The set of all SUBSETS that don't contain themself -> PARADISE
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From: |-|ercules on 2 Jul 2010 11:51 The set of all SETS that don't contain themself -> PARADOX The set of all SUBSETS that don't contain themself -> PARADISE This is the status quo of early 21ST Century mathematics. In the 2nd subset version you can (attempt to) list the subsets and contain refers to the list position being an element of the subset (of natural numbers). Since small examples actually give a new subset not in the list.... TADA... the set of all subsets is BIGGER THAN INFINITY!!! Herc -- "George Greene" <greeneg(a)email.unc.edu> wrote >> Are you really that stupid to assume ...? > > I don't NEED to ASSUME!
From: Transfer Principle on 3 Jul 2010 00:04 On Jul 2, 8:51 am, "|-|ercules" <radgray...(a)yahoo.com> wrote: [drop groups because Google won't let me post to more than five at a time. Naturally, I drop aus.tv since I already know that these posts aren't welcome there, and I also drop comp.ai.philosophy] > The set of all SETS that don't contain themself -> PARADOX > The set of all SUBSETS that don't contain themself -> PARADISE Herc does bring up an interesting point here about the similarity between Russell's Paradox and Cantor's Theorem. The difference? We expect the set of all sets that don't contain themselves to be a _set_ -- and since it isn't (as it would both contain and not contain itself), we conclude that there is no such set. But the set of all subsets that don't contain themselves is still provably a _subset_ -- at least if we're talking about subsets of a set in standard theory. Thus, standard theory proves that if x is a set, then P(x) is a bigger set. But V (a universal set) can't exist, since otherwise P(V) would be bigger. So V isn't a set.
From: |-|ercules on 3 Jul 2010 21:14 "Transfer Principle" <lwalke3(a)lausd.net> wrote > On Jul 2, 8:51 am, "|-|ercules" <radgray...(a)yahoo.com> wrote: > > [drop groups because Google won't let me post to more than > five at a time. Naturally, I drop aus.tv since I already > know that these posts aren't welcome there, and I also > drop comp.ai.philosophy] > >> The set of all SETS that don't contain themself -> PARADOX >> The set of all SUBSETS that don't contain themself -> PARADISE > > Herc does bring up an interesting point here about the > similarity between Russell's Paradox and Cantor's Theorem. > > The difference? We expect the set of all sets that don't > contain themselves to be a _set_ -- and since it isn't > (as it would both contain and not contain itself), we > conclude that there is no such set. > > But the set of all subsets that don't contain themselves > is still provably a _subset_ -- at least if we're talking > about subsets of a set in standard theory. > > Thus, standard theory proves that if x is a set, then P(x) > is a bigger set. But V (a universal set) can't exist, > since otherwise P(V) would be bigger. So V isn't a set. Just add an element and you get a BIGGER set. Is there a bijection from {dog, cat} to {dog, cat, elephant} ? Cantor's proof of transfinite sets was predated 3000 years by Plato's assistant Hermes. Hermes proved that oo+1 is BIGGER than oo! Hey it works on all other examples of sets, just like the powerset! _________________________________________________________________ Herc
From: Dan Christensen on 5 Jul 2010 11:40
On Jul 2, 11:51 am, "|-|ercules" <radgray...(a)yahoo.com> wrote: > The set of all SETS that don't contain themself -> PARADOX > Yes. Assuming the existence of such a set leads to a contradiction. > The set of all SUBSETS that don't contain themself -> PARADISE > Nice. Given any set s, you can construct the set k of all subsets of s that are not elements of themselves if you have power set and subset axioms -- just select those and only those elements of the power set of s that are not elements of themselves. As far as I can tell, you won't arrive at any contradiction though. On the other hand, determining what is and is not an element of k may not always be possible (depending on your set theory). For example, let s be the singleton {x}. Then P(x) = { {x}, { } } Is probably safe to say that { } is an element of k since ~ { } e { }. Is {x} an element of k? That is, can we determine if {x} e {x}? Not without additional axioms, it would seem. Dan Download my DC Proof software at http://www.dcproof.com |