Proof regarding partitions of an infinite cardinal $\kappa$.

Question

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 $\{w, z\} \not \in A$.

Answer 1

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.

Answer 2

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.