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Completeness and denseness

  1. Jun 3, 2010 #1
    I have a little question that's bothering me : Suppose you have a metric space X with a dense subset A. Can we say that X is complete? I mean that if a metric space have a dense subset, is it necessarily complete? I know that the inverse is true (that every complete metric space have a dense subset) so I was wondering about it.

  2. jcsd
  3. Jun 3, 2010 #2
    In general, no, I don't think so because a dense set does not have to contain it's limit points (a complete space would have to).
  4. Jun 3, 2010 #3


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    X is a dense subset of X.
  5. Jun 3, 2010 #4
    Suppose X is the space formed by deleting 0 from the real line. Then the rationals are dense in X, but X is not complete.
  6. Jun 4, 2010 #5
    Right. I didn't think about that, thanks. So it means that having a dense subset doesn't imply a metric space is complete.

    And that's a nice exemple, thanks too!
  7. Jun 6, 2010 #6


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    If X has a dense subset A, then for every point x of X there is sequence in A converging to x. But for X to be complete, every Cauchy-sequence in X has to converge in X.

    If X is not complete, then there are sequences in X which "want to converge" to some element which is not in X. (This might sound vague: in the standard example of a sequence in Q "wanting to converge to sqrt{2}" it is obvious because we know that Q sits in R so we know that sqrt{2} exists. But what if we don't have an obvious "larger space"? This can be made precise: every metric space admits a "completion", which basically amounts to adding all "limits of Cauchy sequences". See here for details.) So for completeness you really need to look at sequences in X, it is not enough to look at sequences in a dense subset.
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