Axiom of Regularity.
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From: zuhair on 12 Jan 2010 21:17 Hi all, Is the following axiom equivalent to the axiom of Regularity? Axiom: For all x ( For all y (y e TC(x) -> ~ y e TC(y)) & For all y (y e TC(x) -> 0 e TC(y)) ). were "TC" stands for "Transitive closure" defined in the standard manner, and "0" stands for the empty set. More specifically is (ZF minus Regularity) + the above axiom equivalent to ZF ? Zuhair
From: Max on 16 Jan 2010 21:25 > Is the following axiom equivalent to the axiom of Regularity? > > Axiom: For all x ( For all y (y e TC(x) -> ~ y e TC(y)) & > For all y (y e TC(x) -> 0 e TC(y)) ). > > were "TC" stands for "Transitive closure" defined in the > standard manner, and "0" stands for the empty set. The second clause seems to demand that 0 belong to Tc(0). The first clause seems to suggest that's not intended. You could fix this by requiring y nonempty in the second line. However, I'm not clear on whether such a principle would rule out a pathology like x0 = {0,x1} x1 = {0,x2} x2 = {0,x3} .... This seems to satisfy both clauses. Correct me if I'm wrong! Best, Max
From: zuhair on 17 Jan 2010 09:52 On Jan 16, 9:25 pm, Max <30f...(a)gmail.com> wrote: > > Is the following axiom equivalent to the axiom of Regularity? > > > Axiom: For all x ( For all y (y e TC(x) -> ~ y e TC(y)) & > > For all y (y e TC(x) -> 0 e TC(y)) ). > > > were "TC" stands for "Transitive closure" defined in the > > standard manner, and "0" stands for the empty set. > > The second clause seems to demand that 0 belong to Tc(0). The first > clause seems to suggest that's not intended. > > You could fix this by requiring y nonempty in the second line. Yes, that is correct. > > However, I'm not clear on whether such a principle would rule out a > pathology like > > x0 = {0,x1} > x1 = {0,x2} > x2 = {0,x3} > ... > > This seems to satisfy both clauses. > > Correct me if I'm wrong! > > Best, You are right! > > Max So the conclusion is that this principle is not equivalent to Regularity, it is strictly weaker than it. Zuhair
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