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Math Help - metric spaces

  1. #1
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    metric spaces

    Hello
    Last edited by rolylane; January 29th 2008 at 11:32 AM. Reason: Trying to fix BB code
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  2. #2
    Senior Member JaneBennet's Avatar
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    The first one is straightforward. If x\in\mathrm{Int}(A)\cup\mathrm{Int}(B), then x is in an open ball S contained in A or one contained in B; in either case, x\in S\subseteq A\cup B and so x\in\mathrm{Int}(A\cup B).

    For the second part, proving \mathrm{Int}(A\cap B)\subseteq \mathrm{Int}(A)\cap\mathrm{Int}(B) is equally straightforward. x\in\mathrm{Int}(A\cap B) \Rightarrow x is in an open ball S contained in A\cap B \Rightarrow \Rightarrow x\in S\subseteq A and x\in S\subseteq B \Rightarrow x\in\mathrm{Int}(A) and x\in\mathrm{Int}(B) \Rightarrow x\in\mathrm{Int}(A)\cap\mathrm{Int}(B).

    For the reverse inclusion, let x\in\mathrm{Int}(A)\cap\mathrm{Int}(B).

    x\in\mathrm{Int}(A) \Rightarrow \exists\,\delta_1>0 such that x\in S_1=\{y:\mathrm{d}(x,y)<\delta_1\}\subseteq A. Similarly x\in\mathrm{Int}(B). \Rightarrow \exists\,\delta_2>0 such that x\in S_2=\{y:\mathrm{d}(x,y)<\delta_2\}\subseteq B.

    Now take \delta=\min\{\delta_1,\delta_2\}.

    Then x\in S=\{y:\mathrm{d}(x,y)<\delta\}\subseteq S_1\cap S_2\subseteq A\cap B proving that x\in \mathrm{Int}(A\cap B).

    For the example, take the closed intervals A=[0,1] and B=[1,2] in \mathbb{R}. Then \mathrm{Int}(A\cup B)=(0,2) but \mathrm{Int}(A)=(0,1) and \mathrm{Int}(B)=(1,2).
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  3. #3
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    Thank you so much for your time and effort in replying to my query. I really am grateful.
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