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

[kaosAD]

If A and B are nonempty convex sets. And C = A + B. How to prove int(C) = int(A) + int(B)?

 

[Hurkyl]

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 "+".

 

[kaosAD]

What I mean is C = \{ x = x_1 + x_2 \ | \ x_1 \in A, x_2 \in B \}. I don't know how to prove it.

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

\alpha x_1^1 + (1 - \alpha) x_2^1 \in A

and

\alpha x_1^2 + (1 - \alpha) x_2^2 \in B.

Hence \alpha (x_1^1 + x_1^2) + (1 - \alpha)(x_2^1 + x_2^2) \in C. But this shows C is convex.

 

[kaosAD]

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 \bar{x} \in \textup{bd}(A). Consider a sequence \{x_k\} belonging to int(A) and converging to a limit point \bar{x}. Pick any y \in B. Hence the sequence \{x_k + y\} converges to some point, say \bar{z}. This \bar{z} may or may not belong to int(C). I reckon if y \in \textup{bd}(B), then the \bar{z} does not belong to int(C).

I don't think it goes well following this line of argument. Need some help.

 

[Hurkyl]

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 \bar{x} \in \textup{bd}(A). Consider a sequence \{x_k\} belonging to int(A) and converging to a limit point \bar{x}. Pick any y \in B. Hence the sequence \{x_k + y\} converges to some point, say \bar{z}. This \bar{z} may or may not belong to int(C). I reckon if y \in \textup{bd}(B), then the \bar{z} does not belong to int(C).
I don't think it goes well following this line of argument. Need some help.

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 x_k and point y, and see how things play out? Sometimes concrete examples help with our intuition.

Also, you make the statement "This \bar{z} 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.

 

[kaosAD]

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

Alright, it is rubbish.

 

[Hurkyl]

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.