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Jun 26, 2013 at 19:17 vote accept Santi Spadaro
May 22, 2013 at 4:24 comment added user10290 Thank you, Joel. I think I understand this a bit better.
May 18, 2013 at 2:10 comment added Joel David Hamkins Well, as I said, $\Delta(\omega_2,\omega_2)$ can already be refuted without KH by considering the tree $2^{\lt\omega_1}$, where one has at least $\omega_2$ many branches of size $\omega_1$, but no $\Delta$ system of size 3, and Kurepa trees have nothing to do with it. But I'm not sure why you think this refutes $\Delta(\omega_2,\omega_1)$, and in particular, I dispute your implication in (A).
May 17, 2013 at 20:15 comment added Eran My question is why KH is not a counter example: A) $\Delta(\omega_2,\omega_1) => \Delta(\omega_2,\omega_2)$ B) KH (using the same arguments as in the proof) provides $\neg \Delta(\omega_2,\omega_2)$ (=>$\neg\Delta(\omega_2,\omega_1)$) C) It is consistent to have ZFC + GCH + KH -------------------------- hence GCH doesn't have to imply $\Delta(\omega_2,\omega_1)$.
May 17, 2013 at 18:48 history edited Joel David Hamkins CC BY-SA 3.0
Fixed confusion of $\lambda$ with $\eta$.
May 17, 2013 at 17:45 comment added Joel David Hamkins Eran, about the Kurepa tree, I don't see how it helps. A Kurepa tree would seem to be about getting a family of $\omega_2$ many subsets of $\omega_1$ without any $\Delta$-system of size 3. But we already have such a family without any KH hypothesis, just by using paths through the tree $2^{\lt\omega_1}$. Or have I misunderstood your idea?
May 17, 2013 at 17:43 comment added Joel David Hamkins Eran, I don't think the implication you mention in your comment is quite right. Although in the former case, we see that every very large family of small sets has a large subfamily forming a $\Delta$-system, perhaps a slightly smaller family of small sets may not have a subfamily forming a $\Delta$ system of that size. Indeed, my argument shows that if $2^\delta=\delta^{++}$, but otherwise the GCH holds, then $\Delta(\delta^{+++++++},\delta^+)$ will hold, but not $\Delta(\delta^{++},\delta^+)$, which contradicts your implication.
May 17, 2013 at 17:39 comment added Joel David Hamkins Erin, my update shows that $\Delta(\delta^{+++++++},\delta^+)$ is equivalent to the assertion that $(\delta^{++++++})^\delta=\delta^{++++++}$, which is equivalent to saying $2^\delta$ is at most $\delta^{++++++}$, a weak form of the GCH.
May 17, 2013 at 17:32 history edited Joel David Hamkins CC BY-SA 3.0
Proved full equivalence
May 17, 2013 at 14:27 comment added Santi Spadaro Very nice, Joel! It's more than I hoped.
May 17, 2013 at 10:07 comment added Eran @Joel - isn't this proof goes just as well assuming ZFC + GCH + KH (Kurepa's Hypothesis)?
May 17, 2013 at 10:00 comment added Eran Erin - $\Delta( \delta ^{+++++++}, \delta^+) => \Delta( \delta ^{++}, \delta^+)$
May 17, 2013 at 7:22 comment added user10290 It is interesting! What happens when $\Delta( \delta ^{+++++++}, \delta^+)$ ?
May 17, 2013 at 0:58 history edited Joel David Hamkins CC BY-SA 3.0
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May 17, 2013 at 0:32 history edited Joel David Hamkins CC BY-SA 3.0
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May 17, 2013 at 0:23 history answered Joel David Hamkins CC BY-SA 3.0