Completeness and Denseness in Metric Spaces

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Discussion Overview

The discussion revolves around the relationship between dense subsets and completeness in metric spaces. Participants explore whether the existence of a dense subset in a metric space implies that the space is complete, examining both theoretical implications and providing examples.

Discussion Character

  • Debate/contested
  • Conceptual clarification
  • Mathematical reasoning

Main Points Raised

  • One participant questions if a metric space with a dense subset must be complete, noting that the inverse is true.
  • Another participant argues that a dense set does not necessarily contain its limit points, which is a requirement for completeness.
  • A participant provides an example of the space formed by deleting 0 from the real line, stating that the rationals are dense in this space but it is not complete.
  • Further clarification is offered that having a dense subset does not imply completeness, reiterating the previous example.
  • Another participant explains that while a dense subset allows for sequences converging to points in the space, completeness requires that all Cauchy sequences converge within the space itself.
  • This participant emphasizes that completeness involves examining sequences in the space, not just those in a dense subset, and mentions the concept of "completion" of a metric space.

Areas of Agreement / Disagreement

Participants generally agree that having a dense subset does not imply that a metric space is complete. Multiple competing views and clarifications are present, particularly regarding the definitions and implications of density and completeness.

Contextual Notes

Some participants note the importance of limit points and Cauchy sequences in the discussion of completeness, highlighting that the relationship between dense subsets and completeness is nuanced and context-dependent.

_DJ_british_?
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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.

Thanks!
 
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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).
 
X is a dense subset of X.
 
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.
 
Hurkyl said:
X is a dense subset of X.

Right. I didn't think about that, thanks. So it means that having a dense subset doesn't imply a metric space is complete.

awkward said:
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.

And that's a nice exemple, thanks too!
 
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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