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

    Prove it or give a counterexample.

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

    Prove it or give a counterexample.

    If the answer is no, then does making 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 X is complete. I feel like that can't happen if Y and A are disjoint.
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    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
    X = \left[ { - 1,0} \right] \cup \left\{ {n^{ - 1} :n \in \mathbb{Z}^ +  } \right\}?
    I think that works. It would be complete because Y is complete and the only Cauchy sequences in A are all the same number or the sequence \{a_n\}=\frac{1}{n} and both of those converge to something in X.

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