Is there a counterexample for the claim in the question subject, that a sum of two closed sets in $\mathbb R$ is closed? If not, how can we prove it?
(By sum of sets $X+Y$ I mean the set of all sums $x+y$ where $x$ is in $X$ and $y$ is in $Y$)
Thanks!
Consider the sets $A=\{ n\mid n=1,2,\ldots\}$ and $B=\{- n+{1\over n}\mid n= 2,3,\ldots\}$. Note that $0$ is not in the sum, but $1\over n$ is for each $n\ge2$.
consider $\mathbb Z$ and $\sqrt 2 \mathbb Z$ both are closed but the sum is not...:) moreover it is dense on $\mathbb R$
It's worth mentioning that :
if one is closed + bounded, another one is closed,then the addition is closed
Since closedness can be charaterized by sequence in $\Bbb{R}^n$,if $(x_n) \in A+B$ we need to show limit of the convergence sequence still lies in it.assume $A$ is compact $B $ is closed.
Since $x_n= a_n +b_n \to x$,compactness implies sequential compactness hence $a_{n_k} \to a\in A$ for some subsequence. now $x_{n_k} \to x$ which means subsequence $b_{n_k}\to x-a$ converge,since $B$ is closed,$x-a \in B$ ,hence $x = a+b \in A+B$,which means the sum is closed.
Take $A=\{(a,0):a\in\mathbb{R}$ and $B=\{(b,\frac{1}{b}):b\in \mathbb{R}-\{0\}\}$. Then both $A,B\subset \mathbb{R}^2$ are closed. But $A+B=\{(a+b,\frac{1}{b}):a\in \mathbb{R},b\in \mathbb{R}-\{0\}\}.$The sequence $\{(0,\frac{1}{n})\}=\{(n-n,\frac{1}{n})\}\subset A+B$ but the limit $(0,0)$ which is not in the sum.
The sum $E +F$ may fail to be closed even if $E$ and $F$ are closed. For instance, set $E = \{(x, y) \in \mathbb R^2 : y > 1/x\text{ and }x > 0\}$ and $F = \{(x, y) \in\mathbb R^2 : y > -1/x\text{ and }x < 0\}$
Then $E$ and $F$ are closed, but $$E + F = \{(x, y) \in \mathbb R^2 : y > 0\}$$ is not closed.
