Does set theory always deal with pure set? My elementary set theory book introduce a one of four rules which consist of axiom of cardinality that 
A ~ {1, 2, 3, ..., K} is equal to card(A) = k, tilde symbol '~' denotes one-to-one correspondence.
Is each element of the set {1, 2, 3, ..., k}(e.g. 1, 2, 3, ..., k) 
actually set? I understand set theory doesn't deal with urelements. These 1, 2, 3, ... help me understand intuitively?
 A: Yes, every element is a set (and usually we start from zero). $0=\emptyset$ is the empty set. $1=\{0\}=\{\emptyset\}$ is the set whose only element is the empty set. $2=\{0,1\}=\{\emptyset,\{\emptyset\}\}$ is the set whose two elements are the empty set and the set whose only element is the empty set. $3=\{0,1,2\} = \{\emptyset, \{\emptyset\}, \{\emptyset,\{\emptyset\}\}\},$ and so on....
A: Indeed, ZFC does not have urelements, therefore the numbers $1$, $2$, $3$ and so on need to be represented as sets.
There are in principle many ways to do this, but there is one way that is pretty standard (because it allows a very useful extension to infinite numbers, the so-called ordinals): Each number $n\in\mathbb N_0$ is given by $\{k\in\mathbb N_0|k<n\}$.
Here $\mathbb N_0$ is the set of natural numbers including $0$ (using this notation because somethimes $\mathbb N$ is defined not to have $0$; $\mathbb N_0$ is unambiguous in that regard).
So you get:


*

*$0 = \{k\in\mathbb N_0|k<0\} = \emptyset$.

*$1 = \{k\in\mathbb N_0|k<1\} = \{0\} = \{\emptyset\}$.

*$2 = \{0,1\} = \{\emptyset,\{\emptyset\}\}$
and so on.
Besides the extensibility to ordinals, it also has the advantage that each set has the cardinality of the number it represents. So instead of saying that $A\sim\{1,2,3,\ldots,n\}$ you can just say $A\sim n$.
Note: Strictly speaking, it is not true that set theory does not deal with urelements; while ZFC doesn't, there are other set theories that do have urelements. However usually the term “set theory” is used to refer to ZFC, where urelements indeed don't exist.
