I honestly don't see what error I am making with elementary set theory. I think the wording of "all" and "every" is the problem.
Not really.
All has two meanings here.
No.
It can mean that all of the natural numbers are in one set, ....
The natural numbers are a set: [imath] \mathbb{N}=\{1,2,3,4,\ldots\} [/imath].
.... or all of the natural numbers have their own set.
This doesn't really make much sense. Every thingy has its own set, namely [imath] \{\text{thingy}\} [/imath]. If thingy is a natural number [imath] n [/imath] then we have [imath] n\in \{n\}\subseteq \mathbb{N}. [/imath] [imath] n [/imath] is an element, the only element of [imath] \{n\} [/imath] which itself is a subset of the natural numbers. These statements do not change if we consider [imath] \{1,2,3,\ldots,n\} [/imath] instead of [imath] \{n\} [/imath] except that [imath] n [/imath] would no longer be the only element in [imath] \{1,2,3,\ldots,n\}. [/imath] The thingies that are in a set are called elements of the set.
Those elements in your initial posts were [imath] \{1\},\{1,2\},\ldots,\{1,2,\ldots n\},\ldots [/imath] You have had sets as elements which is the real reason why things became confusing. You have infinitely many elements in that set [imath] S=\{\{1\},\{1,2\},\ldots,\{1,2,\ldots n\},\ldots\} [/imath], but each element is a set of finitely many numbers. The set [imath] \mathbb{N} [/imath] therefore cannot be an element of the set [imath] S [/imath] because it is not finite. Moreover, the set [imath] \mathbb{N} [/imath] cannot be a subset of [imath] S [/imath] since [imath] S[/imath] consists of sets whereas [imath] \mathbb{N} [/imath] consists of numbers.
Here is a finite example using the set S = {{1}, {1, 2}, {1, 2, 3}, {1, 2, 3, 4}}. Knowing what is in S, the statement for every natural number there is a set (as you said) also means that all of the natural numbers have their own set. As for the other meaning of all, it also means the statement all of the natural numbers have a set .
I don't understand this. Here we have: For every natural number [imath] n [/imath] smaller than five, there is at least one
element [imath] E [/imath] of [imath] S [/imath] such that [imath] n\in E [/imath] is an element of the element [imath] E [/imath].
I would like to know how your statement, "for every natural number there is a set which contains it" ...
Because you constantly confuse elements with sets. Every natural number [imath] n [/imath] is an element of the set [imath] \{1,\ldots,n\} [/imath] which itself is an element of [imath] S. [/imath]
... does not imply that all natural numbers must be in one set.
Because all elements of [imath] S [/imath] are finite sets, whereas [imath] \mathbb{N} [/imath] is an infinite set.
Again, I know that this is a finite set, but I can't find a reason why it wouldn't work for an infinite set too.
There is a difference between [imath] \mathbb{N}\in \{\{1,2,\ldots\}\} [/imath] and [imath] \mathbb{N}\subseteq \{1,2,\ldots\} [/imath]. Unfortunately, the set [imath] \{1,2,\ldots\} [/imath] had never anything to do with [imath] S. [/imath]