Sieve for an infinite array of sets, resulting in an array of the same size of pairwise disjoint sets

Question

Let $\lambda$ be an infinite cardinal. For each $\nu<\lambda$, let $(E_\xi^\nu:\xi<\lambda^+)$ be a family of pairwise disjoint sets each of cardinality at most $\lambda$.

Assume that for every set $T$ of cardinality at most $\lambda$ and for any $\nu<\lambda$, $|\{\xi<\lambda^+:E_\xi^\nu\cap T\ne\emptyset\}|\le\lambda$.

Prove that there are ordinals $\xi(\nu,\eta)<\lambda^+$ ($\nu<\lambda$, $\eta<\lambda^+$) such that the sets $(E_{\xi(\nu,\eta)}^\nu:\nu<\lambda\text{, }\eta<\lambda^+)$ are pairwise disjoint.

This seems to be gist of (and is almost copied from) the top of page 312 of Hajnal, A., "Embedding finite graphs into graphs colored with infinitely many colors," Israel J. Math. 73 (1991), no. 3, 309–319. https://www.researchgate.net/publication/225726350_Embedding_finite_graphs_into_graphs_colored_with_infinitely_many_colors

Answer 1

Consider the reverse lexicographic order $\prec$ on $\lambda\times\lambda^+$ ($(\alpha,\beta)\prec(\gamma,\delta)$ iff $\beta<\delta$ or $\beta=\delta$ and $\alpha<\gamma$); its order type is equal to $\lambda^+$.

Choose the ordinals $\xi(\nu,\eta)$ by recursion with respect to this well-order. We can take $\xi(0,0)=0$, because there are no demands yet. Assume $\xi(\alpha,\beta)$ is found for $(\alpha,\beta)\prec(\nu,\eta)$. Let $T=\bigcup\{E^\alpha_{\xi(\alpha,\beta)}:(\alpha,\beta)\prec(\nu,\eta)\}$. Then $|T|\le\lambda$ and by the assumption/fact above $\{\xi:E^\nu_\xi\cap T=\emptyset\}$ has cardinality $\lambda^+$. Let $\xi(\nu,\eta)$ be the minimum of that set. In this way we guarantee that $E^\alpha_{\xi(\alpha,\beta)}\cap E^\nu_{\xi(\nu,\eta)}=\emptyset$ whenever $(\alpha,\beta)\prec(\nu,\eta)$.