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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
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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 en.wikipedia.org/wiki/Singular_cardinals_hypothesis 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
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3 Answers 3

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
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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" http://www.renyi.hu/~p_erdos/1975-21.pdf and the more recent Kojman, Kubis and Shelah "On Two Problems of Erdos and Hechler: New Methods in Singular MADness" http://arxiv.org/abs/math/0406441

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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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