Does $A^i \cap A^j = \emptyset, $ if $ i \neq j$? I'm doing a bit of set theory and, of course, I'm confused.
How true is it that if we have a series of cartesian products of a set, say $A^n, n< \omega$, then it necessarily holds that $A^i \cap A^j = \emptyset, $ if $ i \neq j$?
I'm sort of imagining that if say, $a \in A^2$, then $a = (\alpha, \beta)$ has the form $\{\{\alpha\}, \{\alpha, \beta\}\}$
If we compare this to something like $A = \{\{\alpha\}, \{\beta\}, \{\gamma\}...\}$, then clearly they dont have any elements in common. 
Does this hold water?
The really crazy thing is that I'm been told forever than $\Bbb R \subset \Bbb R^2$, and so their intersection cannot be empty, then why does it seem like the reasoning above holds?
Thanks in advance.
 A: There are a couple of things to point out here.

*

*It is often easier to define $A^n$ as the set of functions from $n$ into $A$, rather than repeated Cartesian products. The reason is that this way we can extend this notion into infinite sets, not just finite sets.
Not to mention that there are different ways to encode ordered pairs, which may affect the following in quite a substantial way.


*Consider $V_\omega$ which is defined in the following way, $V_0=\varnothing, V_{n+1}=\mathcal P(V_n), V_\omega=\bigcup_{n\in\omega}V_n$. Then it is not hard to show that for every $n\in\omega$ we have that $V_\omega^n\subseteq V_\omega$, and that $V_\omega\times V_\omega\subseteq V_\omega$.
But here comes the kicker, if you consider $V_\omega^n$ as functions from $n$ into $V_\omega$ then for $n\neq k$ we have that $V_\omega^n\cap V_\omega^k=\varnothing$, since the elements of these sets are functions with different domains. On the other hand, if you consider this as repeated Cartesian products then $V_\omega^n\subseteq V_\omega^k$ whenever $0<k\leq n$.


*Regarding $\Bbb{R\subseteq R^2}$, this is yet another delicate issue. You can start by coming up with an object called $\Bbb R^2$ and then define $\Bbb R$ as a subset of that object. You can do that, and then the inclusion holds. Or you can define $\Bbb R$ and then take $\Bbb R^2$, and depending on set theoretical decisions the result may or may not satisfy certain inclusions.
But it ultimately doesn't matter. Sometimes it's easy to think about $\Bbb R$ as a subset of the plane, so we choose one of many ways to embed $\Bbb R$ into the plane, and redefine $\Bbb R$ as that copy. At other times, it's easier to think about them as being totally disjoint (for example when you want to assert that real numbers and points on the plane, or vectors in general, are not the same type of object).
So this is really just about what you want to do with your objects, and the fact that ultimately this doesn't matter since the definition can go from here to there, or from there to here. The same holds if you consider $\Bbb N\subseteq\Bbb R$ or not, if you define $\Bbb R$ using Dedekind cuts this is certainly false, but we identify the natural numbers and redefine that symbol to mean some particular subset of $\Bbb R$. If we choose to do so, anyway.
A: Consider $A= \{a,b, (a,b)\}$. Then $(a,b) \in A \cap A^2$. 
