why is $\{x\} \in \{x\}$ false? I apologise for this simple question. 
So if {x} is a subset of {x, {x}}, then why isn't {x} belong to {x}?
 A: It is a subset. But not an element. $\{x\}\subseteq \{x\}$, but  $\{x\}\notin \{x\}$: if you look in the set containing only $x$, you cannot find an element $\{x\}$ (which is the set containing $x$, not $x$ itself).
A: $\{x\}$ is a subset of $\{x,\{x\}\}$, since every element in $\{x\}$ (i.e $x$) is in $\{x,\{x\}\}$. $\{x\}$ isn't an element of $\{x\}$ since the only element in it is $x$.

I assume $\{x\}\neq x$. This kind of set is called Quine atoms, and in standard set theory (ZF), this kind of set is forbidden because of the axiom of foundation. 
A: A rigorous proof can only be  achieved using the Regularity Axiom:
$$
\forall A(A\ne\varnothing\Longrightarrow \exists z(z\in A \,\&\,z\cap A=\varnothing). 
$$
So in our case: If $\{x\}\in\{x\}$, then that would mean that $\{x\}=x$. But according to the Regularity Axiom, as $A=\{x\}\ne\varnothing$, there exists a $z\in A$, such that $z\cap A=\varnothing$. But, the only element of $A=\{x\}$ is $x$, and hence the Axiom provides that 
$$
A\cap x=\{x\}\cap x=\varnothing,
$$
which implies that if $x=\{x\}$, then $\varnothing=x\cap\{x\}=x\cap x=x$. But $\varnothing\ne\{\varnothing\}$, as $\{\varnothing\}$ is a non-empty set.
A: Given the set $\{x\}$ you can say that $x \in \{x\}$, since $x$ is an element of the set $\{x\}$.
If you have another set, $\{\{x\}\}$, you can say that $\{x\} \in \{\{x\}\}$. In this case the set $\{x\}$ is an element of another set $\{\{x\}\}$.
That is, when you include an object between curly brackets, you are saying that the element belongs to a set. No matter what the object is, it can be also a set itself. The set contains other elements but not itself. Then:
$$\{x\} \not\in \{x\}$$
A: By the definition of singleton set, $z \in \{ x \}$ if and only if $z = x$.
Therefore, by substituting $z \to \{ x \}$, we have that $\{ x \} \in \{ x \}$ if and only if $\{ x \} = x$.
Thus, if we had $\{ x \} \in \{ x \}$ then by substiting $\{ x \} \to x$ twice, we would therefore have $x \in x$. It would be pretty weird for a set to be an element of itself, don't you think? That would be like coming home one day and entering your house, only to find out that your house is in your living room!
(by the axiom of foundation, one cannot have $x \in x$ for any $x$, so we can rest assured that ZFC set theory doesn't have such weird things)
