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If A and B are nonempty convex sets, and C = A + B, then . . . .

  1. Dec 16, 2005 #1
    If A and B are nonempty convex sets. And C = A + B. How to prove int(C) = int(A) + int(B)?
     
    Last edited: Dec 16, 2005
  2. jcsd
  3. Dec 16, 2005 #2

    Hurkyl

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    By starting with the definitions, presumably.

    What have you done on this problem?

    P.S.: You really ought to define terms when there might be confusion. "+", for example, could easily mean "union". In fact, that's what I thought it meant at first, but now that I've thought about the problem, I suspect that's not how you're using "+". :smile:
     
  4. Dec 16, 2005 #3
    What I mean is [tex]C = \{ x = x_1 + x_2 \ | \ x_1 \in A, x_2 \in B \}[/tex]. I don't know how to prove it.

    However I am able to show C is convex. Pick [tex]x_1^1, x_2^1 \in A[/tex] and [tex]x_1^2, x_2^2 \in B[/tex]. Since A, B are convex, then for [tex]0 \leq \alpha \leq 1[/tex],

    [tex]\alpha x_1^1 + (1 - \alpha) x_2^1 \in A[/tex]

    and

    [tex]\alpha x_1^2 + (1 - \alpha) x_2^2 \in B[/tex].

    Hence [tex]\alpha (x_1^1 + x_1^2) + (1 - \alpha)(x_2^1 + x_2^2) \in C[/tex]. But this shows C is convex.
     
    Last edited: Dec 16, 2005
  5. Dec 16, 2005 #4
    This is what I have tried a some what modified problem. But it goes nowhere.

    I will use 'cl' to mean closure and 'bd' to mean boundary. Assume A = cl(A) and B = cl(B) to make life easier for me. Let [tex]\bar{x} \in \textup{bd}(A)[/tex]. Consider a sequence [tex]\{x_k\}[/tex] belonging to int(A) and converging to a limit point [tex]\bar{x}[/tex]. Pick any [tex]y \in B[/tex]. Hence the sequence [tex]\{x_k + y\}[/tex] converges to some point, say [tex]\bar{z}[/tex]. This [tex]\bar{z}[/tex] may or may not belong to int(C). I reckon if [tex]y \in \textup{bd}(B)[/tex], then the [tex]\bar{z}[/tex] does not belong to int(C).

    I don't think it goes well following this line of argument. Need some help.
     
    Last edited: Dec 16, 2005
  6. Dec 17, 2005 #5

    Hurkyl

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    Have you tried some concerete examples?

    Say, maybe work in R and let A = [0, 1] and B = [3, 4]. Then pick an actual sequence [itex]x_k[/itex] and point y, and see how things play out? Sometimes concrete examples help with our intuition.

    Also, you make the statement "This [tex]\bar{z}[/tex] may or may not belong to int(C)." This suggests to me that you ought to try and back it up -- find an actual example where it does belong, and an actual example where it does not belong.
     
  7. Dec 17, 2005 #6
    I was right about it using your simple example. Not sure if it carries to higher dimension.
    Anyway I am still not sure how to go about proving the statement.

    Given A, B convex sets in [tex]\mathbb{R}^n[/tex]. Define [tex]C \triangleq A + B = \{x + y \ | \ x \in A, y \in B \}[/tex]. Consider two sequences [tex]\{x_k\} \subset \textup{int}(A), \{y_k\} \subset \textup{int}(B)[/tex] converging to [tex]\bar{x} \in \textup{bd}(A), \bar{y} \in \textup{bd}(B)[/tex], respectively, as [tex]k \to \infty[/tex]. Hence [tex]x_k + y_k \to x + y \notin \textup{int}(A) + \textup{int}(B)[/tex]. Hence [tex]x + y \notin \textup{int}(C)[/tex]. So [tex]\textup{int}(C) = \textup{int}(A + B) = \textup{int}(A) + \textup{int}(B)[/tex].

    Alright, it is rubbish.
     
    Last edited: Dec 17, 2005
  8. Dec 18, 2005 #7

    Hurkyl

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    I'm slightly confused as to what you're saying. Here's something to consider:

    What if we take a sequence in [0, 1] converging to 1, and a sequence in [2, 3] converging to 2? Doesn't their sum converge to something in the interior of [0, 1] + [2, 3]?

    But since we're dealing with open sets, it might be more fruitful to consider open neighborhoods of points.
     
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