I am working on a proof regarding partitions (Jech 9.7), and I am unfortunately stuck on the last step. I will give all the details up to this point, but I don't think they are all required.

Let $\kappa$ be an infinite cardinal, and let $\{ A, B\} $ be a partition of $[\kappa]^2$. For each $x \in \kappa$, define $B_x := \{ y \in \kappa \mid x < y \text{ and } \{x, y\} \in B \}$.

Now suppose that $\kappa$ is singular, and that there exists a set $S \subseteq \kappa$ such that for all $x \in S$, $|B_x \cap S | < \kappa$. Let $\lambda = \operatorname{cf}{\kappa}$. Consider an increasing sequence of regular cardinals $\langle \kappa_{\xi} \mid \xi < \lambda \rangle$, with the property that each cardinal is $> \lambda$. We also assume that there is no set $H \subseteq \kappa$ of size $\omega$ such that $[H]^2 \subseteq B$, and that for all $\xi \in \lambda$ we have $\kappa_{\xi} \rightarrow (\kappa_{\xi}, \omega)^2$.

Now construct a partition $\{S_{\xi} \mid \xi < \lambda \}$ of $S$, such that $|S_{\xi}| = \kappa_{\xi}$. From the above it follows that there are sets $K_{\xi} \subseteq S_{\xi}$ such that $| K_{\xi}| = \kappa_{\xi}$ and $[K_{\xi}]^2 \subseteq A$.

It follows (details omitted) that for every $\xi < \lambda$, there exists an $\alpha(\xi)$ such that the set $Z_{\xi} = \{x \in K_{\xi} : |B_x \cap S| < \kappa_{\alpha(\xi)}\}$ has cardinality $\kappa_{\epsilon}$. Let $\langle \xi_{\nu} \mid \nu < \lambda \rangle$ be an increasing sequence of ordinals $< \lambda$ such that if $\nu_1 < \nu_2$, then $\alpha(\xi_{\nu_1}) < \large\xi_{\nu_2}.$

We define $H_{\nu} = Z_{\xi_{\nu}} \setminus \bigcup \{B_x \mid x \in \bigcup_{\eta < \nu} Z_{\xi_{\eta}}\}$. Then $|H_{\nu}| = \kappa_{\xi_\nu}$ and $[H_{\nu}]^2 \subseteq A$. Letting $H = \bigcup_{\nu < \lambda}H_{\nu}$, we have that $|H| = \kappa$, but I am stuck trying to show that $[H]^2 \subseteq A$. It's supposed to follow by construction of $H$, but I'm not sure how it does so.

I'm sorry that this post is so long, but any help would be much appreciated.

My initial attempt: My first idea was to take $\{w,z\} \in [H]^2.$ Then there exists $\nu, \nu' < \lambda$ such that $w \in H_{\nu}$ and $z \in H_{\nu '}$. We can assume that $\nu ' < \nu$. It follows by definition of $H$ that $w \not\in B_z$. This means that either $\{w,z\} \in A$ (which is good) or that $w < z$. I am finding it hard to find a contradiction if we assume that $w < z$ and that $\{w, z\} \not \in A$.

  • 1
    $\begingroup$ Could you state the actual goal of the proof for those of us not having Jech handy? $\endgroup$ – Desiato Mar 21 '12 at 17:21
  • $\begingroup$ the point is that I am trying to prove $\kappa \rightarrow (\kappa, \omega)^2$ for any infinite cardinals. He splits the proof up into various cases, and what above is the remaining case. So in my 3rd paragraph, we assume that there are no infinite sets $H$ such that $[H]^2 \subseteq B$, so we want to show there is a set $H$ of size $\kappa$ such that $[H]^2 \subseteq A$. Sorry, it's very confusing $\endgroup$ – Paul Slevin Mar 21 '12 at 17:46

Take any $x < y \in H$. Then there are $\mu , \nu$ such that $x \in H_\mu$ and $y \in H_\nu$. If $\mu = \nu$, we're done (obviously).

  • Assume that $\mu < \nu$. As $y \in H_\nu$, then $y \in Z_{\xi_\nu}$ and $y \notin B_z$ for any $z \in \bigcup_{\eta < \nu} Z_{\xi_\eta}$. Note that $x \in Z_{\xi_\mu}$ and $\mu < \nu$, and so $y \notin B_x$. Therefore either $y < x$ or $x < y$ and $\{ x,y \} \in A$, and we know that the first case is impossible.

I believe that the partition $\{ S_\xi : \xi < \lambda \}$ is $S$ can be set up so that $S_\xi < S_\zeta$ (that is, $x < y$ for all $x \in S_\xi$ and $y \in S_\zeta$) for all $\xi < \zeta < \lambda$.

Inductively take $\delta_\xi \in S$ to be minimal such that $| ( S \setminus \bigcup_{\zeta < \xi} S_\zeta ) \cap \delta_\xi | = \kappa_\xi$, and let $S_\xi$ be this set.

With this alteration, it would be impossible for $\nu < \mu$ to hold in our situation above (since then $K_\xi < K_\zeta$, and therefore $Z_\xi < Z_\zeta$, for all $\xi < \zeta < \lambda$), and the proof would be done.

  • $\begingroup$ Must be right, since we came to the same conclusion. :-) (The system didn’t notify me that an answer had been posted.) $\endgroup$ – Brian M. Scott Mar 21 '12 at 17:56

Just make sure that when the partition $\{S_\xi:\xi<\lambda\}$ of $S$ is constructed, its parts are ordered in the same way as their indices: if $\xi<\zeta<\lambda$, $\alpha\in S_\xi$, and $\beta\in S_\zeta$, then $\alpha<\beta$. This is trivial and ensures that $w<z$ is impossible in your argument.

  • $\begingroup$ Your idea seems to work perfectly. ;) $\endgroup$ – user642796 Mar 21 '12 at 17:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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