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Thread: 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 $\displaystyle x\in y$.
    If $\displaystyle d(x,A)=0$
    then there is a sequence $\displaystyle x_n$ of points in A such that $\displaystyle x_n$ 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 $\displaystyle x\in y$.
    If $\displaystyle d(x,A)=0$
    then there is a sequence $\displaystyle x_n$ of points in A such that $\displaystyle x_n$ converges to x.
    If $\displaystyle x\in A$ what sequence would work? (note the terms do not have to be distinct)

    If $\displaystyle x\notin A$ then $\displaystyle \left( {\exists n \in \mathbb{Z}^ + } \right)\left( {\exists x_n \in A} \right)\left[ {d(x,x_n ) < \frac{1}{n}} \right]$.
    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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