6:35 AM
I thought that it is possible to prove that a maximal almost family disjoint on a set of cardinality $\varkappa$ has cardinality at least $\varkappa^+$ (the cardinal successor).
I thought that the same proof which proves that MAD family on an infinite countable set has cardinality at least $\aleph_1$ works. (In other words, the proof that $\mathfrak a\ge\aleph_1$.)
Was I wrong? Do I need $\varkappa$ to be regular for this?
For example Theorem 790 assumes that $\varkappa$ is regular.
Or Theorem 17.17 in Just-Weese.
The usual proof for $\aleph_0$ is that if I have countably many almost disjoint infinite sets $A_n$, $n<\omega$ then I construct the new set by choosing $b_n\in A_n\setminus \bigcup_{k<n} A_k$.
For more general case I would try something similar; for $\alpha<\varkappa$ I would try to choose $b_\alpha\in A_\alpha \setminus \bigcup_{\beta<\alpha} A_\beta$.
In the other words $b_\alpha \in A_\alpha \setminus \bigcup_{\beta<\alpha} (A_\beta\cap A_\alpha)$.
The sets $A_\beta\cap A_\alpha$ have cardinality $<\varkappa$.
And $A_\alpha$ has cardinality $\varkappa$.
But without regularity, I do cannot say for sure that the difference will be non-empty.
Ok, it seems that I managed to find the problem almost immediately after I asked here...
It seems that this chatroom is useful.

6:59 AM