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Thread: Density of a Set

  1. #1
    Super Member redsoxfan325's Avatar
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    Density of a Set

    Let $\displaystyle Y\subset X$ and let $\displaystyle A\subset X$. If $\displaystyle A$ is countable, $\displaystyle X$ is complete, and $\displaystyle Y\cup A=X$, is $\displaystyle Y$ necessarily dense in $\displaystyle X$?

    Prove it or give a counterexample.

    If the answer is no, then does making $\displaystyle X$ compact change anything?
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  2. #2
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    Quote Originally Posted by redsoxfan325 View Post
    Let $\displaystyle Y\subset X$ and let $\displaystyle A\subset X$. If $\displaystyle A$ is countable, $\displaystyle X$ is complete, and $\displaystyle Y\cup A=X$, is $\displaystyle Y$ necessarily dense in $\displaystyle X$?

    Prove it or give a counterexample.

    If the answer is no, then does making $\displaystyle X$ compact change anything?
    Kind of an obvious counterexample, isn't there? Take A and Y to be completely disjoint!
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  3. #3
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    Kind of an obvious counterexample, isn't there? Take A and Y to be completely disjoint!
    But $\displaystyle X$ is complete. I feel like that can't happen if $\displaystyle Y$ and $\displaystyle A$ are disjoint.
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    $\displaystyle X = \left[ { - 1,0} \right] \cup \left\{ {n^{ - 1} :n \in \mathbb{Z}^ + } \right\}$?
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  5. #5
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Plato View Post
    $\displaystyle X = \left[ { - 1,0} \right] \cup \left\{ {n^{ - 1} :n \in \mathbb{Z}^ + } \right\}$?
    I think that works. It would be complete because $\displaystyle Y$ is complete and the only Cauchy sequences in $\displaystyle A$ are all the same number or the sequence $\displaystyle \{a_n\}=\frac{1}{n}$ and both of those converge to something in $\displaystyle X$.

    It also answers the second part of the question because this set is also compact.
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