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
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.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
Ah, so this is not the correct notation...
Note that we cannot redefine X and that this set is quite distinct from itself. — Tommy
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
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. — EnPassant
Get involved in philosophical discussions about knowledge, truth, language, consciousness, science, politics, religion, logic and mathematics, art, history, and lots more. No ads, no clutter, and very little agreement — just fascinating conversations.