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I was reading Jech's Set theory; there after introducing Russell's Paradox, he asserted:

The safe way to eliminate paradoxes of this type is to abandon the Schema of Comprehension and keep its weak version, the Schema of Separation...

Once we give up the full Comprehension Schema, Russell’s Paradox is no longer a threat...

Replacing full Comprehension by Separation presents us with a new problem. The Separation Axioms are too weak to develop set theory with its usual operations and constructions. Notably, these axioms are not sufficient to prove that, e.g., the union $X \cup Y$ of two sets exists, or to define the notion of a real number.

Why did he say the Separation axioms are too weak 'to develop set theory with its usual operations and constructions'?

Maybe it might be trivial, but I'm not able to comprehend the point. Can anyone shed some light on the author's statement?

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    $\begingroup$ The answer is in the quote you give, "Notably, these axioms are not sufficient to prove that, e.g., the union $X\cup Y$ of two sets exists, or to define the notion of a real number." $\endgroup$ Jul 5, 2016 at 12:39
  • $\begingroup$ I guess it's like the fable of Goldilocks. "This axiom schema is too strong... this axiom schema is too weak... and these axiom schemas are just right!" $\endgroup$
    – hardmath
    Jul 5, 2016 at 16:36

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Suppose that you work with the following naive theory:

  1. Extensionality, so you can say when two sets are equal;
  2. Comprehension, so you can say that if you can describe a collection, then it is a set; and
  3. Choice, if you fancy doing mathematics with the axiom of choice.

Comprehension does not restrict what it means to be a property. So it includes things like "$x$ is a subset of $A$", which gives you the power set axiom; it includes "$x$ is a member of a member of $A$", which gives you union, and so on and so forth.

But oh no, Comprehension is inconsistent. So we replace it with its weaker counterpart, Separation. But now we cannot guarantee the existence of power sets, pairs, unions, and the nice things we had before. Which is why we need to include them in our axioms. To see that, simply note that $\{\varnothing\}$ is a model of Extensionality+Separation+Choice. But not a model of Power. And $\{\varnothing,\{\varnothing\}\}$ is also a model of Extensionality+Separation+Choice, but it is not a model of Pairing.

And you can consider this as an exercise to try and push this further and construct a model of Extensinoality+Separation+Choice, where some $X$ and $Y$ exist in the model, but $X\cup Y$ does not.

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  • $\begingroup$ The reader should note that one has to be a bit careful how to formalize choice in Asaf's post. $\{ \emptyset , \{ \emptyset \} \}$ models the assertion that every set of disjoint sets admits a selecting set, but it doesn't model that every set has a well-order or a choice-function. $\endgroup$ Jul 7, 2016 at 17:47
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    $\begingroup$ @Stefan: Also every set of non-empty sets has a choice function. Mainly because every set is either empty, or all its members are empty. $\endgroup$
    – Asaf Karagila
    Jul 7, 2016 at 17:53
  • $\begingroup$ Right... Also, every set has a strict well-order (; $\endgroup$ Jul 7, 2016 at 17:56
  • $\begingroup$ In fact, $\emptyset$ is a model of Extensionality+Separation+Choice: you can't prove there are any sets at all! $\endgroup$ Jul 15, 2016 at 4:42
  • $\begingroup$ @Eric: Only if you allow empty models. I rather not. $\endgroup$
    – Asaf Karagila
    Jul 15, 2016 at 4:45
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I think he just means that simply replacing Full Comprehension with Restricted Comprehension is not enough. Given sets $X$ and $Y$, you could infer the existence their union $X\cup Y$ using Full Comprehension alone. You cannot do that with Restricted Comprehension, so you would also need an axiom of unions (as in ZFC).

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    $\begingroup$ The "$\cap$" should be a "$\cup$" - we can in fact get intersections from Separation. $\endgroup$ Jul 5, 2016 at 17:30

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