Implication and universal quantifier

Question

Given the statement $P(x,\delta)\implies Q(x,\epsilon),$ is it necessarily assumed that all the variables are properly quantified?

For example is the following definition correct:

Let $A\subset \mathbb R$ and $f:A\to\mathbb R$ be any function. Let $x_0$ be any adherent point of $A$. We say that $\lim_{x\to > x_0}=L$ iff $\forall \epsilon>0 \exists\delta>0$ such that $0<|x-x_0|<\delta\implies|f(x)-L|<\epsilon$.

or is the following definition correct:

Let $A\subset \mathbb R$ and $f:A\to\mathbb R$ be any function. Let $x_0$ be any adherent point of $A$. We say that $\lim_{x\to > x_0}=L$ iff $\forall \epsilon>0 \exists\delta>0$ such that for all $x\in A$ such that $0<|x-x_0|<\delta\implies|f(x)-L|<\epsilon$.

What issue occurs if we omit the "for all $x\in A $"?

Answer 1

Let $A\subset \mathbb R$ and $f:A\to\mathbb R$ be any function. Let $x_0$ be any adherent point of $A$. We say that $\lim_{x\to x_0}=L$ iff $\forall \epsilon>0 \exists\delta>0$ such that $0<|x-x_0|<\delta\implies|f(x)-L|<\epsilon$.

Given the statement $P(x,\delta)\implies Q(x,\epsilon),$ is it necessarily assumed that all the variables are properly quantified?

Adding to Git Gud's comments: while it is fair to assume that $P(x,\delta)\implies Q(x,\epsilon)$ is implicitly universally quantified, this doesn't apply to the above statement, as it exhibits one of its variables as existentially quantified; without knowledge of mathematical analysis, it is not immediately clear where the "for each $x\in A$" is meant to be inserted. With knowledge of mathematical analysis, then it is clear that the correct definition is the second one:

Let $A\subset \mathbb R$ and $f:A\to\mathbb R$ be any function. Let $x_0$ be any adherent point of $A$. We say that $\lim_{x\to x_0}=L$ iff $\forall \epsilon>0 \exists\delta>0$ such that for all $x\in A$ such that $0<|x-x_0|<\delta\implies|f(x)-L|<\epsilon$.