Which symbol do I use for this statement?

Question

I'm trying to write $w \subseteq u \times v$. I've turned the expression into $\forall z \; (z \in w \rightarrow z \in u \times v)$

For $z \in u \times v$, I've rewritten this as $\exists x \; \exists y \; (x \in u \; \land y \in v ** \; z = (x,y))$

The ** is where I don't know which symbol to put. Would it be $\rightarrow or \leftrightarrow$? I'm thinking it should be $\leftrightarrow$ since it seems like $x \in u \; \land y \in v $ are only true if and only if $z = (x,y)$.

Is this correct?

Note: I will be converting $z = (x,y)$ further so it follows first order set theory rules.

Answer 1

The symbol you are wondering about is $\land$.

You want to say that there exists some $(x,y)$ such that $z=(x,y)$, but you also want to make sure that $x\in u$ and $y\in v$, so you want an expression that satisfies all three of these conditions.

If you want to satisfy several conditions at once, you join them with $\land$, so you get

$$\exists x\exists y:(x\in u\land y\in v\land z=(x,y))$$

This is usually shortened to

$$\exists x\in v, y\in u:z=(x,y)$$

Answer 2

To elaborate, what you want is probably a formalization of bounded qunantifiers $$\tag1\forall x\in a\colon \phi(x) $$ and $$\tag2\exists x\in a\colon \phi(x)$$ Those are $$\tag{1'}\forall x(x\in a\color{red}\to \phi(x)) $$ and $$ \tag{2'}\exists x(x\in a\color{red}\land \phi(x))$$ The strange differenes may seem confusing, but that's what it is. Maybe it gets less confusing if you check that the negation of $(x\in a\to \phi(x))$ is $(x\in a\land\neg\phi(x))$ so that the usual rules of $\neg \forall=\exists \neg $ also hold for bounded quantifiers