Showing in $ZFC-Foundation$ that $V\neq R$ In the context of showing that in $ZFC-Foundation$, that $R\neq V$, Kunen, "Set Theory" (2011) states (page 22) that $\{x: x=\{x\}\}$ can be a proper class. (As an aside, does "can be" mean is?) He defines $V:\{x: x=x\}$ and the Russell set as $R:\{x:x\notin x\}$.
I think the members of $\{x: x=\{x\}\}$ violate Foundation as $x\cap\{x\}=\{x\}\ne\emptyset$. 
-- Is $\{x: x=\{x\}\}$ a proper class since without the constraint of Foundation, it would include all sets?
-- Does the context $ZFC-Foundation$ alter what is considered as $V$, or does it effect $R$,(since in general $x\in\{x\}$ but now we can have $\{x\}=x\in\{x\})$? 
(This seems a little strange as wikipedia defines $V$ as the class of hereditary, well-founded sets; so how does Kunen discuss this in the context of $ZFC-Foundation$?)
-- Lastly, how would this be used to show $V\neq R$?
Thanks
 A: Given a class $S$, to say that it can be proper means that it is consistent (with the axioms under consideration) that $S$ is a proper class, that is, there is a model $M$ of these axioms such that the interpretation $S^M$ of $S$ in $M$ is a proper class in the sense of $M$. It does not mean that $S$ is always a proper class. In fact, it could also be consistent that $S$ is empty (of course, the model $M'$ where this happens would be different from $M$).
Things tend to be a bit more delicate, since the theories we are usually discussing here cannot quite be proved consistent unless one assumes a rather strong metatheory (due to issues related to the incompleteness theorems). In that case, one talks of relative consistency. For instance: If $\mathsf{ZF}$ is consistent, then the theory $(\mathsf{ZFC}-{\rm Foundation})+$"$\{x:x=\{x\}\}$ is a proper class" is also consistent. In practice, to establish results of this kind, one starts with a model $N$ of the first theory (in the example, $\mathsf{ZF}$) and somehow modifies it to produce a model $M$ of the second theory (in the example, $(\mathsf{ZFC}-{\rm Foundation})+$"$\{x:x=\{x\}\}$ is a proper class"). Another common approach is to show that there is a way of interpreting one theory inside the other one.  
(Anyway, it seems from your questions that you do not quite understand yet how relative consistency and models of set theory work, and this is causing confusions beyond the difficulties of the actual technical question you posted here. It may be more profitable to read about these topics for a while, and once you have a better grasp on them, to revisit these questions regarding the axiom of foundation.)
Briefly:
Yes, any set in $\{x:x=\{x\}\}$ violates foundation.
No, the class $\{x:x=\{x\}\}$ never includes all sets, regardless of whether we assume foundation. For instance, $\emptyset$ is not in this class (or $\mathbb R$, or $\{1,2\}$, or ...).
You cannot prove in the theory $\mathsf{ZFC}-{\rm Foundation}$ that $\{x:x=\{x\}\}$ is a proper class, because this is false in the stronger theory $\mathsf{ZFC}$. Or rather: If you prove in a theory $A$ that $\phi$ holds, and if $A\subset B$ and the theory $B$ proves that $\lnot\phi$ holds, then $B$ is inconsistent (it proves a contradiction, and therefore it proves everything). So, if we assume that $\mathsf{ZFC}$ is consistent, then no weaker theory can prove that $\{x:x=\{x\}\}$ is a proper class. 
It is in general difficult to produce models of set theory, you are not going to easily guess how to build a model of $(\mathsf{ZFC}-{\rm Foundation})+$"$\{x:x=\{x\}\}$ is a proper class". In any case, (assuming the consistency of $\mathsf{ZF}$) there are models of $(\mathsf{ZFC}-{\rm Foundation})+\lnot{\rm Foundation}$ where $\{x:x=\{x\}\}$ is not a proper class. I am not sure you can currently understand how to build any of these models, it is a non-trivial task. My suggestion is that you postpone learning about how to build these models for a while, until you are more comfortable with the topics I mentioned in my parenthetical paragraph above.
It is not clear what you mean when you ask whether a context alters what $V$ is. You may want to clarify that (it may be harder than you anticipate). It seems to me that you are asking whether not assuming foundation implies that $V$ is larger than if we assume it. This is too vague still, but the answer is no. The point is that if $M$ is a model of a theory $B$ then it is also a model of any weaker subtheory $A$ of $B$. What is true is that there may well be models of $A$ that are not models of $B$. Indeed, (again, assuming the consistency of $\mathsf{ZF}$) there are models of $\mathsf{ZFC}-{\rm Foundation}$ with ill-founded sets. In any such model, we can define the subclass of well-founded sets, and that subclass is in turn a model of $\mathsf{ZFC}$, and the $V$ of this subclass is the class of well-founded sets of the original model. In that sense, $V$ has been altered due to the "context", being "larger" in the model where foundation fails. 
I prefer to call $V$ the universe of all sets, the class $\{x:x=x\}$. Of course, under foundation, this is certainly the class $\mathrm{WF}$ of well-founded sets, but without assuming the axiom of foundation, both classes may well be different. This is not the convention used in the Wikipedia page you visited, but my usage is the standard one.
(By the way, you never told us what $R$ is, and it is not standard notation.)
