Using $p\supset q$ instead of $p\implies q$ I saw that a use for the notation $p\supset q$ instead of $p\implies q$
that got me a bit confused.
One occurrences is in this Wikipedia link.
It seems to me opposite than what it should be, let me explain what
I mean:
If $A,B$ are sets s.t $A\subset B$, $p$ is for $x\in A$, and $q$
is for $x\in B$ then we can identify (in some way) $$A\text{ with }p$$
$$B\text{ with }q$$
We have it that $p\implies q$, since $A\subset B$, but in the above
notation we have $q\subset p$ which to me looks like $B\subset A$
which is the opposite of what we wanted to express.
Can someone please explain to me the logic behind this notation ?
 A: One way to rationalise this notation is to think of a proposition as having a certain information content.
Then $p \supset q$ can be thought of as "the information content of $q$ is contained in that of $p$". A particular piece of information that can be obtained from $q$ is "$q$ is true".
Thus we see that $p \supset q$ naturally gives rise to the statement $p \implies q$, and conversely.

However, I agree with Hagen von Eitzen's comment that $p \to q$ should be the notation of choice, at least in symbolic logic. One could still use $\implies$ in a meta-context, i.e. as more or less a part of the natural language we discuss mathematics in.
A: Notice that in the line in the link numbered 2, they say the symbol $p\supset q$ is often confused for set inclusion. But it is not set inclusion. It's just a symbol used to denote material implication. So, it seems you just exhibited how often it actually happens that this confusion arrises. 
A: Actually, historically, the reason we use $\supset$ is that Peano originally wrote $p C q$ for "$p$ is a consequence of $q$", and wrote a backward "$C$" for "$p$ has as a consequence $q$". Eventually, just as the "$\epsilon$" became "$\in$", so too did the backward "$C$" become "$\supset$".  So it doesn't actually have anything to do with set theory, per se, though maybe he used the same letter for both superset and consequence. 
But as Lord_Farin notes, one way to look at things might be in terms of "informational content" or even (very informally) "provability power" (if "$p \supset q$" is true, then accepting "$p$" requires accepting everything that would go with accepting "$q$"; though what I just said might be very philosophically contentious!). And yes, it seems more natural to think of it semantically, and so think that the "$\supset$" should be a "$\subset$". That's math for you.
