Open set in a topological space

ehrenfest
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Homework Statement


If U is open set in a topological space such that U = A union B, where A and B are disjoint, do both A and B have to both be open?

I think that they do not, but I cannot think of a counterexample...perhaps (-1,0) union [0,-1).

OK. That's a counterexample. So, now the question can both A and B not be open?


Homework Equations





The Attempt at a Solution

 
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What about in R2, trying the same trick only with say, two rectangles, one on (0,1/2)x(0,1) and one on (1/2,1)x(0,1), and then give each half of the boundary between them
 
Or take U to be (0,1), A=rationals in U and B=U-A. Neither is open.
 
No need to get as complicated as "rationals" and "irrationals". Take A= [0, 1], B= [2, 3] so that both are closed. Then the intersection is the empty set which is open!

Or, since the empty set is also closed, you might prefer A= [0, 3/2), B= (1, 2]. Neither is open but their intersection is (1, 3/2) which is open.
 
The union of the two sets is open, not the intersection.
 
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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