Is the Sum of Two Closed Sets in R^n Always Closed?

teacher2love
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Let A, B in R^n be closed sets. Does A+B = {x+y| x in A and y in B} have to be closed?

Here is what I've tried. Let x be in A^c and y in B^c which are both open since A & B are closed. So for each x in A^c there exists epsilon(a)>0 s.t. x in D(x, epsilon(a) is subset of A^c. For each y in B^c there exists epsilon(b)>0 s.t. y in D(y, epsilon(b)) is a subset of B^c.

Can I add the two together to get x+y in (A+B)^c to show that there exist epsilon > 0 s.t. D(x+y, epsilon) is in (A+B)^c. Thus (A+B)^c is open => (A+B) is closed.

Thanks for your help.
 
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It is often much easier to work with sequences when you can!

A set is closed if the limit of every converging sequence of elements of that set lies in the set.
 
teacher2love said:
Can I add the two together to get x+y in (A+B)^c to show that there exist epsilon > 0 s.t. D(x+y, epsilon) is in (A+B)^c. Thus (A+B)^c is open => (A+B) is closed.

It's not true that x+y will be in (A+B)^c. (almost any example you can think of will show this, but to take a simple one, let A=B={0}, x=-y=1).

A good strategy on these types of problems (where you're not sure if the given statement is true or false) is to start by trying to find counterexamples. If you find one, you're done, and if not, try to see what's preventing you from finding one.

For example, you might notice that you can't find any counterexamples when one of the sets is finite. Well, this is just because, as is easy to prove, A+B is closed when one of A or B is finite. So you can continue, now looking only at examples where both A and B are infinite.
 
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