[math-fun] Set of all sets question
We know that the Class of all sets Sets cannot be a set, because this would imply the existence of the set X = {s in Sets | s is not a member of s}, but X belongs to X if and only if it doesn't, hence Contradiction. But we are told that it makes perfect sense to speak of all sets no larger than some fixed cardinal number = beta. Let the set of all sets that are no larger than the cardinal beta be denoted by A(beta): A(beta) = {s in Sets | card(s) <= beta}. QUESTION: --------- How can it be that each A(beta) exists, and for that matter any subset of them exists, but all of them cannot exist at once (without contradiction)? —Dan
They do all exist -- it's just that A(beta) is larger than the cardinal beta, so there's no contradiction.
Sent: Thursday, May 03, 2018 at 9:19 PM From: "Dan Asimov" <dasimov@earthlink.net> To: math-fun@mailman.xmission.com Subject: [math-fun] Set of all sets question
We know that the Class of all sets Sets cannot be a set, because this would imply the existence of the set
X = {s in Sets | s is not a member of s},
but X belongs to X if and only if it doesn't, hence Contradiction.
But we are told that it makes perfect sense to speak of all sets no larger than some fixed cardinal number = beta.
Let the set of all sets that are no larger than the cardinal beta be denoted by A(beta):
A(beta) = {s in Sets | card(s) <= beta}.
QUESTION: --------- How can it be that each A(beta) exists, and for that matter any subset of them exists, but all of them cannot exist at once (without contradiction)?
—Dan
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On Thu, May 3, 2018 at 4:19 PM, Dan Asimov <dasimov@earthlink.net> wrote:
QUESTION: --------- How can it be that each A(beta) exists, and for that matter any subset of them exists, but all of them cannot exist at once (without contradiction)?
I don't understand what you think is a contradiction. All of them can and do exist at once. Their union does not exist (if using ZF) or is a class but not a set (if using Goedel-Bernay), but this is not a contradiction, because the union axiom only says that given a *set* of sets, the union set also exists. It does not guarantee the existence of the union of any class of sets exists. I often find it helpful to think of "set" in the axiomization as meaning a "nice" set. The union axiom doesn't say that unions exist, it says that the union of a nice set of nice sets is nice. (or more precisely, a set whose nice elements are exactly the nice elements of the nice elements of the original set). The power set axiom doesn't say that the set of all subsets of a set exists; it says that the set of all nice subsets of a nice set is itself nice. This makes the fact that there are countable models of ZF seem much less mysterious. There are ways to define "nice" (nonconstructively) so that there are only countably many nice sets, yet all the axioms are still satisfied. The proof that the power set of the integers is uncountable is still valid, but all that says is that there is no "nice" 1-1 correspondence between the integers and the set of nice sets of integers: they are both countable sets, but none of the one-to-one correspondences between the is itself a "nice" set. Andy
—Dan
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Very nice-ly put, Andy! Jim On Thursday, May 3, 2018, Andy Latto <andy.latto@pobox.com> wrote:
On Thu, May 3, 2018 at 4:19 PM, Dan Asimov <dasimov@earthlink.net> wrote:
QUESTION: --------- How can it be that each A(beta) exists, and for that matter any subset
of them exists,
but all of them cannot exist at once (without contradiction)?
I don't understand what you think is a contradiction. All of them can and do exist at once. Their union does not exist (if using ZF) or is a class but not a set (if using Goedel-Bernay), but this is not a contradiction, because the union axiom only says that given a *set* of sets, the union set also exists. It does not guarantee the existence of the union of any class of sets exists.
I often find it helpful to think of "set" in the axiomization as meaning a "nice" set. The union axiom doesn't say that unions exist, it says that the union of a nice set of nice sets is nice. (or more precisely, a set whose nice elements are exactly the nice elements of the nice elements of the original set). The power set axiom doesn't say that the set of all subsets of a set exists; it says that the set of all nice subsets of a nice set is itself nice. This makes the fact that there are countable models of ZF seem much less mysterious. There are ways to define "nice" (nonconstructively) so that there are only countably many nice sets, yet all the axioms are still satisfied. The proof that the power set of the integers is uncountable is still valid, but all that says is that there is no "nice" 1-1 correspondence between the integers and the set of nice sets of integers: they are both countable sets, but none of the one-to-one correspondences between the is itself a "nice" set.
Andy
—Dan
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participants (4)
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Adam P. Goucher -
Andy Latto -
Dan Asimov -
James Propp