5
$\begingroup$

I've been making my way through the new Kunen and I've come across an exercise that I can't work out. The question is this:

Let $\kappa$ be a singular cardinal. Show that there is a collection $A$ of $\kappa$ many two-element subsets of $\kappa$ such that no element of $[A]^\kappa$ forms a $\Delta-$ system. Where $[A]^\kappa$ is the set of subsets of A of size $\kappa$.

Any help would be appreciated (i.e. hints welcome).

  • 0
    @AsafKaragila From context I presume it refers to the (otherwise unnamed) collection of two-element subsets of $\kappa$; I've edited the post slightly to reflect that. cody, if my edits are incorrect, please feel free to revert.2012-10-19

1 Answers 1

3

$\newcommand{\cf}{\operatorname{cf}}$Let $\cf\kappa=\lambda<\kappa$, and let $\langle\alpha_\xi:\xi<\lambda\rangle$ be a strictly increasing sequence cofinal in $\kappa$ such that $\alpha_0=0$. For $\xi<\lambda$ let $K_\xi=[\alpha_\xi,\alpha_{\xi+1})$; then $\kappa=\bigcup_{\xi<\lambda}K_\xi$, and $|K_\xi|<\kappa$ for each $\xi<\lambda$. Let $A=\bigcup_{\xi<\lambda}[K_\xi]^2\;;$ clearly $|A|=\kappa$, and I leave to you the straightforward verification that $A$ has no $\Delta$-system of power $\kappa$.

  • 0
    Sorry for the random comment on an old post, but I am starting Chapter 3 in Kunen's text and this is the exercise I'm currently working on. How would you show that no $B \in [A]^{\kappa}$ forms a delta-system? I'm trying to assume that there is a $B \in [A]^{\kappa}$ that forms a delta system and somehow derive a contradiction.2014-02-11