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I was reading Halmos Naive Set Theory, Chapter 2 The Axiom of Specification, in which after stating the Russel Paradox, he goes on to say "The moral is that it is impossible, especially in mathematics, TO GET SOMETHING FOR NOTHING. To specify a set it is not only enough to pronounce some magic words; it is also necessary also to have at hand a set to whose elements the magic words apply."

This is after he shows that a universal set cannot exist ( in the naive perspective of set theory of which this book is about).

What is bothering me, is the statement GET SOMETHING FOR NOTHING. I cannot seem to understand what he means there. If anyone has read the book and could explain it it would be a great help .

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I haven't read the book. But the context is quite clear.

It might be a naive belief that stating a property will grant you a set of all objects with that property. Like the property if being a set. Or being a set which is not contained in itself.

This very naive belief turns out to be inconsistent, as shown by Russell and others in the early days of set theory. So the solution was to restrict this process.

Specifying a property does not grant there is a set of all objects with the property. But having a pre-existing set $A$ and a property does grant us a set of all the elements which are in $A$ and also have said property.

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    $\begingroup$ Thank you very much. But I am afraid I cannot appreciate the subtle distinction here. Doesn't the Axiom of Specification state that for every sentence P there corresponds a set B of whose elements have the property P ? $\endgroup$ – SaitamaSensei Jan 14 '16 at 13:16
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    $\begingroup$ No. It says that given a set $A$ and a property $P$, there is a set $B$ whose elements are exactly the elements of $A$ which satisfy $P$. So there is no set of all sets, but given a set $A$, there is a set of all the sets which belong to $A$. $\endgroup$ – Asaf Karagila Jan 14 '16 at 13:19
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    $\begingroup$ Ahhhh.....now I think I get it.....the fact that a set has to be first given and then a statement of its elements which satisfy the statement can constitute a set...am I right? $\endgroup$ – SaitamaSensei Jan 14 '16 at 13:44
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    $\begingroup$ Yes, that is the point. You can't create a set out of nothing; you can only create sets from the axioms and previously existing sets. $\endgroup$ – Asaf Karagila Jan 14 '16 at 13:45
  • $\begingroup$ Thanks a lot my friend.....you helped me greatly :) $\endgroup$ – SaitamaSensei Jan 14 '16 at 13:45

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