Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

Zermelo set theory question!?

Okay...so, how would I show that every set other than the empty set contains another set as an element? i.e. there are no "set-theoretic axioms" (I don't think that is the correct word, but it's hard to read the handout!). Why do I know this is true?

Thank you!

3 Answers

Relevance
  • Favorite Answer

    My guess is that the handout says that there are no "set-theoretic ATOMS". There were, historically, variations of set theory that contained things called "atomic elements" -- objects that were not themselves regarded as sets, but just "things". They comprised the bottom-most stratum of those theories, with sets of atoms at the second level, sets of sets of atoms at the third, and so on.

    The ZF axioms do away with those elements -- they turned out to be unnecessary complications to the theory. ZF begins with one set only -- a set with no members -- and builds everything up from that. The entire edifice of mathematics is thus quite literally built on nothing. It's all very zen.

    That, at any rate, is what I think the question is ABOUT.

    Now, as to the proof -- it pretty much is a tautology, as a previous answerer proposed. Suppose that there are two sets, both of which have no elements. Then each one is a proper subset of the other -- this is because the subset requirement is vacuously satisfied (there are no elements that belong to one set, but not to the other). Hence they are the same set. So there is only one element-less set. Any other set thus contains an element. That element, in turn, must be a set: either it is empty, or it contains an element. That's about all there is to it.

    Source(s): Practice, baby, practice.
  • 1 decade ago

    Well there are two parts to this proof as far as I can tell.

    The first is to show that the empty set is unique.

    To do this you just need to cite extensionality, that any two sets with the same members are equal (two sets with no members have the same members...none).

    Second you state that if a set isn't empty then it must have a member...there really is no formal justification for this that I can think of, more common sense.

    So what this gets you is that any set other than the empty set contains an element. You might have worried "What if there are two empty sets?" since that is the only time this statement would be false. But that is why you proved the uniqueness of the empty set.

    Source(s): My Brain.
  • 1 decade ago

    When you say that there are no axioms, do you mean you can't use any of the ZF axioms? How are you expected to prove anything without them?

    In case I'm misunderstanding you, this should be a direct consequence of the axiom of regularity (your book might call it the axiom of foundation). If you aren't allowed to use that, then I need to know what tools/facts/theorems/etc. you are allowed to use before I can help you.

Still have questions? Get your answers by asking now.