$A(c) \to \forall x A(x) $ not valid even though $A(c)$ can be used to prove $\forall x A(x)$ 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.
 A: 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".
A: 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 $c$.
