Axiomatic set theory My question is about Axiomatic set theory. Concerning formulas with free variables, we adopt the notational convention that all free variables of a formula
$\phi(u_1, . . . , u_n)$:
What's this $\phi(u_1, . . . , u_n)$ and $ u_1, . . . , u_n$?
I do not understand. 
And that Axiomatic set theory : Do the same Classes
 is set??   explain more!!
thank you very much
 A: When we write $\phi(u_1, \ldots, u_n)$ what we mean is that $\phi$ is a well-formed formula in your given language and that at most the variables $u_1, \ldots, u_n$ appear as free variables in $\phi$. Intuitively this mean that we can plug values in $u_1, \ldots, u_n$ and then evaluate the resulting statement. More precisely, we can take any sequence $(x_1, \ldots, x_n)$ of elements of our universe and then replace every free occurrence of $u_i$ in $\phi$ with $x_i$, for $i = 1, \ldots, n$. We write, depending on the author,


*

*$\phi(u_1, \ldots, u_n)[x_1, \ldots, x_n]$,

*$\phi(u_1, \ldots, u_n)[x_1/u_1, \ldots, x_n/u_n]$,

*$\phi[x_1/u_1, \ldots, x_n/u_n]$,

*$\phi[x_1, \ldots, x_n]$, or

*...
for the resulting statement. Consult your textbook for the precise notation it uses.
Let us consider some examples:


*

*Let $\phi \equiv u_1 = u_2$. Then we would write $\phi(u_1,u_2)$, since both $u_1$ and $u_2$ occur freely in $\phi$ and those are the only free variables. If $x_1, x_2$ are elements of our universe, then
$$\phi(u_1,u_2)[x_1,x_2] \equiv x_1 = x_2. $$
This statement is true if $x_1 = x_2$ and false otherwise.

*Let $\phi  \equiv \exists u_1 \colon u_1 = u_2$. This formula has two variables, $u_1$ and $u_2$, but only $u_2$ occurs freely. So we would write $\phi(u_2)$. Now for any $x_2$ in our universe we obtain the statement
$$
\phi(u_2)[x_2] \equiv \exists u_1 \colon u_1 = x_2
$$
and this statement is always true (let $u_1 := x_2$).

*A variable can occur both freely and bounded. Let $\phi \equiv u_1 = u_2 \vee \exists u_1 \colon u_1 = u_2$. $u_1$ occurs both freely and bounded and $u_2$ occurs freely. No matter whether it occurs also bounded, since $u_1$ occurs freely we write $\phi(u_1, u_2)$. In this case, however, there is some annoying detail we have to look at. Given $x_1, x_2$ in our background universe, we want to evaluate $\phi(u_1,u_2)[x_1,x_2]$. To do so, we only replace the free occurrences of $u_1$ with $x_1$, so the resulting statement is
$$
\phi(u_1, u_2)[x_1, x_2] \equiv x_1 = x_2 \vee \exists u_1 \colon u_1 = x_2.
$$
We would now like to replace the remaining $u_1$ with $x_2$ to see that this statement is always true. However, this seems ambiguous. Should we replace the instances of $u_1$ that we already plugged $x_1$ into? So should the result be $x_2 = x_2 \vee \exists x_2 = x_2$ or should it be $x_1 = x_2 \vee \exists x_2 = x_2$? The latter is correct. More precisely, when a given variable $u_i$ occurs both freely and bounded, replace all the bounded instances of it with a variable that hasn't been used thus far. So instead of
$$
\phi(u_1,u_2) \equiv u_1 = u_2 \vee \exists u_1 \colon u_1 = u_2
$$
we let
$$
\phi(u_1,u_2) \equiv u_1 = u_2 \vee \exists u_3 \colon u_3 = u_2.
$$
Regarding your second question: With every formula $\phi(u_1, \ldots, u_2)$ we can associate a (virtual) class
$$
C_{\phi(u_1, \ldots, u_n)} := \{ (x_1, \ldots, x_n) \mid \phi(u_1, \ldots, u_n)[x_1, \ldots, x_n] \text{ is true} \}
$$
(we usually omit the "is true").
For example, let $\phi(u_1) \equiv u_1 = u_1$. Then
$$
C_{\phi(u_1)} = \{ x_1 \mid x_1 = x_1 \}
$$
is the virtual class of all sets. This (in most commonly used set theories) is not a set. On the other hand, if $x$ is a set, then we can consider $\phi(u_1) \equiv u_1 \in \dot c$, where $\dot c$ is a constant symbol we added to our language. If we now let $\dot c = x$ in our interpretation, then
$$
\begin{align*}
C_{\phi(u_1)} &= \{ x_1 \mid \phi(u_1)[x_1] \} \\
&= \{ x_1 \mid x_1 \in \dot c \} \\
&= \{ x_1 \mid x_1 \in x \} \\
& = x,
\end{align*}
$$
so every set can be represented as a virtual class in a canonical way. (Here I assume that extensionality is an axiom of our set theory, but this is the case for any set theory that you should encounter.)
