Set Theory: Conventions Dealing with Quantifiers and Set Builder

Question

Let's say I have a set $A=\lbrace x^2|x\in \mathbb{R}\rbrace$. What is more correct to say: option $(1)$, $(2)$, or both are fine?

$(1)$ If $y\in A$, then $y=x^2$ for some $x\in \mathbb{R}$.

$(2)$ If $y\in A$, then $y=x^2$ and $x\in \mathbb{R}$.

I personally write option $(1)$ all the time in my proofs, but I wonder if that is a bad habit.

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I know that the set $A$ can also be rewritten as $A=\lbrace y| y=x^2 \text{ and } x\in \mathbb{R}\rbrace$ too, but I still am unsure what is more correct.

In logic terms, is the following correct?

$y \in A \leftrightarrow \exists x \in \mathbb R (y = x^2)\leftrightarrow x \in \mathbb R \text{ and } y=x^2$

Answer 1

In answer to your edited question, you cannot simply ignore the existential quantifier. While this is a bit picky, it winds up being important down the road to explicitly distinguish between free and bound variables. "$y=x^2\wedge x\in\mathbb{R}$" is not an assertion about just $y$, it's an assertion about $y$ and $x$; by contrast, "$\exists x(y=x^2\wedge x\in\mathbb{R})$" (or "$\exists x\in\mathbb{R}(y=x^2)$") is just an assertion about $y$ (as desired) since the new variable $x$ is bound by the existential quantifier.

Now it is true that in this specific context dropping the "$\exists x$" will probably not lead to confusion, but it is in fact not correct to do so.

Answer 2

If I understand correctly you are basically asking on whether $$ y \in A \iff \exists x \in \mathbb R (y = x^2)$$ or $$y \in A \iff \exists x (y=x^2 \land x \in \mathbb R)\ .$$

If that is the case you are in luck, because, as long as you are working in the framework of (untyped) classical logic the two statements are equivalent, indeed in general a formula of the form $$\exists x \in A \phi(x)$$ is actually a short hand for $$\exists x (x \in A \land \phi(x))\ .$$

Answer 3

All of them are correct.

Your $(1)$

If $y∈A$, then $y=x^2$ for some $x∈R.$ is my preference.