Let $A,B,C \subset \mathbb{R}^n$ be convex, bounded and closed set, then $A+C=B+C \Rightarrow A=B$

Question

Here a question and I would like to know if my solution is correct.
I think that it is not needed to use the fact that those sets are convexe, bounded & closed in order to solve the exercice, anyways below is my solution.

Question:
Let $A,B,C \subset \mathbb{R}^n$ convexe, bounded, not empty & closed set then $A+C=B+C \Rightarrow A=B$
Reminder: $A+C=\left \{ a+c : a \in A, c \in C \right \}$

EDIT: This answer is wrong please see my first post for the answer

My solution:
1-First lets prove that $A\subseteq B$, this is equivalent of proving the following contrapositive statement: $\forall x \notin B \Rightarrow x \notin A$
Now lets suppose by absurd that $x\notin B $ but $x\in A$. In such a case we will have that for any fixed $c' \in C $ that $x+c' \in A+C$. On the other hand by definition $x+c' \notin B+C$ and this is a clearly contradiction to the question because in such a case we get $B+C \subset A+C$.
2-The prove that $B\subseteq A$ is very similar.

Q.E.D.

Is this correct?

Answer 1

Fix $a \in A$. We wish to show $a \in B$.

Fix $c_0 \in C$. Then $a + c_0 \in A + C \subseteq B + C$. Thus, there exists some $b_1 \in B$ and $c_1 \in C$ such that $a + c_0 = b_1 + c_1$. Further, $a + c_1 \in A + C \subseteq B + C$, so there exist $b_2 \in B$ and $c_2 \in C$ such that $a + c_1 = b_2 + c_2$. Continuing this, we can form a sequence $(c_n)_{n=0}^\infty$ and $(b_n)_{n=1}^\infty$ in $C$ and $B$ respectively, satisfying: $$a + c_n = b_{n+1} + c_{n+1}.$$ Summing the first $n$ of these equations, $$na + \sum_{i=0}^{n-1} c_i = \sum_{i=1}^n b_i + \sum_{i=1}^n c_i \iff a = \frac{1}{n}\left(c_n - c_0 + \sum_{i=1}^n b_i\right).$$ Since $C$ is bounded, $(c_n - c_0) / n \to 0$. Thus, $$\frac{1}{n} \sum_{i=1}^n b_i \to a.$$ As $B$ is convex, this sequence lies in $B$. As $B$ is closed, the limit, $a$, lies in $B$. Thus $a \in B$, as required.

Note: not all premises are necessary here. We actually use no assumptions on $A$; literally any set will work (including the empty set, vacuously). We only require $C$ to be non-empty and bounded (convexity and closedness are optional). We do require $B$ to be convex and closed. It could be unbounded, and it could be empty (though the $B = \emptyset$ case is fairly trivial).

Answer 2

I found a brief answer (not from me) that I develop below with more details. In all the case if someone has an other solution (shorter) I will be happy to read it.

1-
We define the sub linear function $p_A(x)=Sup \left \{z\cdot x|\forall z\in A \right \}$ (which is a convex function). Moreover by definition: $ \partial p_A (x)=\left \{ d \in \mathbb{R}^n|p_A(y)\geq p_A(x)+d\cdot (y-x), \forall y \in \mathbb{R}^n \right \}$ and as $p_A(x)$ is a sub linear function we get: $\partial p_A (x)=\left \{ d \in \mathbb{R}^n| p_A(x) \geq d\cdot x, \forall x \in \mathbb{R}^n \right \}$. This means that $\partial p_A (x)$ is the set of all the vectors $d \in \mathbb{R}^n$ that verify $Sup \left \{z\cdot x|\forall z\in A \right \} \geq d\cdot x$.
It is obvious that $\forall d \in A \subseteq \mathbb{R}^n$ this inequality is verify $\Rightarrow A \subseteq \partial p_A$
On the other hand if $d \notin A \Rightarrow Sup \left \{z\cdot x|\forall z\in A \right \}$ is not define and hence $d \notin \partial p_A \Rightarrow \partial p_A \subseteq A $
We just proved that $A = \partial p_A, B = \partial p_B, C = \partial p_C$.

2-
By property and what we wrotte above, we know that: $\partial (p_A + p_C)= \partial p_A + \partial p_C= A+C=B+C= \partial p_B + \partial p_C =\partial (p_B + p_C) \Rightarrow \partial (p_A + p_C) = \partial (p_B + p_C) \Rightarrow p_A+p_C=p_B+p_C \Rightarrow p_A=p_B \Rightarrow A=\partial p_A= \partial p_B= B$
Q.E.D.

It is nearly ok but I need help to justify correctly $\partial (p_A + p_C) = \partial (p_B + p_C) \Rightarrow p_A+p_C=p_B+p_C$ is it a property for sub derivative that I don't know?

EDIT: The prove of this step is here and has been given to me by @Theo Bendit