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Math Help - Doubt on the Axioma of Separation (Russel's Paradox)

  1. #1
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    Doubt on the Axioma of Separation (Russel's Paradox)

    How can this evade Russel's Paradox?


    If p is a property, for any X, there exists a set Y = {x E X;P(x)}

    If P(x) = x ~E x (a set x cannot contain itself) ^ Y = x

    Then

    Y E X ^ Y ~E Y

    A paradox:

    Y E X and Y ~E Y cannot coexist
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  2. #2
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    Re: Doubt on the Axioma of Separation (Russel's Paradox)

    If $P(x) \stackrel{\text{def}}{=} x \not\in x$, then since this holds for EVERY set $x$, we have that $Y = \{x \in X: P(x)\} = X$.

    Therefore, we have $Y \not\in X$, since this is the same as saying: $X \not\in X$, which is true. There is no paradox, here. $Y$ is a subset of $X$, not an element.
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