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Let $\mu$ be any infinite cardinal, and define a collection $N\subset[\mu]^\mu$ to be, maximal almost disjoint (MAD) over $\mu$, iff

  1. $\forall\{A,B\}\in[N]^2$ $( A\cap B \in [\mu]^{<\mu})$
  2. $\forall X\in[\mu]^\mu \exists A\in N$ $( X \cap A \in [\mu]^\mu)$

My questions are as follows: when $\mu$ is singular

  1. What is known about MAD families (or any other combinatorial structures, like, towers, SFIP families without pseudo-intersection, etc) over $\mu$?
  2. Are such families degenerate in the sense that an infinite family can have cardinality below $\mu$?
  3. Is there any connection between such constructs on $\mu$ and the corresponding constructs on $cf(\mu)$?

The main point I really want to know is this: Is it possible to add new subsets to an arbitrary singular cardinal without adding new subsets the the cardinals below it?

Side Request: I've been told there are forcing constructions which will add an order type $\omega$ cofinal sequence to a cardinal with cofinality $\omega$, can anyone point me in the correct direction with a book or article?

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In response to your side request, I think this will work. Take a countable unbounded $A \subseteq \mu$ and then force with perfect trees $T \subseteq A^{{<}\omega}$ with the property that for every $t \in T$, there is a $t^{\prime} \supseteq t$ such that $t^{\prime} {}^{\frown} \langle a\rangle \in T$ for every $a \in A$ ordered by (ordinary) inclusion. Then if $G$ is $V$-generic for this forcing, $\bigcap G$ will be a new unbounded $\omega$ sequence through $\mu$, and I believe all cardinals will be preserved in $V[G]$. This is an adaptation of Sacks forcing (See Jech). – Jason Jan 23 '11 at 8:02
Actually, also look up Namba forcing in Jech as this may be a better comparison. – Jason Jan 23 '11 at 8:50
Please disregard my comments about using more complex forcing to achieve the goal that Chris accomplished by adding a Cohen Real, which does indeed preserve all cardinals. Nevertheless, I'm keeping the comments there in case they prove to be useful. Also, see for some restrictions on adding subsets to singular strong limit cardinals (i.e., when $0^{\sharp}$ does not exist or if $\mu$ has uncountable cofinality and the GCH holds below it, then $2^{\mu} = \mu^+$). – Jason Jan 23 '11 at 12:00
You define the MAD family as $M$ but use $N$ thereafter. – Asaf Karagila Jan 23 '11 at 12:24
@Jason thank you @Karagila I've corrected the statement of the question, thanks for noticing. – Michael Blackmon Jan 23 '11 at 23:02
up vote 5 down vote accepted

Theorem If $0^{\sharp}$ does not exist and $\lambda$ is a singular cardinal, then any forcing adding subsets to $\lambda$ necessarily adds subsets to a cardinal below $\lambda$.

Proof: Let $\mathbb{P}$ be a partial order in the ground model and $G \subseteq \mathbb{P}$ be $V$-generic. Without loss of generality, we may assume that $\mathbb{P}$ is a partial order on a cardinal so that its elements are all ordinals. Also let $\vec{s} = \langle s_{\alpha}| \alpha < \text{cof}(\lambda)\rangle$ be a cofinal sequence in $\lambda$ in the ground model and $\dot{A}$ a $\mathbb{P}$-name for a subset of $\lambda$ in $V[G]$. Now suppose that $V$ and $V[G]$ agree on the bounded subsets of $\lambda$. Then for all $\alpha$, we have $a_{\alpha} = s_{\alpha} \cap \dot{A}_G \in \mathcal{P}(\lambda)^{V}$ so for every $\alpha < \text{cof}(\lambda)$, there will be:

$p_{\alpha} \in G$ such that $p_{\alpha} \Vdash \dot{A} \cap \check{s_{\alpha}} = \check{a_{\alpha}}$

Because $V$ is a definable class in $V[G]$, the forcing relation for $V$ is definable in $V[G]$, and we may therefore choose such a $\text{cof}(\lambda)$-sequence of conditions $p_{\alpha}$ below a condition forcing that $\dot{A}$ is a name for a subset of $\lambda$. Let $S_A = \{p_{\alpha}| \alpha < \text{cof}(\lambda)\} \in V[G]$ be such a set of ordinals. Now $S_A$ is a set of ordinals having size $\text{cof}(\lambda)$ in $V[G]$ so by the nonexistence of $0^{\sharp}$, it follows from Jensen's covering lemma that there is a constructible set $C$ of size $\theta = \max\{\omega_1, \textrm{cof}(\lambda)\}$ in $V[G]$ covering $S_A$. Then since $\lambda$ is singular, $\theta < \lambda$ whereby $C \in L \subseteq V$ will also have size $\theta$ in $V$ by virtue of the fact that a poset adding no new subsets to $\theta$ cannot collapse any cardinals below $\theta^{++}$. But now $S_{A} \subseteq C$ must also be in $V$ because otherwise $V[G]$ would be adding a subset of $\theta$ induced by $f''S_{A}$ where $f: C \rightarrow \theta$ is a bijection in the ground model. However, then $V$ can construct $A$ from $S_A$ in the ground model since

$A = \bigcup\{a \in \mathcal{P}(\lambda)| p \in S_{A} \land s_{\alpha} \in \text{range}(\vec{s}) \land p \Vdash \dot{A} \cap \check{s_{\alpha}} = \check{a}\}$. $\Box$

In particular, this shows that if $V$ is a forcing extension of $L$, then we cannot add a subset to a singular cardinal without adding a subset to a cardinal below it. I don't have an answer for what happens when $0^{\sharp}$ does exist, but at least this shows that your very interesting question is closely tied to the existence of certain large cardinals.

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In question 2, perhaps I am missing something, but it seems that such families can have cardinality below $\mu$: just take $M = \{\mu\}$. That satisfies your definition of MAD and has 1 element. (Generally, it's easy to get small MAD families--the trick is constructing big ones, right? Or were you wondering if there are $\mu$ for which every MAD family is smaller than $\mu$?)

On your main point, I don't have an answer, but here is at least an observation. Let $\mu$ be singular and let $\kappa = cf(\mu)$. Let $P$ be a notion of forcing that adds new subsets to $\mu$ without adding subsets to any cardinals below $\mu$. Then $P$ is not $\kappa$-distributive (and hence not $\kappa$-closed). (Here, $\kappa$-closed is in the sense of Jech, not Kunen: $P$ is $\kappa$-closed if for $\lambda\leq\kappa$, descending $\lambda$-sequences in $P$ have lower bounds. See p. 228 of Jech 3rd millennium edition.) Proof: Let $\lambda_\alpha$, $\alpha<\kappa$, be cofinal in $\mu$. Let $x\subseteq\mu$ be in $V[G]$ but not $V$. Define $f:\kappa\to V$ by $f(\alpha) = x\cap \aleph_{\lambda_\alpha}$. We cannot have $f\in V$, otherwise $x$ would be in $V$. By Theorem 15.6 of Jech, then, $P$ cannot be $\kappa$-distributive.

As for your side request, if $\mu$ has cofinality $\omega$, there is at least one easy way to add order type $\omega$ cofinal sequences into $\mu$. Let $\lambda_n$, $n<\omega$, be a cofinal increasing sequence in $\mu$. Use any notion of forcing that adds new reals without collapsing any cardinals you care about (like adding Cohen reals). If $x\subseteq \omega$ is new, then $\lambda_n$ for $n\in x$ is a new cofinal sequence in $\mu$.

I'm not a set theorist and I haven't eaten in several hours, so that should all be taken with a grain of salt.

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Question 2 is poorly stated, I meant for them to be infinite, I will edit it now, I'm sorry. Also, Cohen will collapse anything that is not regular. – Michael Blackmon Jan 23 '11 at 5:59
Cohen forcing doesn't collapse any cardinals... – Justin Palumbo Jan 23 '11 at 8:24
@Justin: Forcing to add a Cohen subset of a regular cardinal $\lambda$ for which $2^{{<}\lambda} = \lambda$ doesn't collapse cardinals because then the forcing is ${<}\lambda$-closed and $\lambda^+$-c.c. However, forcing to add a Cohen subset of a regular cardinal $\lambda$ will always force $2^{{<}\lambda} = \lambda$ in the extension so if $2^{{<}\lambda} > \lambda$, it will necessarily collapse all cardinals between $\lambda^+$ and $2^{{<}\lambda}$, inclusively. This is because it is dense that the bounded subsets of $\lambda$ from $V$ will be coded in blocks on the newly added subset. – Jason Jan 23 '11 at 8:48
@Jason: By Cohen forcing, I meant adding a Cohen real (as Chris uses it in his answer). Since $2^<\omega=\omega$, it won't collapse any cardinals, for the reasons you've stated. – Justin Palumbo Jan 23 '11 at 8:54
@Justin: Sorry for that. I should've read the post before commenting. – Jason Jan 23 '11 at 9:09

With regards to 2 and 3, if you have a MAD family of size $\kappa$ on $cf(\mu)$ then you have a MAD family of size $\kappa$ on $\mu$. (So, for example there would be a 'degenerate family' of the form you mention in 2 on $\aleph_\omega$ if there is a MAD family of size $\aleph_1$ on $\omega$).

Let me give the argument for $\mu=\aleph_\omega$. Suppose $\mathcal{A}$ is a MAD family on $\omega$. For each $A\subseteq\omega$, let $B(A)\subseteq\aleph_\omega$ be the union of the intervals $I_n=[\aleph_n,\aleph_{n+1})$ such that $n\in A$. Let $\mathcal{B}$ be all the $B(A)$, for $A\in\mathcal{A}$. We claim that $\mathcal{B}$ is MAD.

The only interesting thing to check is the maximality. Suppose that $C\subseteq\aleph_\omega$ has cardinality $\aleph_\omega$. Let $n_0 < n_1 < n_2 < \ldots $ be a sequence such that $|I_{n_{k+1}}\cap C|\geq\aleph_{{n_k}+2}$. (The +2 is there to make sure the $n_{k+1}$ we find must be bigger than $n_k$). Letting $X=\{n_k:k\in\omega\}$, there is $Y\in\mathcal{A}$ with $X\cap Y$ infinite. Then $B(Y)\cap C$ has cardinality $\aleph_\omega$.

As for general references to MAD families on singular cardinals, it looks like you should check out Erdos, Hechler's "On Maximal Almost-Disjoint Families over Singular Cardinals" and the more recent Kojman, Kubis and Shelah "On Two Problems of Erdos and Hechler: New Methods in Singular MADness"

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Very cool, I had a feeling there was some connection with the confinality of $\mu$. Thank you. – Michael Blackmon Jan 23 '11 at 23:01

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