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Math Help - Prove that a set cant be a element of itself

  1. #1
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    Prove that a set cant be a element of itself

    Prove that a set cant be a element of itself.

    I have spent hours researching this but i simply cant understand how this is impossible! How can the fact that something is true make it false? I seriously just cant understand this concept.

    Online i keep getting people referencing " the Axiom of Regularity" etc. and Honestly this is adding to my confusion.
    so if you want to help, please try not to make reference to other theorems, but if it is absolutely necessary please also add a proof for the theory.
    lastly please provide maths with English explanations.

    thanks to anyone that is willing to help !
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    Re: Prove that a set cant be a element of itself

    Quote Originally Posted by iokio View Post
    Prove that a set cant be a element of itself.
    Online i keep getting people referencing " the Axiom of Regularity"
    I will follow James M Henle's An Outline of Set Theory.
    The Axiom of Regularity: If $D$ is a set, then either $D=\emptyset$ or else & $(\exists B\in D)[D\cap B=\emptyset]$

    Now suppose that $(\exists A)[A\in A]$ Define $D=\{A\}$. Clearly that is a non-empty set.

    So by the axiom $(\exists B\in D)[D\cap B=\emptyset]$ but $D$ contains only one element so $B=A$.

    But that means $A\in (A\cap B) \text{ or }A\in (D\cap B)\ne\emptyset$.
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    Re: Prove that a set cant be a element of itself

    Thanks for your reply... but im still a little hesitant.

    in the 3rd line you state that: D∩B=∅ but then in the 4th line you say that A∈(D∩B)≠∅.

    Im confused by this... basically your saying A∈∅≠∅. what is the reasoning behind this?
    if A is an element of nothing then should A not also be nothing?
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    Re: Prove that a set cant be a element of itself

    Quote Originally Posted by iokio View Post
    in the 3rd line you state that: D∩B=∅ but then in the 4th line you say that A∈(D∩B)≠∅.
    Im confused by this... basically your saying A∈∅≠∅. what is the reasoning behind this?
    if A is an element of nothing then should A not also be nothing?
    The first line of the proof says that suppose that $A$ is a set having the property that $A\in A$.

    If $D=\{A\}$ and $B\in D$ then $B=A$ so $A\in B$ and that means $B\cap D\ne\emptyset$.

    I hope that you understand that $\emptyset\ne \{\emptyset\}$.
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    Re: Prove that a set cant be a element of itself

    Ok... so because you successfully proved that B∩D≠∅, you have opposed the Axiom of Regularity.

    So only because this defies the Axiom of Regularity, set A is not an element of A.

    is this the Logic?
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  6. #6
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    Re: Prove that a set cant be a element of itself

    Naively, if A ∈ A was possible, then we would have a LOT of sets with:

    A = {A}, so that:

    A = {A} = {{A}} = {{{A}}} =.....

    Such a set is said to be "not well founded" because we keep "taking it apart" and finding our original set inside (in other words, it has no "foundation").

    The axiom of regularity is cleverly designed specifically to prevent that; in lay terms: the buck stops "somewhere".

    Now by the axiom Plato so patiently is trying to explain, we have that A ∩ {A} = ∅, which rules out A = {A}, unless A = ∅.

    Which leaves us with just "one case" to settle: Is ∅ ∈ ∅? We can safely answer "no" to this, because the null set has no elements.
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