Is the Union of Intersecting Connected Sets Always Connected?

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Let A, B be two connected subsets of a topological space X such that A intersects the closure of B .
Prove that A ∪ B is connected.

I can prove that the union of A and the closure of B is connected, but I don't know what to do next. Could anyone give me some hints or is there another way to treat this problem?

Also, for the specific case, the union of the open ball B((−1, 0), 1) and closed ball B((1, 0), 1) should be connected, right? I can see it, but I'm not sure how to word the proof. Any help is greatly appreciated!
 
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I would assume AuB is disconnected and try to find a contradiction with A intersecting the closure of B.
 
Let A \cup B = E_1 \cup E_2, with \overline{E_1} \cap E_2 = \emptyset and \overline{E_2} \cap E_1 = \emptyset
(the overline denoting closure). You need to prove one of E_1, E_2 is the empty set.
 
A space is disconnected means that it's the union of two disjoint nonempty open sets. Since we're given that A,B are connected, the only way we could possibly write A u B as a union of disjoint open sets is if A and B are in fact open and disjoint.

You can see this can't be true since A intersects the closure of B, i.e. there's a point a in A such that any neighborhood of a intersects B. But if A is open, then A contains a neighborhood of a, so that A intersects B.
 
Thanks a lot for your guys. I wasn't that clear about the concept in the first place, but now I know how to handle this type of problems.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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