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I have been reading Halmos' book, Naive Set Theory and while reading the part about Russell's paradox I had the following question: Halmos shows that nothing can contain everything, as he puts it, but what if the only thing that cannot be contained is this one problematic set and the set containing this set and so on and so forth? What I am asking then is, do we have any proof showing what is not containable or do we just know that not containable things exist?

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  • $\begingroup$ The proof is based on the Axiom of Specification and mimicks the well-known argument used for the Russell’s Paradox. But now the conclusion is that the "set $A$" to which Separation is applied does not contain the "Russell's set" $B$. This means that : for very set $A$ there is at least a subset $B$ of it that does not belongs to $A$. $\endgroup$ – Mauro ALLEGRANZA Mar 26 at 15:11
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    $\begingroup$ You have to take care about the difference between the two relation element : $\in$ and subset : $\subseteq$. What the proof shows is that $B \subseteq A$ but $B \notin A$. $\endgroup$ – Mauro ALLEGRANZA Mar 26 at 15:12
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    $\begingroup$ The proof is exactly : for every $A$, the subset $B = \{ x \in A \mid x \notin x \}$ is not an element of $A$. This means that $B$ is not "contained into" (an element of) $A$. $\endgroup$ – Mauro ALLEGRANZA Mar 26 at 15:14
  • $\begingroup$ But there are different versions of set theory where the Universal set does exist. $\endgroup$ – Mauro ALLEGRANZA Mar 26 at 15:18
  • $\begingroup$ See my answer here. (And generally the questions and answers on that page.) $\endgroup$ – Asaf Karagila Mar 26 at 15:33
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The approach of modern set theory is not merely to say, "I can prove this set can't be done, but I can't prove this one can't be", or even, "I can prove no contradiction would result from this set". Unfortunately, some "non-paradoxical" sets contradict each other, like an irresistible force meeting an immovable object.

Instead we start from some axioms that say either "this set exists" or "sets have these properties". Which axioms you start from varies by mathematician, but ZF or ZFC is the most popular choice. These axioms are sufficient to refute some hypothetical sets.

What would be an example of the force/object analogy? ZFC, if consistent, can neither prove nor disprove either of the following claims:

  • $\Bbb R$ is the same size as $\omega_1$;
  • $\Bbb R$ is the same size as $\omega_2$.

Needless to say, "a set the same size as both $\Bbb R$ and $\omega_1$" and the $\omega_2$ counterpart to that can't both exist.

ZFC can show, however, that some ordinal $\alpha$ is the same size as $\Bbb R$. For that ordinal, there is no set of sets of both that size and the size of $\Bbb R$, because there are "too many" such sets for one set to contain them. On the other hand, for any other ordinal this construction would just give us the empty set, which is fine. So ZFC implies there is exactly one ordinal $\alpha$ for which $\{x|x\approx\alpha\land x\approx\Bbb R\}$ is not a set. (Which ordinal that is depends on the model of ZFC.)

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    $\begingroup$ Forces are usually unstoppable, not irresistible. $\endgroup$ – Asaf Karagila Mar 26 at 15:31
  • $\begingroup$ @AsafKaragila Either's fine. $\endgroup$ – J.G. Mar 26 at 16:06

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