Axiom of Regularity I am having difficulty understanding Axiom of Regularity :

Every non-empty set $\rm A$ contains an element $\rm B$ which is disjoint from $\rm A.$

So from my understanding if I have a set like: 
$$\{1, 2, 3, 4, 5\}$$
Then one of $[1,2,3,4,5]$ is not an element in $\rm A?$ huh?
 A: In axiomatic set theory everything is a set, you don't work with other objects.
So even if you denote some things by $1$, $2$, $3$, $4$, $5$, they are in fact sets.
In fact, if we look at your example and use the standard construction of positive integers in ZFC, then we have
$0=\emptyset$,
$1=\{0\}$,
$2=\{0,1\}$,
$3=\{0,1,2\}$,
$4=\{0,1,2,3\}$ and
$5=\{0,1,2,3,4\}$. 
Axiom of regularity says that one of the elements of the set $A=\{1,2,3,4,5\}$ is a set, which is disjoint with $A$. Indeed, $1$ is such set -- the only element of $1$ is $0=\emptyset$, which is not an element of $A$; hence $1\cap A=\emptyset$.

This might be unusual viewpoint for someone used to work in naive set theory, but once again, the basic idea is that: Everything is a set. We have some axioms, which allow us to create new sets from the sets we have already constructed. To work in this setting we try to create models of various things using these axioms. So each integer, rational number, real number will be modeled in ZFC as some set.

Let me also say that you probably don't need to worry too much about Axiom of Regularity if you are just beginning to study axiomatic set theory.
You will only need this axiom much later (perhaps when you encounter cumulative hierarchy, where this axiom ensures that every set can be obtained by this repeated process of taking unions and power sets or when you encounter inductive sets in Axiom of Infinity and construction of natural numbers.) A lot of stuff can be done without this axiom.
A: Terry Tao gives the following example in his Analysis I book (section 3.2).
Consider the following set of sets
$$A=\big\{\; \{3,4\},\;\{3,4,\{3,4\}\}\;\big\}.$$
Then take $x=\{3,4\}$.
We know that $x\in A$ and that $x$ is a set.
Since $x$ is a set, we can ask: what is $x\cap A$?
Computing
$$
x\cap A = \{y: y\in x\text{ and } y\in A\} = \{y: (y=3\text{ xor } y=4) \text{ and } y\in A\} = \emptyset
$$
since neither $3$ nor $4$ are elements of $A$.
Thus we've found an element $x\in A$ that is disjoint from $A$.
A: The axiom of regularity says that one of $1,2,3,4,5$ is disjoint from $\{1,2,3,4,5\}$, there is some $x\in\{1,2,3,4,5\}$ such that $x\cap\{1,2,3,4,5\}=\varnothing$.
It may sounds a bit weird, but in modern set theory everything is a set.
For example take the set $\{\varnothing\}$, it has one element and indeed $\varnothing\cap\{\varnothing\}=\varnothing$. It does not imply that $\varnothing\notin\{\varnothing\}$.
So the axiom tells us that every set $A$ either has $\varnothing\in A$, or it has some element $x$ which is not a subset of $A$ (in fact $x\cap A=\varnothing$, which is a stronger requirement).
