I am reading An Introduction To Set Theory by Professor William A. R. Weiss, and during chapter 3, on the proof of $\forall x \forall y \exists z \ z = x\times y$ (the euclidean product exists), there's the following:
We obtain: $$\forall t \in y \ \exists q \ q=\left\{ v : \exists u \in x \ v=\langle u,t \rangle \right\}$$ By Extensionality, in fact, $\forall t \in y \ \exists ! q \ q=\left\{ v : \exists u \in x \ v=\langle u,t \rangle \right\}$.
This step seems intuitively clear, but it also doesn't seem like an special case, it seems like this logic could be applied to any set built using builder notation.
So my question is: For any arbitrary set $x$ and any abritrary formula $\Omega$, does $\exists q \ q=\left\{ v:\Omega \right\}$ imply, by Extensionality, that $\exists ! q \ q=\left\{ v:\Omega \right\}$?