Axiom of Regularity and infinite sequences I'm wrestling with the elementary implications of the Axiom of Regularity.
The axiom:
$∀A(A≠∅→(∃x∈A)(A∩x=∅))$
implies that every set A either has $∅∈A$, or it has some element $x$ such that $x∩A=∅$.
My questions:
(1) If $x∈A$, and $x$ is not a subset (or, x is a subset that consists of only one element, i.e. {1}), then does $x∩A=∅$? I'm trying to understand how $x∩\{1,2,3\}=∅$ 
(2)How does this guarantee no infinitely descending sequences? 
Thank you for your time. 
 A: You’ve stated the axiom of regularity correctly,
$$\forall A\Big(A\ne\varnothing\to\exists x\in A(A\cap x=\varnothing)\Big)\;,\tag{1}$$
but your verbal paraphrase isn’t quite right: $(1)$ says that either $A=\varnothing$, or some element $x$ of $A$ is disjoint from $A$. That is, your first alternative should be $A=\varnothing$, not $\varnothing\in A$.
Your first question isn’t entirely clear. In the specific example, are you asking what element of $A=\{1,2,3\}$ is disjoint from $A$? In order to answer that, you have to have some definition of $1,2$, and $3$ as sets. The usual one is that $1=\{0\},2=\{0,1\}$, and $3=\{0,1,2\}$. If you’re using that definition, then $x=1$ works: 
$$x\cap A=1\cap\{1,2,3\}=\{0\}\cap\{1,2,3\}=\varnothing\;,$$
because the only element of $1$ is $0$, and $0\notin\{1,2,3\}$.
For your second question, suppose that $A=\langle a_n:n\in\Bbb N\}$ is a family of sets with the property that $a_{n+1}\in a_n$ for each $n\in\Bbb N$:
$$\ldots\in a_4\in a_3\in a_2\in a_1\in a_0\;.$$
Let $x\in A$. Then $x=a_n$ for some $n\in\Bbb N$, and $a_{n+1}\in x\cap A$, so $a_{n+1}\ne\varnothing$. Since $A$ is clearly non-empty, this contradicts $(1)$.
A: First, the alternative isn't $\emptyset \in A$, as you have it, but $\emptyset=A$.  Hence every nonempty $A$ has an element that isn't a subset.
Suppose now that $$A=\{1,2,\{1\}, \{\{1\}\},\{1,2,4\},3\}$$
If we take $x=\{1\}$, then $x\cap A=\{1\}\neq \emptyset$.  If we take $x=\{1,2,4\}$, then $x\cap A=\{1,2\}\neq\emptyset$.  However if we take $x=1$, then $x\cap A=\emptyset$.
Suppose now that there were an infinitely descending subsequence $$x_1\ni x_2 \ni x_3 \ni \cdots$$
We could then take $$A=\{x_1,x_2,x_3,\ldots\}$$ which would voilate the axiom, because for each $i$, $A\cap x_i$ is a set which contains $x_{i+1}$.
