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Proof: (related to "There is no set of all sets")

  • We have seen in the above explanation how the axiom of separation avoids the Russel’s Paradox.
  • In particular, we have seen that for every set $X$ the set $\{z\in X\mid z\not\in z\}$ is well-defined (exists).
  • Therefore, if a “set of all sets” existed, it would have to contain $\{z\in X\mid z\not\in z\}$ as an element.
  • But for all $X$ we have $\{z\in X\mid z\not\in z\}\not\in X.$
  • Therefore, a “set of all sets” does not exist.
q.e.d

| | | | Contributors: bookofproofs | References: [656], [983]


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Bibliography (further reading)

[983] Ebbinghaus, H.-D.: “Einf├╝hrung in die Mengenlehre”, BI Wisschenschaftsverlag, 1994, 3

[656] Hoffmann, Dirk W.: “Grenzen der Mathematik – Eine Reise durch die Kerngebiete der mathematischen Logik”, Spektrum Akademischer Verlag, 2011

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