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On page $13$ of his book Set Theory and the Continuum Hypothesis Paul Cohen writes:

The point of these definitions is the following obvious fact:

THEOREM $1$. ... If a set of statements $S$ has a model then it is consistent.

Sadly, I'm unable to prove this obvious fact... Thank you very much in advance for your help!

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Pierre-Yves Gaillard
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Suppose that $S$ was not consistent, let $\psi$ a statement which is both provable and refutable from $S$. And let $M$ be a model of $S$.

By the definition of the satisfaction relation, $M\models\varphi$, we know that if $S$ proves $\varphi$, then $M\models\varphi$. In particular $M\models\psi$ and $M\models\lnot\psi$. But this is a contradiction to the definition of the truth.

What have we used here? We used the fact that $M\models\varphi$ is defined in a certain way, and that $M$ cannot satisfy both a statement and its negation, which follows from the definition by induction.

Asaf Karagila
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  • Dear Asaf: Thanks! I don't understand why "$M$ cannot satisfy both a statement and its negation", and why this "follows from the definition by induction". For instance, in Part $1$ of Cohen's Definition p. $12$, why can't we have $\overline x_i=\overline x_j$ and $\overline x_i\neq\overline x_j$? – Pierre-Yves Gaillard Oct 25 '14 at 05:45
  • Pierre, I don't have Cohen's book, so I can't quite help you by looking at it. But I can tell you that you have to assume at least that the logic you are working with is sound, and that no object is both equal and different from itself. You know, the little stuff. – Asaf Karagila Oct 25 '14 at 05:50
  • If you want to work in a logic where $x_i=x_j\land x_i\neq x_j$, then perhaps classical first-order logic is not the right choice. – Asaf Karagila Oct 25 '14 at 05:54
  • Dear Asaf: Thanks for your comments! Do you think it might be possible that Cohen assumes implicitly the consistency of Zermelo-Fraenkel Set Theory? – Pierre-Yves Gaillard Oct 25 '14 at 06:04
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    @Pierre-YvesGaillard: By definition, $M\vDash \neg\varphi$ exactly when $M\vDash\varphi$ is not true. So we cant have $M\vDash\neg\varphi$ at the same time as $M\vDash\varphi$, because that would mean that $M\vDash\varphi$ is both true and not-true at the metalevel. Of course we're assuming that whatever reasoning we're using at the metalevel (which need not be ZF) is sound. – hmakholm left over Monica Oct 25 '14 at 06:09
  • @Henning - Thank you dear Henning! It is also my understanding of Cohen's wording that ZF is not the only option, but since it is an option I'd rather stick to it. (Do you use the word sound as a synonym for consistent?) On p. 71 Cohen writes: "THEOREM. If ZF is consistent, so is ZF + Axiom of Regularity." Can't this suggest that, in his mind, the consistency of ZF is not always viewed as an automatic assumption? – Pierre-Yves Gaillard Oct 25 '14 at 07:03
  • @Pierre: Since the axiom of regularity is part of $\sf ZF$, I'm not quite sure what you mean by that. In either case, the consistency of set theory is not automatically assumed when working in set theory, since if $\sf ZFC$ is consistent then there are models of $\sf ZFC+\lnot\operatorname{Con}(ZFC)$, and inside those models $\sf ZFC$ is not consistent (although the universe of sets, and this model will disagree on what exactly constitute as the axioms of $\sf ZFC$). – Asaf Karagila Oct 25 '14 at 07:14
  • Dear Asaf - Sorry if my comment wasn't clear. I'm happy that you wrote "the consistency of set theory is not automatically assumed when working in set theory". I was under the (false) impression that you were making an implicit consistency assumption in your answer. Again, I'm most grateful for your answer, I don't doubt that it is outstanding, but I'm afraid I'm not clever enough to understand it... – Pierre-Yves Gaillard Oct 25 '14 at 07:52
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    @Pierre: The consistency of ZF in particular is not an automatic assumption, but like any other mathematician Cohen implicitly assumes that mathematical reasoning is possible at all, and then offers an informal argument in the hope that the reader will consider it valid. Some readers will base their validity judgment on whether Cohen's argument can be formalized in their foundational system of choice, for others it will be a more intuitive matter. (After all, mathematicians made decisions about whether they found arguments valid for many centuries before Cantor and Zermelo ... (contd) – hmakholm left over Monica Oct 25 '14 at 13:23
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    (contd) ... came along, so having a formal system in mind is clearly not a prerequisite for this). In any case, which foundational system the reader prefers is not something Cohen decides. If someone reads his book with a narrow conception of what's a valid mathematical argument, and therefore rejects it, that's just too bad. – hmakholm left over Monica Oct 25 '14 at 13:26
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    Slight correction to the above: Cohen does pledge (on p. 54) to stick to reasoning that can be expressed in ZF unless otherwise noted, but that's more of a convenient shorthand for which demands on the reader's credulity he's going to make. It still makes sense to distinguish between where he's assuming Consis(ZF) as a technical statement and where he's just implicitly depending on his own mathematical reasoning being possible. – hmakholm left over Monica Oct 25 '14 at 13:33
  • @Asaf: Cohen's formulation of ZF does include Regularity, but on p. 70 he temporarily excludes it: "By ZF we shall now mean ZF without Regularity and either with or withour the Axiom of Choice since this does not affect the discussion". – hmakholm left over Monica Oct 25 '14 at 13:39
  • @Pierre: By "sound" I mean that we can trust the result of the reasoning -- say, if our reasoning proves a statement about the integers, then it will also be Platonically true. Being consistent is a necessary but not sufficient condition for this. – hmakholm left over Monica Oct 25 '14 at 13:41
  • @Henning: Thanks, and thanks for all the great comments! – Asaf Karagila Oct 25 '14 at 13:45
  • @Henning - Dear Henning: Thank you very much for your comments. I'll think quietly about them. For no good reason, here is one more quotation (p. 41): "Now a statement which is provable in $Z_1$ is of course true in the model of integers. Also since $Z_1$ is consistent not both $A$ and $\sim A$ are provable." – Pierre-Yves Gaillard Oct 25 '14 at 16:31
  • @Henning - Dear Henning: The above quotation suggests that Cohen viewed Peano arithmetic as obviously consistent. But (as I'm sure you know) Edward Nelson tried (in 2011) to prove the inconsistency of Peano arithmetic. I have some trouble understanding how these two facts can be compatible. – Pierre-Yves Gaillard Oct 26 '14 at 10:05
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    @Pierre: Can you make sense in the fact that some people believe in one god, and other people believe in another, and some people deny the existence of any deity whatsoever? – Asaf Karagila Oct 26 '14 at 11:12
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    @Pierre: Um, Cohen and Nelson were different persons. They had different opinions about whether or not PA is obviously true. What is there to understand about that? -- And kindly stop "dear"ing me. – hmakholm left over Monica Oct 26 '14 at 11:16