Are these two venn diagrams valid for $(A-B)\cup (B\cap C)$?

When I'm told to represent a Venn diagram for $(A-B)\cup (B\cap C)$ are these two valid?

The first one:

The second one:

I don't seem to understand clearly whether an union implies having both sets "touch" each other in the diagram or if it doesn't matter at all as long as I color them red as I did in the second one.

Also, note that the exercise doesn't actually tell me if $A,B,C$ are really intersecting each other (only $A$ with $B$ and $B$ with $C$ but never $A$ with $C$), is that supposed to make a difference in the way I display the diagrams?

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While both diagrams represent the set $(A \setminus B ) \cup ( B \cap C )$, the second is done under the additional assumption that $A \cap C = \emptyset$. Generally Venn diagrams are supposed to represent all of the possible interactions between the sets they represent, and if you do not know beforehand that $A \cap C = \emptyset$, then the second diagram loses information (in my opinion, anyway).

As such, I would be hesitant to provide the second as an answer to the question, as you could have equally done the following: represent the sets $A , B , C$ as discs which do not overlap at all (this is the situation $A \cap B = \emptyset$, $A \cap C = \emptyset$ and $B \cap C = \emptyset$. Then $( A \setminus B ) \cup ( B \cap C )$ would be represented by filling in the $A$ circle, and leaving the rest blank.

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They are both valid. The sets need not "touch" each other or be connected in any manner.

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Both diagrams are correct. The set of $(A-B)\cup (B\cap C)$ is, by definition, all of the elements that are either in $A-B$ or in $B$ AND $C$ (not an exclusive or, the element could be in both). Hence, whether the sets "touch" each other or not is actually irrelevant. I hope you find this somewhat enlightening!

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Wait, if the element could be in both at the same time, the second drawing would make no sense, or would it? I mean, since the two red areas are visually separated, it would be impossible for an element to be in both, no? – Zol Tun Kul Dec 3 '12 at 6:05
In fact, since we're considering $A-B$ and $B\cap C$, it couldn't be in both, but in general, a union is not an exclusive "or". To clarify, in this setting, if it could be in both, then $A\cap B^c\cap B\cap C$ would have to be nonempty ($A-B=A\cap B^c$), but in this case, it couldn't be nonempty. – Clayton Dec 3 '12 at 6:09