermore, all of these non-equivalent generalizations of the notion of a $P_{<\kappa}$-space all look like satisfactory notions. However, with everything being said, these generalizations of the notion of a $P_{<\kappa}$-space have barely been studied in the context of point-free topology, so much research still needs to be done in this area.

Baire and dense open elements

Every frame $L$ has a smallest dense sublocale, namely $B_{L}=\{x|x^{**}=x\}$. The set $B_{L}$ is a complete Boolean algebra and $B_{L}$ is closed under arbitrary meets. If $L$ is a frame, then let $\mathfrak{B}(L)$ denote the Boolean subalgebra of complemented elements in $L.$

If $\kappa$ is a regular cardinal and $L$ is a $<\kappa$-distributive frame, then we say that $L$ is a $<\kappa$-Baire frame if whenever $|I|<\kappa$ and $x_{i}$ is dense in $L$ for $i\in I$, then $\bigwedge_{i\in I}x_{i}$ is dense in $L$.

Theorem: Suppose that $L$ is a regular frame and $\kappa$ is an uncountable regular cardinal.

  1. If $B_{L}$ is $<\kappa$-distributive, then whenever $|I|<\kappa$ and $x_{i}$ is dense in $L$ for $i\in I$, then $\bigwedge_{i\in I}x_{i}$ is also dense in $L$.

  2. Suppose $L$ is a $<\kappa$-distributive regular frame. Then $L$ is a $<\kappa$-Baire frame if and only if $B_{L}$ is $<\kappa$-distributive.

Proof: 1. Suppose that $B_{L}$ is $<\kappa$-distributive, $|I|<\kappa$, and $x_{i}$ is dense in $L$ for each $i\in I$. Then for each $i\in I$, there is some partition $p_{i}$ of $B_{L}$ where $\bigvee^{L}p_{i}\leq x_{i}$. Since $B_{L}$ is $<\kappa$, distributive, there is a partition $p$ of $B_{L}$ that refines each $p_{i}$. However, if $a\in p$ and $i\in I$, there there is some $b\in p_{i}$ where $a\leq b\leq x_{i}$. Therefore, $a\leq\bigwedge_{i\in I}x_{i}$. Thus, $\bigvee^{L}p\leq\bigwedge_{i\in I}x_{i}$. Since $p$ is a partition of $B_{L}$, $\bigvee^{L}p$ is dense in $L$, so $\bigwedge_{i\in I}x_{i}$ is also dense in $L$.

  1. Suppose that $L$ is $<\kappa$-Baire and $<\kappa$-distributive. Let $|I|<\kappa$ and let $p_{i}$ be a partition of $B_{L}$ for each $i\in I$. Let $x_{i}=\bigvee^{L}p_{i}$ for each $i\in I$. Then $\bigwedge_{i\in I}x_{i}$ is dense. Now, $$\bigwedge_{i\in I}x_{i}=\bigwedge_{i\in I}\bigvee^{L}p_{i}=\bigvee^{L}\{\bigwedge_{i\in I}r_{i}|r_{i}\in p_{i}\,\text{for}\,i\in I\}$$. Since $$\bigvee^{L}\{\bigwedge_{i\in I}r_{i}|r_{i}\in p_{i}\,\text{for}\,i\in I\}$$ is dense in $L$, we conclude that $$\{\bigwedge_{i\in I}r_{i}|r_{i}\in p_{i}\,\text{for}\,i\in I\}\setminus\{0\}$$ is a partition of $L$ that refines each $p_{i}$. Therefore, $B_{L}$ is $<\kappa$-distributive. Q.E.D.

We say that $x\in L$ is regularly dense open if there are $r,s\in B_{L}$ where $r\wedge s=0,r\vee^{B_{L}}s=1$ and where $r\vee^{L}s\leq x$.

Theorem: Suppose that $L$ is a frame.

  1. If $B_{L}$ is $(<\kappa,2)$-distributive, then whenever $x_{i}$ is regularly dense open for $i\in I$ and $|I|<\kappa$, then $\bigwedge_{i\in I}x_{i}$ is dense.

  2. If $L$ is $(<\kappa,2)$-distributive, then $B_{L}$ is $(<\kappa,2)$-distributive if and only if whenever $x_{i}$ is regularly dense open for $i\in I$ and $|I|<\kappa$, then $\bigwedge_{i\in I}x_{i}$ is dense.

Proof: 1. Suppose that $B_{L}$ is $(<\kappa,2)$-distributive and assume that $x_{i}$ is strongly dense for $i\in I$ and $|I|<\kappa$. Then for $i\in I$, there are $r_{i,0},r_{i,1}\in B_{L}$ which are complements in $B_{L}$ but where $r_{i,0}\vee^{L}r_{i,1}\leq x_{i}$. By the $(<\kappa,2)$-distributivity of $B_{L}$, we have $$1= \bigvee^{B_{L}}\{\bigwedge_{i\in I}r_{i,a(i)}|a:I\rightarrow 2\}$$ and therefore $$\bigvee^{L}\{\bigwedge_{i\in I}r_{i,a(i)}|a:I\rightarrow 2\}$$ is dense in $L$. However, $$\bigvee^{L}\{\bigwedge_{i\in I}r_{i,a(i)}|a:I\rightarrow 2\}\leq \bigwedge_{i\in I}x_{i},$$ so $\bigwedge_{i\in I}x_{i}$ is dense in $L$.

  1. We have already proven the direction $\rightarrow$, so let's embark on a proof of the direction $\leftarrow$. Suppose that whenever $|I|<\kappa$ and $x_{i}$ is regularly dense open for $i\in I$, then $\bigwedge_{i\in I}x_{i}$ is regularly dense open. Now, let $r_{i,0},r_{i,1}$ be complements in $B_{L}$ for $i\in I$. Then let $x_{i}=r_{i,0}\vee^{L}r_{i,1}$. Then $ \bigwedge_{i\in I}x_{i}$ is dense. Therefore, by the $(<\kappa,2)$-distributivity of $L$, we have $$\bigwedge_{i\in I}x_{i}=\bigwedge_{i\in I}(r_{i,0}\vee^{L}r_{i,1})=\bigvee^{L}\{\bigwedge_{i\in I}R_{i,a(i)}|a:I\rightarrow 2\},$$ so since $$\bigvee^{L}\{\bigwedge_{i\in I}r_{i,a(i)}|a:I\rightarrow 2\}$$ is dense in $L$, we conclude that $$\bigvee^{B_{L}}\{\bigwedge_{i\in I}r_{i,a(i)}|a:I\rightarrow 2\}=1.$$ Therefore, $B_{L}$ is $(<\kappa,2)$-distributive. Q.E.D.

I will leave it an an exercise to generalize the above result to $(<\kappa,\lambda)$-distributivity.

Corollary: Suppose that $L$ is an extremally disconnected frame. If $L$ is $(<\kappa,2)$-distributive, then $B_{L}$ is also $(<\kappa,2)$-distributive.

Proof: Observe that each element in $B_{L}$ is complemented in $L$ and if $r,s\in B_{L}$ are complements, then $r\vee^{L}s=1$. Therefore, the only regularly dense open element in $L$ is $1$. Therefore, by the above result, if $R$ is a collection of regularly dense open elements in $L$, then $\bigwedge R\in L$ as well. Q.E.D.

Example: Suppose that $\mu$ is a measurable cardinal and $\mathcal{U}$ is a normal ultrafilter on $\mu$. If $A\in[\mu]^{<\omega},R\in\mathcal{U}$, then let $$C_{A,R}=\{B\in[R]^{<\omega}\mid A\subseteq B\}.$$ Then give $[\mu]^{<\omega}$ the topology where the basic open sets are the sets of the form $C_{A,R}$ where $$\max(A)<\min(R),A\in[\mu]^{<\omega},R\in\mathcal{U}.$$ The space $[\mu]^{<\omega}$ is a $P_{<\mu}$-space.

We say that a topological space is extremally disconnected if the closure of every open set is open. Equivalently, a frame $L$ is extremally disconnected if each element of $B_{L}$ is complemented in $L$ and in this case $\mathfrak{B}(L)=B_{L}$. The motivation behind extremal disconnectedness is that a frame $L$ is extremally disconnected precisely when the Boolean algebra $\mathfrak{B}(L)$ of complemented elements of $L$ is a complete Boolean algebra. As observed in this answer, the space $[\mu]^{<\omega}$ is extremally disconnected, so $\mathrm{Ro}([\mu]^{<\omega})$ is precisely the set of all clopen subsets of $[\mu]^{<\omega}$. The Boolean algebra $\mathrm{Ro}([\mu]^{\omega})$ is the complete Boolean algebra for Prikry forcing. Furthermore, the extremal disconnectedness of $\mathrm{Ro}([\mu]^{\omega})$ is a reformulation of the lemma known as the Prikry condition which states that if $P_{\mathcal{U}}$ is the poset for Prikry forcing (i.e. $(x,R)\in P_{\mathcal{U}}$ if and only if $x\in[\mu]^{<\omega},R\in\mathcal{U},\max(x)<\min(R)$) , then whenever $(x,R)\in P$ and $\theta$ is a statement in the forcing language, then there is some $S\subseteq R,S\in\mathcal{U}$ where $(x,S)$ decides $\theta$.

We observe that each $[\mu]^{\leq n}$ is closed and $[\mu]^{\leq n}$ is nowhere dense in $[\mu]^{\leq n+1}$. Therefore, $[\mu]^{<\omega}$ is the union of countably many nowhere dense sets and therefore $[\mu]^{<\omega}$ is not a Baire space. Therefore, since $[\mu]^{<\omega}$ is not a Baire space, $\mathrm{Ro}([\mu]^{<\omega})$ is not $<\aleph_{1}$-distributive, so forcing with $\mathrm{Ro}([\mu]^{<\omega})$ adds a new sequence $(x_{n})_{n\in\omega}$. On the other hand, since $\mathrm{Ro}([\mu]^{<\omega})$ is extremally disconnected, the complete Boolean algebra $\mathrm{Ro}([\mu]^{<\omega})$ is $(<\mu,2)$-distributive and therefore $\mathrm{Ro}([\mu]^{<\omega})$ adds no new subsets to $P(\alpha)$ for $\alpha<\mu$.

  • 6
    $\begingroup$ I downvoted this answer; it's a confused and overly complicated exposition of the elementary result that $\kappa$-representable $\kappa$-complete Boolean Algebras are $(\mu, \lambda)$-distributive for every $\mu, \lambda < \kappa$ (a Boolean Algebra is $\kappa$-representable, iff, it can be realized as, $CO(X)$ for some regular $P_{<\kappa}$-space $X$, iff, it can be realized as a $\kappa$-complete field of sets.) $\endgroup$ – Not Mike Apr 9 '19 at 13:55
  • 3
    $\begingroup$ (See Proposition 14.4, p. 214; The Handbook of Boolean Algebras, Vol. 1. ) $\endgroup$ – Not Mike Apr 9 '19 at 14:09
  • $\begingroup$ @NotMike. If you have a simplification of these results, then please post another answer. $\endgroup$ – Joseph Van Name Apr 9 '19 at 14:14
  • $\begingroup$ @NotMike. You are incorrect about your claim about representability. So $<\kappa$-representability is typically defined as being isomorphic as a Boolean algebra to a $<\kappa$-field of sets. This is not equivalent to being $\mathrm{CO}(X)$ for a $P_{<\kappa}$-space (here I am assuming that $\mathrm{CO}(X)$ denotes the algebra of clopen sets). $\endgroup$ – Joseph Van Name Apr 9 '19 at 23:59
  • $\begingroup$ If $B$ is any $<\kappa$-complete Boolean algebra which is not both complete and atomic and where every $<\kappa$-complete ultrafilter on $B$ is principal, then $B$ is not isomorphic to the algebra of all clopen sets of any $P_{<\kappa}$-space. If $X$ is a $P_{<\kappa}$-space and $M$ is the $<\kappa$-complete algebra of clopen sets and every $<\kappa$-complete ultrafilter on $M$ is principal, then for each $x\in X$, the set $\{R\in M|x\in R\}$ is a $<\kappa$-complete ultrafilter on $M$ and hence principal. Therefore, $X$ is the discrete space, and hence $B\simeq M=P(X)$. $\endgroup$ – Joseph Van Name Apr 9 '19 at 23:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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