Is ∃x∈S(P (x)) shorthand for ∃x(x ∈ S ∧ P (x)) or ∃x(x ∈ S → P (x))?

Question

In Discrete mathematics and its applications Book

by Kenneth H. Rosen

In chapter 2 setction 2.1(page no 124 in pdf)

"Using Set Notation with Quantifiers

Sometimes we restrict the domain of a quantified statement explicitly by making use of a particular notation. For example, ∀x∈S(P (x)) denotes the universal quantification of P (x) over all elements in the set S. In other words, ∀x∈S(P (x)) is shorthand for ∀x(x ∈ S → P (x)). Similarly, ∃x∈S(P (x)) denotes the existential quantification of P (x) over all elements in S. That is, ∃x∈S(P (x)) is shorthand for ∃x(x ∈ S ∧ P (x)). "

But isn't ∃x∈S(P (x)) shorthand for ∃x(x ∈ S → P (x)) ?

If it is ∃x∈S(P (x)) shorthand for ∃x(x ∈ S ∧ P (x)),then why? Doesn't '∧'(and) has to br replaced by '→'(impiles that) ?

Answer 1

The original description seems to be correct. The difference is due to the nature of "limiting" the quantified statement to a smaller set. I will just rephrase the logic in English: For universal quantification, you want to say that every element in S satisfies P, in other words, for any x, if x is in S, then x must satisfy P. For existential quantification, you want to say that there is SOME element in S that satisfies P, in other words, there is some x such that x is in S, AND x satisfies P.

On the other hand, your proposal, ∃x(x ∈ S → P (x)) means something else. It means that there is some x such that "if x is in S, then x satisfies P". In particular, any x outside of S will satisfy the statement.

Eg, take the set S = {1, 2, 3} and the condition P(x) = x > 4. Now ∃x∈S(P (x)) should be false because there is no x in S that satisfies P. And sure enough, ∃x(x ∈ S ∧ P (x)) is false. But ∃x(x ∈ S → P (x)) is true because of the number 5. The number 5 is not in S, therefore, it satisfies "if x is in S then x satisfies P". See the truth table of implication if this is surprising to you. It's just how "implication" is defined in logic.