Clarification regarding domain definition

Question

I have been reading the fourth chapter of Velleman's How to prove book and this is the definition for domain which I have encountered:

Suppose $R$ is a relation from $A$ to $B$. The the domain of $R$ is the set

$ Dom(R) = \{ a \in A | \exists b \in B ((a,b) \in R)\}$

Now what is $ \exists b \in B ((a,b) \in R)$ in the above logical expression ?

Is it equivalent to $ \exists b \in B \land (a,b) \in R$ ?

Answer 1

For a "formal" treatment, we have to start from the concept of ordered couple : $\langle x, y \rangle$.

The we have the definition of relation :

$R$ is a relation iff $\, \forall z(z \in R \rightarrow \exists x \exists y (z= \langle x, y \rangle))$.

From it, we can define the domain $Dom(R)$ of a relation $R$ :

$Dom(R) = \{ x \mid ∃y (\langle x, y \rangle \in R) \}$.

From this, it follows that :

$a \in Dom(R)$ iff $\, ∃z∃b(z= \langle a,b \rangle \land z \in R)$.

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In general, we have to note that the "set abstraction" symbol $\{ \, \, \mid \, \, \}$ has the property :

$y \in \{ x \mid \varphi(x) \}$ iff $\varphi(y)$.

Thus, if $R$ is a relation from $A$ to $B$, i.e. $R \subseteq A \times B$, we have that :

$a \in Dom(R)$ iff $a \in A \land ∃b(b \in B \land \langle a,b \rangle \in R)$.

Answer 2

$\exists\,B\in B (a,b)\in R$. $\text{dom }R$ is the set of $a\in A$ such that for some $b\in B$, the ordered pair $(a,b)\in R$.

Answer 3

These are called "bounded quantifiers", since we wish to bound the context of the objects over which we quantify. In reality, these are shorthands:

$$(\exists x\in X)(\varphi(x))\iff\exists x(x\in X\land\varphi(x))\\ (\forall x\in X)(\varphi(x))\iff\forall x(x\in X\rightarrow x\in X)$$

Recall that $\exists x\varphi(x)$ is true if we can find $x$ for which $\varphi$ holds; so by bounding the quantifier, $(\exists x\in X)(\varphi(x))$ we require that the witness $x$ came from $X$. And similarly for $\forall x\varphi(x)$.

So here, $\exists b\in B((a,b)\in R)$ means that we are limiting our search for a witness to the set $B$.