• Tommy
    13
    Suppose set A = {a, b} the subsets of this set are:

    {{0}, {a}, {b}, {a, b}} Now redefine set A as:

    A\A. That is, A\{a, b}

    Now the subsets of A are:

    {{0}, {a}, {b}}

    This is what I mean by X\X.
    EnPassant

    Ah, so this is not the correct notation. You have asked us to consider the set , where is the power set of . Note that we cannot redefine X and that this set is quite distinct from itself.
  • ssu
    7.9k
    I don't know enough to say but as far as I can see this question can be resolved with simple set theory alone - if my ideas are coherent that is.EnPassant
    Well, Russell himself use Type Theory and basically Zermelo-Fraenkel Set Theory (ZF) was made basically to avoid Russell's paradox. If you find your ideas resembling theirs, you can be proud of yourself.
  • EnPassant
    665
    Ah, so this is not the correct notation...
    Note that we cannot redefine X and that this set is quite distinct from itself.
    Tommy

    Yes, this is the difficulty with notation but it is really just a notational wrinkle. You can rename the set X' and you have it. All that is required is a set without X as a subset but with all the other sets. The same logic can then be applied to X'2 and so on. It is really a question of getting the notation right but the concept seems to be coherent.

    Well, Russell himself use Type Theory and basically Zermelo-Fraenkel Set Theory (ZF) was made basically to avoid Russell's paradox. If you find your ideas resembling theirs, you can be proud of yourself.ssu

    There are immense complexities with infinite sets so I'm very unsure if my logic holds...
  • ssu
    7.9k
    Well, start from the fact that NOBODY knows what actually infinity is, so don't be too harsh on yourself. :up:
  • EnPassant
    665
    Russell suggested we consider the collection "the set of all sets which are not subsets of themselves". Note that, by Frege's abstraction principle, this is necessarily a set. Asking that we, in effect, look the other way and consider instead another set, as you've proposed, doesn't prevent us from considering Russell's set.Tommy

    If I understand you correctly, you are saying "the set of all sets which are not subsets of themselves" is necessarily a set. But it turns out that it is a pathological set or not really a set at all. My argument is that it is an entity of some kind and I'm attempting to define what this entity is. It seems that there can be things that are not simple sets but are infinities of sets. Maybe all sets are really subsets of infinite sets...
  • Tommy
    13
    If I understand you correctly, you are saying "the set of all sets which are not subsets of themselves" is necessarily a set.EnPassant

    If we accept Frege's abstraction principle, then yes, we are committed to the position that this is a set. As you have pointed out, Russell's Paradox illustrates that such a notion is inconsistent, which is why no one accepts Frege's abstraction principle.

    You seem quite motivated to learn more about this topic (which is a very good thing! :up: ). I really recommend you check out the book I mentioned in my first post. It is actually quite good and the first few chapters are quite approachable. It's Dover so its also extremely affordable. A worthwhile investment.

    If the mathematical formalism is difficult at first, then you might want to try Halmos' Naive Set Theory. While I own a copy, I can't honestly say I've sat down and spent any real time with it, but I understand it is a very good place to start.

    Best of luck! Hope to see more of you in the forums.
  • EnPassant
    665
    I really recommend you check out the book I mentioned in my first post. It is actually quite good and the first few chapters are quite approachable.Tommy

    I'll definitely check that one out. Thanks.
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