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Thread: proper map. Distributions

  1. #1
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    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
    Last edited by yemino; Oct 2nd 2009 at 03:59 AM.
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  2. #2
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    And what does "A+B proper" means?
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  3. #3
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    Quote Originally Posted by InvisibleMan View Post
    And what does "A+B proper" means?
    ups,
    I was wrong, it should say closed instead of proper.


    Thanks InvisibleMan
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