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

1. Dec 16, 2005

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. Dec 16, 2005

### Hurkyl

Staff Emeritus
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 "+".

3. Dec 16, 2005

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.

Last edited: Dec 16, 2005
4. Dec 16, 2005

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.

Last edited: Dec 16, 2005
5. Dec 17, 2005

### Hurkyl

Staff Emeritus
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.

6. Dec 17, 2005

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.

Last edited: Dec 17, 2005
7. Dec 18, 2005

### Hurkyl

Staff Emeritus
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.