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 $"?

If $x$ is arbitrary, then it is $\forall x$, in this case, it should be "$\forall x\in A$ such that..." — Javi, Feb 18 '18 at 10:22
The second definition is the formally correct one. Often in such situations the "$\forall x\in A$" i omitted for some reason (the reason may be the following: you have "$p\implies q$" with an $x$ appearing in $p$ and $q$ but you have no information on $x$ so you "feel like it should apply to any $x$"). Most of the time it's harmless (like here, there's barely any doubt as to where $\forall x$ should go, but from time to time there's an ambiguity — Maxime Ramzi, Feb 18 '18 at 10:26
I was thinking that if we use the first definition then why do we need $x_0$ to be an adherent point as part of the definition. With the second definition it is clear that if adherent point is not part of the definition then every $L\in\mathbb R$ would become a limit for $x_0$ not an adherent point. — , Feb 18 '18 at 10:30
The first one is incorrect. I absolutely disagree with Max that there isn't any doubt about where $\forall x$ should go, just think about uniform continuity and you'll see a familiar example of a common statement with a (perhaps) unintuitively placement of universal quantifiers. In the case of simple continuity it is only "obvious" because people already known a-priori what is meant, people only know because they already knew... — Git Gud, Feb 18 '18 at 11:36
In fact, there's even no reason to assume it should be a universal quantifier (other than that abhorrent pseudo-convention that when a variable isn't quantified, it should be assumed as universally quantified). See Is "A and B imply C" equivalent to "For all A such that B, C"? for a very similar (perhaps duplicate?) question. — Git Gud, Feb 18 '18 at 11:36
Reminds me of https://math.stackexchange.com/questions/4448168/does-if-turn-a-free-variable-into-a-bound-variable — mr_e_man, Mar 08 '23 at 16:59

ryang · Answer 1 · 2023-03-08T16:40:29.547

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 $0<|x-x_0|<\delta\implies|f(x)-L|<\epsilon$.

Implication and universal quantifier

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