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 $\mathfrak a\ge\aleph_1$.)
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 $b_n\in A_n\setminus \bigcup_{k<n} A_k$.
6:59 AM
@MartinSleziak The correct link there should have been: books.google.com/books?id=nx0SCgAAQBAJ&pg=PA83
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