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Math Help - How to prove such a sequence exists

  1. #1
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    How to prove such a sequence exists

    Let A be a non-empty subset of Y, and let x\in y.
    If d(x,A)=0
    then there is a sequence x_n of points in A such that x_n converges to x.

    Can anyone give me a starting point here? Should I try to prove this by contradiction? I'm just unsure of how to prove a sequence exists...
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  2. #2
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    Re: How to prove such a sequence exists

    Quote Originally Posted by paupsers View Post
    Let A be a non-empty subset of Y, and let x\in y.
    If d(x,A)=0
    then there is a sequence x_n of points in A such that x_n converges to x.
    If x\in A what sequence would work? (note the terms do not have to be distinct)

    If x\notin A then \left( {\exists n \in \mathbb{Z}^ +  } \right)\left( {\exists x_n  \in A} \right)\left[ {d(x,x_n ) < \frac{1}{n}} \right].
    Now you need to explain why that is true.
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  3. #3
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    Re: How to prove such a sequence exists

    Thanks... I had actually found a proof that works right after I posted this! I was coming back to say that but you had already commented. :-)
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