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Math Help - Question about open extensions of sets in metric spaces.

  1. #1
    Junior Member RaisinBread's Avatar
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    Question about open extensions of sets in metric spaces.

    Let (X,d) be an arbitrary metric space. For every A\subset X and t>0, we define the open t-extension of A as the set

    A_t=\{x\in X:d(x,A)<t\},

    where d(x,A)=\inf\{d(x,y):y\in A\}.

    It seems intuitive that, given A\subset X and s,t>0, we should have that A_{t+s}=(A_t)_s=(A_s)_t. However, I haven't been able to even begin to construct a rigorous argument to prove it. Any hint or proposed direction would be greatly appreciated (or counterexample in the event that it is false).
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  2. #2
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    Re: Question about open extensions of sets in metric spaces.

    Quote Originally Posted by RaisinBread View Post
    Let (X,d) be an arbitrary metric space. For every A\subset X and t>0, we define the open t-extension of A as the set
    A_t=\{x\in X:d(x,A)<t\},
    where d(x,A)=\inf\{d(x,y):y\in A\}.
    It seems intuitive that, given A\subset X and s,t>0, we should have that A_{t+s}=(A_t)_s=(A_s)_t. However, I haven't been able to even begin to construct a rigorous argument to prove it. Any hint or proposed direction would be greatly appreciated (or counterexample in the event that it is false).
    When you say that you are unable "to construct a rigorous argument", does that mean that you cannot even start?
    Can you show that (A_t)_s\subseteq A_{t+s}~?

    Before I work on this, please post what you have been able to do.
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