Singular Cardinals, and A Strange Question. - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T10:18:20Z http://mathoverflow.net/feeds/question/52897 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/52897/singular-cardinals-and-a-strange-question Singular Cardinals, and A Strange Question. Michael Blackmon 2011-01-23T02:42:31Z 2011-02-07T04:03:55Z <p>Let $\mu$ be any infinite cardinal, and define a collection $N\subset[\mu]^\mu$ to be, maximal almost disjoint (MAD) over $\mu$, iff </p> <ol> <li>$\forall\{A,B\}\in[N]^2$ $( A\cap B \in [\mu]^{&lt;\mu})$</li> <li>$\forall X\in[\mu]^\mu \exists A\in N$ $( X \cap A \in [\mu]^\mu)$</li> </ol> <p>My questions are as follows: when $\mu$ is <strong>singular</strong></p> <ol> <li>What is known about MAD families (or any other combinatorial structures, like, towers, SFIP families without pseudo-intersection, etc) over $\mu$?</li> <li>Are such families degenerate in the sense that an infinite family can have cardinality below $\mu$? </li> <li>Is there any connection between such constructs on $\mu$ and the corresponding constructs on $cf(\mu)$?</li> </ol> <p>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? </p> <p>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?</p> http://mathoverflow.net/questions/52897/singular-cardinals-and-a-strange-question/52916#52916 Answer by Chris Hardin for Singular Cardinals, and A Strange Question. Chris Hardin 2011-01-23T05:13:09Z 2011-01-23T05:13:09Z <p>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 $\mu$?)</p> <p>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 $\lambda_\alpha$, $\alpha&lt;\kappa$, 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.</p> <p>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 $\lambda_n$, $n&lt;\omega$, 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 $\mu$.</p> <p>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.</p> http://mathoverflow.net/questions/52897/singular-cardinals-and-a-strange-question/52933#52933 Answer by Justin Palumbo for Singular Cardinals, and A Strange Question. Justin Palumbo 2011-01-23T09:18:49Z 2011-01-23T18:20:24Z <p>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 $\omega$).</p> <p>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.</p> <p>The only interesting thing to check is the maximality. Suppose that $C\subseteq\aleph_\omega$ has cardinality $\aleph_\omega$. Let $n_0 &lt; n_1 &lt; n_2 &lt; \ldots$ 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 $\aleph_\omega$.</p> <p>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" <a href="http://www.renyi.hu/~p_erdos/1975-21.pdf" rel="nofollow">http://www.renyi.hu/~p_erdos/1975-21.pdf</a> and the more recent Kojman, Kubis and Shelah "On Two Problems of Erdos and Hechler: New Methods in Singular MADness" <a href="http://arxiv.org/abs/math/0406441" rel="nofollow">http://arxiv.org/abs/math/0406441</a></p> http://mathoverflow.net/questions/52897/singular-cardinals-and-a-strange-question/54597#54597 Answer by Jason for Singular Cardinals, and A Strange Question. Jason 2011-02-07T02:59:24Z 2011-02-07T04:03:55Z <blockquote> <p><B>Theorem</B> 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 $\lambda$.</p> </blockquote> <p>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 $\vec{s} = \langle s_{\alpha}| \alpha &lt; \text{cof}(\lambda)\rangle$ 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 $\alpha &lt; \text{cof}(\lambda)$, there will be: </p> <p>$p_{\alpha} \in G$ such that $p_{\alpha} \Vdash \dot{A} \cap \check{s_{\alpha}} = \check{a_{\alpha}}$</p> <p>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 $S_A = \{p_{\alpha}| \alpha &lt; \text{cof}(\lambda)\} \in V[G]$ 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, $\theta &lt; \lambda$ 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 </p> <p>$A = \bigcup\{a \in \mathcal{P}(\lambda)| p \in S_{A} \land s_{\alpha} \in \text{range}(\vec{s}) \land p \Vdash \dot{A} \cap \check{s_{\alpha}} = \check{a}\}$. $\Box$</p> <p>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.</p>