Set theoretic definition of cartesian product of two sets I am trying to study mathematics rigorously.It is mentioned in my book in the chapter of set theory that the symbol "$:$" stands for the phrase "such that". 
Then in the next chapter of relations, the cartesian product of two sets $A$ and $B$ is defined as follows:
If $A$ and $B$ be two sets,
$A\times B =\{(a,b):\forall (a\in A , b\in B) \}$.
If this symbolic definition is translated into words, it reads:"The cartesian product of two sets $A$ and $B$ is defined as the set of all ordered pairs $(a,b)$ such that for all $a$ belongs to $A$ and $b$ belongs to to $B$. 
My concern is about the last part in the symbolic definition, that is, "$:\forall (a\in A , b\in B)$". If this is translated into words it reads "such that for all $a$ belongs to $A$ and all $b$ belongs to to $B$". This sounds incomplete to me.
Is this correct both mathematically and grammatically? Would not it be better to define the cartesian product of $A$ and $B$ as $A\times B =\{(x,y):(x,y)=(a,b)\forall (a\in A , b\in B) \}$ or simply as $A\times B =\{(a,b):a\in A , b\in B) \}$. Among these three which is correct grammatically as well as rigorous mathematically?
 A: $$A \times B = \{(a,b): a \in A, b \in B\}$$
$A \times B$ is the set of all ordered pairs $(a,b)$ where $a$ belongs to $A$ and $b$ belongs to $B$.
A: Formally speaking (since you ask for a "set theoretic definition") the cartesian product cannot be defined: it needs its own axiom in set theory. 
You cannot define sets with general formulas, otherwise you run the risk of running into Russell's paradox:
$$
R = \{x\colon x \not \in x\}.
$$
To avoid these kinds of open definitions one can only define subsets with a formula, as in:
$$
A = \{x \in B\colon \phi(x)\}.
$$
So the proposed definition (which is correct in naive set theory)
$$
A \times B := \{(a,b) \colon a \in A, b \in B\}
$$
is not actually a valid definition.
The solution is to say that $A \times B$ exists by assumption (precisely the axiom of pairing in ZF and most importantly, as pointed out in the comments, the axiom of power set) and satisfies the following relation (updated):
$$
x \in A \times B \iff \exists a \in A, \exists b\in B\colon x = (a,b).
$$
(added) I write here the nice formal definition given by @daniel-schepler in the comments: assuming you define $(a,b):=\{\{a\},\{a,b\}\}$, then actually you would say 
$$
A \times B := \{ p \in \mathcal P(\mathcal P(A \cup B))\colon \exists a\in A, \exists b\in B\colon p=(a,b)\}.
$$
A: The issue is of that of free variables. When you write $\{x : \varphi\}$, it is expected that $x$ is the only free variable of $\varphi$ (there might be parameters fixed in advance).
Once you put a quantifier on $a$ and $b$, they are no longer free, and the formula is now a sentence which is either "always true" or "always false". So the set is either everything or empty.
To add insult to injury, $\forall(a\in A, b\in B)$ is not a well-formed formula to begin with. The comma, while not a valid logical symbol is often taken to mean "and" in the set-builder context.
So indeed the correct way to write this would be, as others and yourself have suggested $$A\times B=\{(a,b): a\in A, b\in B\}.$$
A: This looks a bit like a typo.
$A\times B =\{(a,b):\forall (a\in A , b\in B) \}$ does not make sense, your suggested solution
$$
A\times B =\{(a,b): a\in A , b\in B \}
$$
is the right solution.
Note that
$A\times B =\{(x,y):(x,y)=(a,b)\forall a\in A , b\in B \}$
is not the same!
