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Math Help - IS it complete?

  1. #1
    Junior Member sorv1986's Avatar
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    Exclamation IS it complete?

    Is the set D={(x,y)∈R:x+y<1} complete?

    how?
    actually i am in dark how to do it.

    thanks in advance
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by sorv1986 View Post
    Is the set D={(x,y)∈R:x+y<1} complete?

    how?
    actually i am in dark how to do it.

    thanks in advance
    There is a common theorem which says that for subspaces of a complete space completeness and closedness are synonomous. To be more explicit, if D were complete the sequence \displaystyle 1-\frac{1}{n} (I'm thinking in a+bi notation) would converge to some point of D. Does it?
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  3. #3
    Junior Member sorv1986's Avatar
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    Quote Originally Posted by Drexel28 View Post
    There is a common theorem which says that for subspaces of a complete space completeness and closedness are synonomous. To be more explicit, if D were complete the sequence \displaystyle 1-\frac{1}{n} (I'm thinking in a+bi notation) would converge to some point of D. Does it?
    a lot of thanx. i got the concept what you wanna said.But (1,-1/n) belongs to that D? hey!i am a fan of yours.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by sorv1986 View Post
    a lot of thanx. i got the concept what you wanna said.But (1,-1/n) belongs to that D? hey!i am a fan of yours.
    Haha, thanks. I'm not quite sure what you mean. The sequence I mentioned as actually \left(1-\frac{1}{n},0\right) which is Cauchy in D yet converges to (1,0)\notin D.
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  5. #5
    Junior Member sorv1986's Avatar
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    Quote Originally Posted by Drexel28 View Post
    Haha, thanks. I'm not quite sure what you mean. The sequence I mentioned as actually \left(1-\frac{1}{n},0\right) which is Cauchy in D yet converges to (1,0)\notin D.
    got it. Then any subset of R which is closed, is complete and which is not closed, is not complete. is it right?
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by sorv1986 View Post
    got it. Then any subset of R which is closed, is complete and which is not closed, is not complete. is it right?
    Right--it holds more generally for any complete metric space.
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