Disclaimer: Yes I know, this question has been asked multiple times before here, and if you want, you can flag this question as duplicate. I only chose to make a new question because of the kind of backwards comment-reputation system.
We all know that the Axiom of Regularity states:
For every non-empty set X, there exists an element y in X, so that X and y are disjoint.
and that Zermelo-Fraenkel set theory (with the Axiom of choice, hence ZFC) doesn't need urelements to exist, thus every element is also a set. But that poses two questions for me:
- For a set that doesn't have the empty set as an element, how does this work?
For example:
X = {1,2,3}, every element in X is also a set, so y can be {1}, {2} or {3}, but all of these are not disjunct with X, so what is disjoint with x? I know that every set contains the empty set as a subset, but "containing" and "element of" are not equal statements, or did I understand something wrong?
and if that works somehow:
- Take the Power Set of that X, remove the empty set and repeat the procedure. Since every subset of X is included, again, how does it work? I'm confused.
X ={1,2}, so P(X) = {{},{1},{2},{1,2}} and P(X)/{} = {{1},{2},{1,2}}.
And just to clarify, I don't want anybody to tell me if that axiom is "not useful". I don't care. I'm here to learn about set theory, not to learn if it's worth it. Sorry, but I've seen that under the other questions too and it's just such a dodgy answer.
$1$ and $2$ and $3$ are sets, but since you are thinking about them in an abstract capacity of "three natural numbers" or even just "three distinct objects", there is no reason to treat them as one set or another. These could be the standard ordinal-based construction of $\Bbb N$; or you can use the fact that there is a canonical copy of $\Bbb N$ in any construction of $\Bbb R$, and maybe you meant these. And so on and so forth.
But in any case, let's take the standard ordinal based approach. So $0=\varnothing, 1=\{0\}, 2=\{0,1\}$ and $3=\{0,1,2\}$. In that case it is not hard to check that $1\cap\{1,2,3\}=\varnothing$. So indeed $1$ is the element guarnateed by the axiom of regularity. If you choose a different coding, then the answer could be different. So your mileage may vary when it comes to this answer.
With the power set example this is also not hard. Since $\{1\}$ is disjoint from $\mathcal P(X)$, as there is no element of $\mathcal P(X)$ which is the set $1$. Mainly because $1\neq\{1\}$. And again, your mileage may vary depending on your coding, but I'm assuming the standard coding as above. But any other coding schema will have an answer as one of these sets if not more.