1. proper map. Distributions

Hi, I have the follow question:

Probe that, if $\displaystyle A$ and $\displaystyle B$ are closet sets in $\displaystyle \mathbb{R}^n$, and the restriction of the map $\displaystyle (x,y)\mapsto x+y$ to $\displaystyle A+B$ is proper, then $\displaystyle A+B$ is closed.

Def: Let $\displaystyle A_1,\dots A_m$ be closet subsets of $\displaystyle \mathbb{R}^n$, We shall say the restriction of the map

$\displaystyle \mu:\mathbb{R}^{nm}\to\mathbb{R},\mu(x^{(1)},\dots x^{(m)})=x^{(1)}+\dots +x^{(m)}$

to $\displaystyle A_1\times\dots\times A_m$ is proper if, for any $\displaystyle \delta >0$ , there is $\displaystyle \delta '>0$ such that $\displaystyle x^{(j)}\in A_j,j=1,\dots m$ and $\displaystyle |x^{(1)}+x^{(m)}|\leq \delta$ imply that $\displaystyle |x^{(j)}|\leq\delta'$ for $\displaystyle j=1,\dots m$.

Thanks for you help.

PS: edited: in the first line, change: proper to closed. Tnaks InvisibleMan

2. And what does "A+B proper" means?

3. Originally Posted by InvisibleMan
And what does "A+B proper" means?
ups,
I was wrong, it should say closed instead of proper.

Thanks InvisibleMan