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Math Help - few Properties in normed linear space

  1. #1
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    few Properties in normed linear space

    Let A,B be subsets of a normed linear space X.
    Show that 1) If A or B is open then A+B is open.
    2) If A and B both are compact then A+B is compact..
    Any help appreciated..
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  2. #2
    Senior Member jakncoke's Avatar
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    1)are you sure it's not if A and B is open?
    2) Assuming both A and B are both compact, then for any open cover  {O_a}| a \in S covering A we can find a finite subset G of S s.t  {O_p} | p\in G is also an open cover for A. same goes for B... Is the union of two finite sets finite?
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  3. #3
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    Quote Originally Posted by math.dj View Post
    Let A,B be subsets of a normed linear space X.
    Show that 1) If A or B is open then A+B is open.
    Suppose that A is open. Let a+b\in A+B and d\in X with \|d-(a+b)\|<\delta. Then \|(d-b)-a\|<\delta, which implies that d-b\in A if \delta is sufficiently small (because A is open). Then d = (d-b)+b\in A+B. That shows that A+B is open.
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    Hey for the Part 2) if I define a map f:A*B-->A+B by f(a,b) = a+b then f is a continous map..Also A*B is compact..all i need to show is f is surjective..then i can say that A+B is also compact..
    Please help me in the surjection part!
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  5. #5
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    Quote Originally Posted by math.dj View Post
    Hey for the Part 2) if I define a map f:A*B-->A+B by f(a,b) = a+b then f is a continous map..Also A*B is compact..all i need to show is f is surjective..then i can say that A+B is also compact..
    Please help me in the surjection part!
    Yes, that's a neat way to do part 2), and the surjectivity is really obvious! In fact, by definition any element of A+B is of the form a+b, for some a in A and b in B. So it is equal to f(a,b). That's all there is to it.
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