How can a set contain itself im reading a little about the Russell Paradox , and i dont really get what they mean by a set that contains itself , can anybody please explain ?
 A: There are two different issues here with two different answers.  
One is: Can a set have itself as an element?  This would be a pretty strange set.  One could imagine something like $\{\{ \cdots \{1 \}  \cdots \} \}$, with infinitely many brackets.  But just because I write this down doesn't mean it makes sense.  Whether or not you allow something like this depends on your choice of axioms of set theory.  The most common formalization of set theory, ZFC, does not allow this due to the Axiom of Regularity.  Almost all of math research can be formalized in ZFC, so for the most part you will never see sets that contain themselves.  (Maybe they are useful somehow in other axiom systems - I don't know.)
The other issue is: Does the set $$\{S \,|\, S \notin S\}$$ make sense?  Evidently, the answer is no, because here is where we run into Russell's Paradox.  But the issue is perhaps not what you might think it is.  For example, if you don't believe in complex numbers, then there is no number $x$ with $x^2 = -1$.  However, you can still write the set $\{x \in \mathbb{R} | x^2 = -1\}$.  You just get the empty set.  So it's not a problem to write things like $S \in S$ or $S \notin S$.  It's just the first one will always be false, and the second always true, if you're in the context of ZFC.
The problem is that there is no restriction to what sets the variable $S$ is allowed to run over.  If we start with a given set of sets $\mathcal{F}$ and write $\{ S \in \mathcal{F} | S \notin S\}$ we don't have any problems.  It's okay that we are writing $S \notin S$.  For example let $\mathcal{F} = \{ \{1,2\}, \{3,4\}, \{5,6\} \}$.  There's no element of $\mathcal{F}$ that's an element of itself.  So, $\{ S \in \mathcal{F} | S \notin S\}  = \{ \{1,2\}, \{3,4\}, \{5,6\} \}$.  You have $\mathcal{F} \notin \mathcal{F}$, but this does not mean that $\mathcal{F} \in \{S \in \mathcal{F}| S \notin S\}$.  Why?  Because the elements $S$ of $\{ S \in \mathcal{F} |S \notin S\}$ are specifically restricted to be elements of $\mathcal{F}$, so $\mathcal{F}$ is not under consideration in the definition - just like $i$ is not an element of $\{x \in \mathbb{R}: x^4 = 1\}$ since by definition every element of $\{x \in \mathbb{R}: x^4 = 1\}$ is real.
The paradox arises where we allow a set $U$ to be the "set of all sets". Then since $U$ is a set, $U \in U$.  Furthermore, if such a set $U$ exists, then the set $\{S \in U | S \notin S\}$ makes perfect sense and we get Russell's Paradox. Suddenly the world implodes: every logical statement can be proven true.
tl;dr version: In the usual formulation of set theory, sets can't contain themselves, and you can't have a set of all sets, but restricted sets like $\{S \in \mathcal{F}| S \notin S\}$ are fine.
