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I've been reading this paper by Shelah for about a week and still do not understand on the page 11, on line 9, how it follows from $w \subseteq u_1 \wedge w \nsubseteq u_0$ then we have either $\emptyset \cap \emptyset =\emptyset$ mod $\cal D$ or else $A_{u_0} \cap \emptyset =\emptyset$ mod $\cal D$.

How did he come up with the 5 empty sets $\emptyset$ and "$A_{u_0}$" there from the assumptions $w \subseteq u_1 \wedge w \nsubseteq u_0$,i.e. how these two assumptions are used? I think that it is not needed to read everything to answer the question, just perhaps pages 10 and 11.

Martin Sleziak
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user122424
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1 Answers1

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I frankly don’t understand the breakdown. It seems to me that one can simply argue as follows. First, if $w=u$, then $A_{u_0}'=A_{u_0}$, $A_{u_1}'=A_{u_1}$, and $A_u'=A_u$, so so $A_{u_0}'\cap A_{u_1}'=A_u'\pmod{\mathcal{D}}$. (This does seem to correspond to his first bullet point.) Otherwise, $u\nsubseteq w$, so $A_u'=\varnothing$, and either $u_0\nsubseteq w$ or $u_1\nsubseteq w$, so either $A_{u_0}'=\varnothing$ or $A_{u_1}'=\varnothing$. Thus, $A_{u_0}'\cap A_{u_1}'$ is one of $A_{u_0}\cap\varnothing$, $\varnothing\cap A_{u_1}'$, or $\varnothing\cap\varnothing$, all of which are indeed equal to $A_u'=\varnothing$ modulo $\mathcal{D}$. In other words, it seems to me that we can argue as he does in $(2)$. However, I have not tried to read more than Section $4$ and could well be missing something.

Martin Sleziak
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Brian M. Scott
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  • Does his condition $w \subseteq u_1 \wedge w \nsubseteq u_0$ express $u \nsubseteq w$? – user122424 Dec 01 '16 at 17:45
  • @user122424: It implies that $u\nsubseteq w$, but it’s certainly not equivalent, even conjoined with the condition in the third bullet point. – Brian M. Scott Dec 01 '16 at 19:00
  • So if he added a fourth bullet saying $w \nsubseteq u_1 \wedge w \nsubseteq u_0 $ then it would exhaust all possibilities and this would yield a completely correct breakdown? Or would it be better and correct to use a kind of reversal such as $u_1 \subseteq w \wedge u_0 \nsubseteq w$ ? – user122424 Dec 02 '16 at 16:39
  • @user122424: It would cover the possibilities, but it wouldn’t fix the conclusions that he drew for his second and third cases. If I were reading the paper, I’d simply substitute my own argument and move on. – Brian M. Scott Dec 02 '16 at 18:47