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Homework Help: General Topology

  1. Aug 29, 2007 #1
    I have the following [itex]A\subset\mathbb{R}^{n}[/itex] is dense then [itex]A[/itex] isn't bounded. Is this true? I know that [itex]A[/itex] is dense iff [itex]\bar{A}=\mathbb{R}^{n}[/itex] and that [itex]A[/itex] is bounded iff [itex]\exists \epsilon>0\mid B_{\epsilon}(0)\supset A[/itex]. How to proof it? Or there is an counterexample?
     
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  3. Aug 29, 2007 #2

    George Jones

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    Try reductio ad absurdum (proof by contradiction).
     
  4. Aug 30, 2007 #3

    matt grime

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    It's slightly nicer to show that a bounded set in R cannot be dense. Many proofs by contradiction are unnecessary - i.e. you wish to show A implies B, so you assume A and not B and show not B implies not A, without any use of the assumption of A.
     
  5. Aug 30, 2007 #4

    HallsofIvy

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    But you just said "Assume A"!! Anyway, many people would consider proving the contrapositive to be "proof by contradiction".
     
  6. Aug 30, 2007 #5

    matt grime

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    What? The point was just make a constructive proof of the contrapositive statement without making an unnecessary preliminary assumption.
     
  7. Aug 30, 2007 #6

    George Jones

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    Yes, I see. As I read the original post, I said to myself "Well, they both can't be true (being dense and bounded)," and this placed proof by contradiction in my mind. Bot all that is needed is ~B (bounded), so contapositive gives a direct proof.
     
  8. Aug 30, 2007 #7
    Thank you for the help!!
     
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