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Is a sum of closed sets closed?

  1. Mar 11, 2008 #1
    We have two closed sets A,B in R^n. Does A+B= {x+y | x is an element of A, y is an element of B} have to be closed?

    I know that both the union and intersection of two closed sets have to be closed. I'm guessing from the question that the answer is no, but I've been playing around with different intervals or R^2 sets that could represent A and B and can't come up with any counterexample to show that A+B need not be closed.

    Thanks for any help you can offer!
    Last edited: Mar 11, 2008
  2. jcsd
  3. Mar 11, 2008 #2
    Have you tried choosing open intervals as your set. Try A={x|0<x<1} and B={y|3<y<4}
  4. Mar 11, 2008 #3
    Sorry, I thought i had included that A and B have to be closed but I forgot to write that in...
  5. Mar 11, 2008 #4
    A and B were not said to be closed in the OP, but judging by the topic, I suppose that was supposed to be the case. (edit: I was typing this before the previous comment by benjamin111)

    I noticed that I couldn't get the result very straightly with basic topology. A disturbing thing...

    If you assume that A and B are compact, I think it seems quite clear that you can get the compactness of A+B with sequences and subsequences. It could be that one way to solve your problem is to start diving those sets into bounded subsets, but I'm not sure... If you don't have any better ideas, why not give it a try?
  6. Mar 11, 2008 #5
    Here's a counter example:

  7. Mar 11, 2008 #6
    I guess I don't quite see how that parallels this sum of closed sets. Isn't it more like just dividing closed sets over and over? If you could run me through how to interpret the function in terms of the problem I'm looking at I'd be most appreciative
  8. Mar 11, 2008 #7
    I must be asleep today. I was thinking intersection. Sorry.
  9. Mar 11, 2008 #8
    I believe I got this done. This is my outline for the proof:

    Use Heine-Borel theorem, and the fact that in metric spaces compactness is equivalent to the sequential compactness, to prove that compactness of A+B follows from the compactness of A and B.

    Then for arbitrary closed A and B, consider sets

    A_m := A \cap [-m,m]^n,\quad\quad B_m := B \cap [-m,m]^n

    and some arbitrary point [itex]x\in\overline{A+B}[/itex].
  10. Mar 11, 2008 #9
    Well, consider the collection of sets [tex] \{i\}_i[/tex] where i is an integer. Now if we consider the set which is a sum of all these sets then it should be open.
    Last edited: Mar 11, 2008
  11. Mar 11, 2008 #10
    Opps sorry for my latex mistake.

    Any each set [tex]\{i\}[/tex] is closed isn't it? all limits withing [tex]\{i\}[/tex] contain i so it contains it's limit points. The set of natural numbers is not closed because it does not contain the upper limit right. Forgive me if my concepts of set theory are week.
  12. May 24, 2008 #11
    A+B need not to be closed!

    The sum of compact sets K_1 and K_2 is compact, and hence closed, as
    K_1 x K_2 is compact,
    Sum: R^n x R^n ---> R^n , Sum(x,y)=x+y is continuous,
    and the image of compact sets is compact under continuous functions.

    Thus, if you want to have a counterexample, you have
    to deal with unbounded sets, as:
    set X compact <===> X bounded and closed
    in R^n

    Why not take this in R^2:

    A= y-axis
    B= graph of function f:[0,1) ---> R, f(x) = 1/(1-x)

    A and B are closed.

    A + B = [0,1) x R,
    which is not closed.
  13. May 24, 2008 #12
    Very nice, I see my mistake.
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