Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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).

share|improve this question
1  
What is $A$ in this context? –  Asaf Karagila Oct 19 '12 at 0:50
    
@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. –  Steven Stadnicki Oct 19 '12 at 1:10

1 Answer 1

$\newcommand{\cf}{\operatorname{cf}}$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 $\kappa$.

share|improve this answer
    
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. –  josh Feb 11 at 17:19

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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