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Math Help - Closure Proof

  1. #1
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    Closure Proof

    Hi all,
    I'm just wondering if you could show me how to prove this statement. Obviously, it's intuitively clear but I just can't get a proof working for it:

    Closure(B(1,a)) = { x in R^n: |x-a| <= 1 }

    Thanks a lot!
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: Closure Proof

    Quote Originally Posted by AKTilted View Post
    Hi all,
    I'm just wondering if you could show me how to prove this statement. Obviously, it's intuitively clear but I just can't get a proof working for it:

    Closure(B(1,a)) = { x in R^n: |x-a| <= 1 }

    Thanks a lot!
    You know that RHS is a closed set containing B_1(a) and so \overline{B_1(a)}\subseteq RHS. That said, since each x with \|x-a\|=1 is a limit point of B_1(a) (why?) one must have that RHS\subseteq\overline{B_1(a)}, can you conclude?


    Oh, and it's not as intuitively obvious as one would hope. For example, in a discrete space (X,d) one has that \overline{B_1(x)}=B_1(x) but \left\{y\in X:d(x,y)=1\right\}=X for all x\in X
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  3. #3
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    Re: Closure Proof

    Does this line of logic work?

    If I assume that the closure of the ball, B(1,a) = { x in R^n: |x-a| <= 1 } and then go to prove that this is closed?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Re: Closure Proof

    Quote Originally Posted by AKTilted View Post
    Does this line of logic work?

    If I assume that the closure of the ball, B(1,a) = { x in R^n: |x-a| <= 1 } and then go to prove that this is closed?
    I'm sorry, I can't understand what you mean there.
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  5. #5
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    Re: Closure Proof

    Basically this:

    If I can prove that { x in R^n: |x-a| <= 1 } is a closed set, can I use this to state that Closure(B(1,a)) = { x in R^n: |x-a| <= 1 }?
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Re: Closure Proof

    Quote Originally Posted by AKTilted View Post
    Basically this:

    If I can prove that { x in R^n: |x-a| <= 1 } is a closed set, can I use this to state that Closure(B(1,a)) = { x in R^n: |x-a| <= 1 }?
    No..you can use it to state that \overline{B_1(a)}\subseteq\left\{x\in\mathbb{R}^n:  \|x-a\|\leqslant 1\right\}.
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