re-reading some of the Shoenfield's book, "Mathematical logic", I noticed a statement incomprehensible. The offending statement is on page 89, where it says:

With the result mentioned in (iv), this give a new proof of completness of ACF.

If, kindly, someone in possession of the text could take a look (pages 86-89), should detect the strangeness that I mentioned. For this I am curious to hear independent opinions.

(Note: just look at the pages I mentioned taking the definitions and results presented, i.e. there is no need to go into details of the demonstrations.)

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    $\begingroup$ In both the statement on p. 89 and (iv) on p. 88 you should interpret $ACF$ as each $ACF(n)$. $\endgroup$ – Brian M. Scott Jan 17 '13 at 18:35
  • $\begingroup$ thank you very much for confirming my doubt and clarification of the way in which the expression of the text to be read (and should have been written - I think it is a mistake of writing the text, ACF is one thing, ACF(n) is another one). That could not be ACF was clear, and I had thought of ACF(n), but the fact that there was ACF in (iv) and after the theorem, gave me serious doubts (a double misprint seemed unlikely). So thank you again. Mr Blass has obviously assimilated ACF and ACF(n), and therefore did not detect the underlying problem to my question: it is impossible completeness of ACF. $\endgroup$ – Bento Jan 17 '13 at 19:30

I see nothing strange here. The result [about ACF] mentioned in (iv) [on page 88] is that ACF is categorical in all uncountable powers but not in $\aleph_0$. Then on page 89, the Los-Vaught theorem says that categoricity in even a single infinite power implies completeness provided the theory is consistent and has only infinite models. Since ACF is certainly consistent (the complex numbers form a model) and has only infinite models (well known from algebra), this gives a proof of completeness of ACF. It's a new proof in the sense that it's different from the proof by quantifier elimination already given on page 86.


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