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Let $X=(X,|\cdot|$) be a Banach space and let $A,B \subset X$. Then $$ \overline{A+B} \subset \overline{A}+\overline{B}\:?$$ Where, $\overline{A+B}$, $\overline{A}$ and $\overline{B}$ denotes the closure of the sets $A+B$, $A$ and $B$, respectively.

I do not know if this is true. But I tried to prove it as follows:

Let $x \in \overline{A+B}$ then exist $(x_n)_{n \in\mathbb{N}} \subset A+B$ such that $x_n \longrightarrow x$. In addition also, exists $(a_n)_{n \in\mathbb{N}} \subset A$ and $(b_n)_{n \in\mathbb{N}} \subset B$. So, $$a_n+b_n=x_n \longrightarrow x.$$ Hence, exists $y \in A$ and $z \in B$ such that $$ a_n \longrightarrow y \:\: \text{and} \:\: b_n \longrightarrow z.$$ Therefore, $y+z=x$ and $y+z \in \overline{A}+\overline{B}.$ Therefore, $ \overline{A+B} \subset \overline{A}+\overline{B}$.

I was in doubt if I can guarantee the existence of the elements $ y $ and $ z $, in $ A $ and $ B $, respectively.

Is that correct?

Guilherme
  • 1,717

2 Answers2

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The result is not true, even in finite-dimension. Consider $X=\mathbb R^2$ with any norm you want (the topology is the important part, and they all give the same one). Consider $A=\mathbb R\times\{0\}$, and $B=\{(x,\frac{1}{x}):x>0\}$. It shouldn't be too hard to convince yourself that $A$ and $B$ are closed, and that $$A+B=\{(x,y)\in\mathbb R^2:y>0\}$$ is not closed. Thus $\overline{A+B}$ is not a subset of $A+B=\overline{A}+\overline B$.

Aweygan
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Let $X=\ell_2$. Let $(e_n)_{n=1}^\infty$ be the canonical $\ell_2$ basis. Let $A=\{e_n: 1<n\in \mathbb{N}\}$ and let $B=\{-e_n+\frac{1}{n}e_1: 1<n\in \mathbb{N}\}$. Then $A,B$ are closed sets (they are norm discrete, each point of $A$ is a distance at least $\sqrt{2}$ to any other point of $A$, and similarly for $B$). Moreover, $0\notin A+B=\overline{A}+\overline{B}$. But $e_n+(-e_n+\frac{1}{n}e_1)\in A+B$, so $0\in \overline{A+B}$.