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I would like some advice on a few sentences, although I realize they might be too far removed from their context. This is the statement, from page 11 in Paul Cohen’s book “Set theory and the continuum Hypothesis”:

"Assume we know that $A(c)$ is a valid statement. We shall show that $\forall x A(x)$ is a valid statement (This is not the same as saying $A(c) \to \forall x A(x) $ is valid which it is not in general)."

Intuitively I make the mistake he is talking about. What is the explanation?

Let me add that I also appreciate some advice on books in set theory with more examples and extensive comments. I saw that the book was highly recommended, but it is mainly a facsimile of a typewritten historical text, very compact.

Asaf Karagila
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Mikael Jensen
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  • Without context, this is impossible to answer. I can hazard I guess, though. "Valid" might mean "true in every model". "$c$" might refer to a constant symbol in the language. "$A$" then is a formula with a single free variable; given a model (which includes an interpretation $\gamma$ of $c$), $A(c)$ then has a truth value, i.e., $![A]!$ is true. But of course, given a model with some interpretation of $\gamma$ of $c$, you can look at a nearly identical model, where only the interpretation of $c$ is different. Since $A(c)$ is valid, $![A]!$ is true as well. – Magdiragdag Nov 24 '13 at 19:42
  • So, given a model $M$ without an interpretation of $c$ (in which you cannot talk about validity of $A(c)$), you can look at the meaning of $\forall x. A(x)$. Picking an element $\gamma$ of $M$, you can build a new model $M'$ exactly like $M$ but with an interpretation of $c$, namely $\gamma$. In $M'$, $[![A(\gamma)]!]$ is true, but that claim is just as well a statement about $M$. So, whatever $\gamma$ in $M$ you pick, $[![A(\gamma)]!]$ is true, i.e., $\forall x.A(x)$ is true in $M$. – Magdiragdag Nov 24 '13 at 19:49
  • Allright, this has become an answer, rather than a comment - assuming that I guessed the context correctly. If you (@Mikael Jensen) can confirm, I'll copy it as an answer. – Magdiragdag Nov 24 '13 at 19:50
  • The concept of a model enters (only 2 pages) later, but it sounds like this must be the right type of explanation. He mentions a few pages before (as I interpret it) that for A→B to be valid requires that they are valid for any axiom that holds regarding the context given. I would also be happy to hear about a modern textbook, perhaps less advanced. This one proves Gödels theorem (for completeness and incompleteness) in chapter one and it has 4 chapters. – Mikael Jensen Nov 24 '13 at 21:20
  • I am not familiar with the particular book and am in general no expert in this area, so I can't help you with a reference. I may have terminology wrong: maybe what I call a model is properly called a structure (and a model for a given set of formulae is then a structure in which those formulae are true). – Magdiragdag Nov 24 '13 at 21:29

2 Answers2

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What is going on here is that there is some language $L$ which has, among other thing, a constant symbol $c$. $A(x)$ is a formula with a single free variable $x$ over this language $L$ and $A(c)$ is the same formula with the constant symbol $c$ substituted for $x$.

Now "$A(c)$ is valid" means that for every structure ${\mathfrak A}$ for the language $L$, the interpretation of $A(c)$ in ${\mathfrak A}$ is true, i.e., ${\mathfrak A} \models A(c)$. Note that such a structure ${\mathfrak A}$ includes an interpretation for the constant $c$.

Now look at the language $K$ which is exactly the same as $L$ except it does not have the constant symbol $c$.

In a structure ${\mathfrak B}$ for $K$, you cannot interpret $A(c)$, but you can interpret $A(x)$ if you're also given a valuation $\eta$ of all free variables of $A(x)$ (i.e., of just $x$) in ${\mathfrak B}$. It is customary to say that $A(x)$ is true in ${\mathfrak B}$ (i.e., $\mathfrak B \models A(x)$) if it is true for every valuation of (i.e., for every $\eta \colon \{x\} \to {\mathfrak B}, {\mathfrak B} \models_\eta A(x)$.) This is exactly the same as saying ${\mathfrak B} \models \forall x.A(x)$.

So "$\forall x.A(x)$ is valid" means that for every structure ${\mathfrak B}$ for $K$ and for every element $\gamma$ of ${\mathfrak B}$, $\mathfrak B \models_{(x \mapsto \gamma)} A(x)$.

Now a structure for $K$ together with a particular element $\gamma$ is exactly the same things as a structure for $L$ (using $\gamma$ to interpret $c$). This is why "$A(c)$ is valid" implies "$\forall x.A(x)$ is valid".

Magdiragdag
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The issue is this: if $c$ is a new constant symbol, then $A(c)$ being valid means that you can prove $A(c)$. That means, in turn, that for every model $M$ and every $z \in M$, $M$ will satisfy $A(z)$. Thus every model will satisfy $(\forall x)A(x)$, and so that is a valid formula if $A(c)$ is valid.

On the other hand, $A(c) \to (\forall x)A(x)$ will fail if we take any model $M$ in which $(\forall x)A(x)$ does not hold but which has some $z$ for which $A(z)$ is true. Then, if we interpret $c$ as that element $z$, the compound statement $A(c) \to (\forall x)A(x)$ will fail in $M$.

The key point is that there is a difference between $A(c)$ being provable (i.e. valid) compared to $A(c)$ merely happening to be true in some model for some particular interpretation of $c$.

Carl Mummert
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  • I'm not sure to understand your answer. If $\forall x A(x)$ does not hold in $M$ and there is a $z$ for which $A(z)$ is false, and if we interpret $c$ as that element $z$, $A(c)$ in turn will be false, and so the conditional $A(c) \rightarrow \forall xA(x)$ will be true. – Mauro ALLEGRANZA Nov 25 '13 at 13:38
  • Thank you for noticing that. I meant to say "true"; I edited the answer to fix it. – Carl Mummert Nov 25 '13 at 13:45
  • I am looking for examples and comments similar to those you have submitted. I am sure you must know some more reader-forbearing book than this compact typewritten facsimile! – Mikael Jensen Nov 27 '13 at 17:03