Talk:Complete metric space
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I removed this link:
from the discussion of topologically complete spaces, since it's malformed and didn't fit into the sentence. If anybody knows what it's about, please put it back in as a legible sentence and with a link that might be the title of an article (perhaps it should be two).
I also moved the definition of Cauchy net to its own page.
-- Toby 23:40 Feb 20, 2003 (UTC)
Inner Product Spaces
What's all this about inner product spaces? There's nothing in Inner product space about completeness -- except completeness in a metric! If anything's wrong, then it's that this article should have something to say about how the ideas are applied to certain types of metric spaces -- such as inner product spaces. -- Toby 00:50 Feb 21, 2003 (UTC)
OK ... I think I was hasty and actually had in mind the idea of "complete" orthonormal set in an inner product space. Such an "orthonormal basis" is not always a "basis" in the sense of "Hamel basis", i.e., not every vector in such a space is a finite linear combination of basis vectors. And in infinite linear combinations, there are various different notions of convergence, and the one that matters in this case is the one defined by the metric. Michael Hardy 01:05 Feb 21, 2003 (UTC)
Yes, and there's a link to Completeness from Hermitian that's about this. So we should add it to Completeness (I just did). But this is complete orthonormal set, not complete space. -- Toby 03:46 Feb 21, 2003 (UTC)
complete spaces vs. closed set
from this article i don't really understand what is the difference between a closed set and a complete space. maybe someone can give an example of a closed space which ins't complete? —The preceding unsigned comment was added by 184.108.40.206 (talk) 09:38, 8 February 2007 (UTC).
Spaces aren't closed or open; subspaces are/aren't. Consider the rational numbers, Q, thought of as a topological space with the topology induced from the usual Borel topology on the real line R. Q is a closed subset of itself, since the empty set is an open subset of any topological space. However, Q is not complete; consider the sequence of successive decimal approximations to the square root of 2, which is Cauchy but not convergent. Sullivan.t.j 14:14, 8 February 2007 (UTC)
Other formulations of the completeness axiom
Is there a reason this specific formulation of the Completeness axiom was chosen over, say, the Supremum formulation, or that of Monotonic Convergence? Should these not at least be mentioned with a discussion of the benefits / downfalls of different formulations? I ask this specifically as the Supremum (or Infimum) formulation is often easier for the non-mathematician to understand. In a sense the most complex formulation is that represented here, and the question is whether this is where Wikipedia articles should be aimed.
-Jdr [13/3/2007 - 10:06pm GMT]
- "Supremum" makes sense when there is an ordering, but in metric spaces generally there is no supremum. Which "monotic convergence" formultation are you talking about? Does it work in metric spaces generally? "Monotonic" seems to connote an ordering. Michael Hardy 22:54, 13 March 2007 (UTC)
In the text there is the example "The topological vector space R^ω of sequences of real numbers which have finitely many nonzero terms (the topology of this space can alternatively be defined as the limit topology of the Rn or as the coproduct of infinitely many copies of R) is not complete (even though its underlying field is). The completion of this space is the product ΠR of infinitely many copies of R. If instead we endow the space with the lp norm, its completion is the space lp(N)."
This contains many false statements, like
1.) The beginning "The topological vector space R^ω of sequences of real numbers which have finitely many nonzero terms" is already a bit strange, since so far no topology is given and therefore R^ω is just a vector space and not yet a topological one. There are many totally different topologies on this vector space.
2.) "the topology of this space can alternatively be defined as the limit topology of the Rn or as the coproduct of infinitely many copies of R" Yes, it is possible to endow this vector space with this particular topology, but then is not metrizable (and therefore not at all a good example for a "Complete Metric Space"). By the way, the space R^ω endowed with this topology is complete but not with respect to a metric (since there is none defining this topology) but with respect to the uniform strucutre as a topological vector space.
3.) The completion of this coproduct in number (2.) is NOT the product ΠR, since the coproduct is already complete as mentioned above. It is true that the product can be the completion of R^ω, but with another topology (which that in fact will be again metrizable)
4.) I admit that the last sentence is correct. "If instead we endow the space with the lp norm, its completion is the space lp(N)."
I would suggest that we remove this example or at least restrict it to the case where we put an lp-norm on it. --220.127.116.11 08:09, 7 August 2007 (UTC)
- I cut out the example per your item 1 (I am not competent enough to judge items 2 and 3. :) There are many good examples in the text that we can get by without this very technical example, especially that it is not well-written and possibly incorrect. Oleg Alexandrov (talk) 19:43, 7 August 2007 (UTC)
Small correction is needed
The article says, in the section Some theorems:
- If X is a set and M is a complete metric space, then the set B(X,M) of all bounded functions f from X to M is a complete metric space
I don't think one can make sense of "bounded functions" to a metric space. Either one takes all functions and equip M with a bounded metric, or take M a Banach space. --Bdmy (talk) 18:11, 12 January 2009 (UTC)
- This looks fine to me, with perhaps one minor clarification. Take F to be the space of all functions from X to M and equip it with the uniform structure of uniform convergence. This uniform structure can be defined with a single pseudometric d(f,g) = sup d(f(x),g'(x)) where x ranges over X. (This is in general not a finite pseudometric.) As the uniform space F is clearly Hausdorff, it is also metrizable. And if M is complete, the the uniform space F is complete.
- Now call a function f from X to M bounded if the image of X by f is bounded.
- I don't remember of any interesting application of the notion of bounded set in a general metric space (not even sure that it is defined in books, although I agree that there's no much choice), and in my opinion a definition of bounded function should be given. Other than that, I agree with your preceding paragraph: taking this uniform structure with infinite values, you will define a complete space of all functions, without this rather awkward restriction of bounded. Of course I understand that one wants a finite metric, but this could be achieved by replacing d by min(d, 1) without changing much of the interesting properties. --Bdmy (talk) 22:39, 12 January 2009 (UTC)
- Bonjour. Agree, bounded functions should be defined in the article before the statement. And exactly as you write, the min(d,1) is a metric that produces the same uniform structure. As for these concepts being defined in books, they certainly found their way into Topologie Générale of Bourbaki (Ch. X). In fact, Bourbaki treats the somewhat more general set-up: let S be a set of subsets of X. Then one defines in the obvious way the uniformity and topology of uniform convergence in the sets of S (or simply S-convergence). If S consists of just X we're back to uniform convergence, of course, and we can produce the uniformities of compact convergece (for X a topological space), of simple convergence etc. this way. Now carry out all of the above in this more general set-up: the set F equipped with the uniformity of S convergence is a complete uniform space for any choice of S if just M is complete (M does not have to be even metric for this). Also, with M metric, define BS as the set of functions which are bounded in each of the sets in S. Equip also BS with the uniformity of S-convergence, and it is true that it is a complete uniform space; indeed, it is again closed in F equipped with the topology of S-convergence. Amitiés, Stca74 (talk) 07:35, 13 January 2009 (UTC)
- The set B of these bounded functions is a subset of F and the restriction of the supremum pseudometric above to B is by construction finite, hence a metric. It defines on B the topology of uniform convergence (making B a uniform subspace of F). Now clearly a uniform limit of functions in B is bounded, and hence B is closed inside F. But then as a closed subset of a complete (metrizable) uniform space B is itself complete. So to be fully explicit the article should make it explicit that it is the sup metric defining the topology and uniformity of uniform convergence with respect to which B is a complete metric space. Stca74 (talk) 22:20, 12 January 2009 (UTC)
Characterization of complete spaces through nested balls
I think the following fact (from A. A. Kolmogorov, S. V. Fomin "Introductory real analysis") should be added in the section "Some Theorems" (or even under a new title „A Completeness Criterion of a Metric Space“) or in the as of now poor article "Nested intervals":
Page 60 THEOREM 2 (Nested sphere theorem). A metric space R is complete if and only if every nested sequence of closed spheres in R such that as has a nonempty intersection.
- I think this has to do with expansion constants, so I added a bit. -- Taku (talk) 21:57, 23 February 2009 (UTC)