Does Every Set in the Axiom of Choice Include an Empty Set?

[kidsasd987]

Following theorems are congruent(a) Axiom of Choice

(b) if ∀i:i∈I: <Yi | i∈I > → Yi≠Ø

(c) Ø∉S → ∃f: f is on a set S
s.t. f(X)∈X for all X∈S. where f is choice function of S.
I am confused with the theorem (c), as how the Collection S does not include empty set.
I believe every set needs to include an empty set as its element?

Can anyone please help me figure out this?

 

[TeethWhitener]

I believe every set needs to include an empty set as its element?

The empty set is a subset of every set, but not necessarily an element.

 

[math771]

Hi there,

The theorem (c) is referring to the Axiom of Choice, which states that given a collection of non-empty sets, there exists a function that can choose one element from each set in the collection. This function is called a "choice function."

In this context, the statement is saying that if the collection S does not include the empty set, then there exists a choice function that can choose one element from each set in S. This means that even if S does not include the empty set as an element, the Axiom of Choice still holds true.

To understand why this is the case, it's important to remember that the Axiom of Choice is a fundamental principle in set theory, and it allows us to make certain assumptions and conclusions about collections of sets. It does not necessarily mean that every set must include the empty set as an element.

I hope this clarifies things for you. Let me know if you have any further questions.