Do any two imply the other: $(X_1,X_2)$ is uniform, $X_1$ and $X_2$ are uniform, and $X_1$ and $X_2$ are independent?

There are three statements:

1. $(X_1,X_2)$ is a uniform random vector over some subset $S \subset \mathbb{R}^2$,
2. $X_1$ and $X_2$ are two uniform random variables over some $S_1 \subset \mathbb{R}$ and $S_2 \subset \mathbb{R}$ respectively,
3. $X_1$ and $X_2$ are two independent random variables.

where $S_1$ and $S_2$ are projections of $S$ onto the two axes for $X_1$ and $X_2$ respectively.

1. I was wondering if any two of the three statements imply the other one? I know that 3 and 2 imply 1, and suspect other implications are also true.
2. What is the necessary (and sufficient) condition for 2 and 1 to imply each other? Is 3? Or some requirements on $S$?
3. Also I wonder if the above true statements can be generalized to cases when there are finite random variables $(X_1, X_2, \dots, X_n)$, or even infinitely many (countably or uncountably) random variables? Is 3 required to be modified to be mutually independent, or pairwise independence will work?

Thanks!

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If 3. is true, then I think 1. and 2. are equivalent. Otherwise neither 1. nor 2. implies the other, nor do they (even taken together) imply 3. –  mjqxxxx Oct 25 '12 at 14:33
@mjqxxxx: How do you show that 3 and 1 imply 2? Why 2 and 1 do not imply 3? –  Tim Oct 25 '12 at 14:36
@Tim if they are independent random variables, then the joint density is the product of marginals. In other words, if a joint density is given with the independence conditions they should be generated by two uniform marignals. –  Seyhmus Güngören Oct 25 '12 at 14:47

3 + either 1 or 2 implies the other. If you also have $S=S_1$x$S_2$, 1 and 2 imply 3.
Independence of random variables is equivalent to the statement that the joint density function $f_{X,Y}(x,y)$ is equal to product of the marginal densities $f_X(x)\cdot f_Y(y)$. If you have 1) and 2) and $S=S_1$x$S_2$, then you can just work out both expressions and you'll see they're the same.
You already noted that 2 and 3 imply 1 so let's assume we have 1 and 3 now. Write $f_{X,Y}(x,y)=f_X(x)\cdot f_Y(y)$; considering that equation for fixed $y\in S_2$ we see that $f_X(x)$ must be uniform over $S_1$ and similarly $f_Y(y)$ must be uniform over $S_2$.
Sure about that? Consider $X_1=X_2$ uniform on $(0,1)$ (more sophisticated examples exist, in particular such that the joint distribution of $(X_1,X_2)$ has a density). –  Did Oct 25 '12 at 15:39
Whoops. Misread the part about $S_1$ and $S_2$ being projections. Fixed. –  anonymous Oct 25 '12 at 15:43