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Metrization question

  1. Jul 31, 2007 #1
    I'm trying to understand Munkres' proof of the Nagata-Smirnov metrization theorem : A space X is metrizable iff X is regular and has a basis that is countably locally finite.

    He takes the union of all the Bn collections of the countably locally finite basis. Roughly, he then constructs a function F: X -> [0,1]^J where J is a subset of ZxB where B is all the collection of all elements in the union of all the Bn's.

    Then he concludes that this imbedding implies X is metrizable.

    All this I follow. But I thought that if J was uncountable, then [0,1]^J isn't metrizable. And hence his result would only follow if we could show that J is countable.

    Am I on the right track? Do I need to show that J is always countable? Or is [0,1]^J with uncountable J metrizable?

    thanks
     
  2. jcsd
  3. Aug 1, 2007 #2

    mathwonk

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    i would think you want J countable. can you tell if it is by looking more carefully at the definition of J?
     
  4. Aug 1, 2007 #3
    Unfortunately, I don't see any obvious ways of insuring J is countable.

    In the construction, J is the subset of Z+ x B consisting of all pairs (n,B) such that B is an element of Bn.

    So as far as I can tell, this all boils down to the size of B. And I don't see any reason why a countably locally finite basis is necessarily countable.
     
  5. Aug 2, 2007 #4

    mathwonk

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    well i don't know the definitions of your sets Bn, or whether they are countable. there are many other proofs available on the web.
     
  6. Aug 2, 2007 #5

    mathwonk

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  7. Aug 2, 2007 #6

    mathwonk

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    he seems to be summing over an uncountable index set, but the family of functions is mostly zero at each point, so yiou get a gloibally defined function.
     
  8. Aug 2, 2007 #7
    I also found an alternate proof at:

    http://at.yorku.ca/p/a/c/a/04.pdf

    So I'm trying to work thru this as well.

    Its interesting that what I can see of your link, as well as the proof from yorku do not attempt to follow Munkres' approach. Of course, I can't conclude from that that Munkres' proof is broken.

    For the Bn:

    X has a countably locally finite basis B. B = the countable union of the Bn's for each n in Z.

    thanks for your help.
     
    Last edited: Aug 2, 2007
  9. Aug 2, 2007 #8

    mathwonk

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    so the Bn's are not countable, but since they are locally finite, the sum over n of the values of some associated functions at each point is a countable sum.

    this may mean munkres may be mapping into some metrizable subspace of his big product. i.e. maybe most of the entries in his vectors are zero.

    but the simpelst thing is if his set J is countable.
     
  10. Aug 2, 2007 #9

    mathwonk

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    maybe you have not got the defn of J just right. waht you wrote was little imprecise at least, too many "B's".
     
  11. Aug 2, 2007 #10

    mathwonk

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    ok what you want is on opage 251 of munkres.

    he is not using the proiduct topology on the space [0,1]^J at all, but the uniform metric.

    i.e. by definition [0,1]^J is a function space, the space of all functions J-->[0,1], and since these functions are all bounded between 0 and 1, there is a natural metric on them called the uniform metric. in fact it is a normed vector space, where the norm of a function, or size, is the sup of its values.

    then the metric defines the distnace between f and g as the norm of f-g, i.e., the furthest apart they ever get.

    this is always a COMPLETE METRIC SPACE, I believe, as long as the target is compact interval. have you read the earlier parts of the book, on metric spaces, and norms and so on?

    anyway he spends the whole page 251 showing the map he gave is an embeding in the sense of this metric.

    so thats your mystery.

    if the indexs set were countable then, you could define a euclidean metric by summing the squares of the coordinates, suitably shrunk down, or some sort of sum metric, but it is interesting that the sup metric works mroe generally.

    note for function spaces one defiens metrics also by integration, not summation, unless the domain is countable.

    but your problem just apparently comes from not grasping earlier material in the book. that happens when you jump into a book at the end, or in the middle.

    when you see the words "uniform metric" you need to search back and find what they mean.

    but i also forgot that [0,1]^J is a uniformly bounded function space and has a uniform norm. i have notes on this from my freshman honors calc course though i may try to post.

    good luck.
     
    Last edited: Aug 2, 2007
  12. Aug 2, 2007 #11
    I had read up on the uniform metric, but it seemed that his proof R^J in the product metric was unmetrizable (p. 133) would imply the [0,1]^J in the uniform metric was similarly unmetrizable. Now I need to go back to that proof and see why it fails for the uniform metric, which is a finer topology than the product metric.

    It is also very interesting that we need to invoke complete metric spaces, and a result that Munkres doesn't prove until the chapter after nagata-smirnov to make sense of this proof.

    Thanks a lot. That clears it up. This is tricky stuff to self-teach.
     
  13. Aug 7, 2007 #12

    WWGD

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    Well, (0,1)^J is not, tho I am not sure for [0,1]^J .


    Maybe you can check wether the space is 1st-countable, or if it satisfies
    the sequence lemma.

    This is the way you show R^R (aka, (0,1)^J ) is
    not metrizable, i.e, you show that the sequence lemma (if x is in the closure of a set A, then there is a sequence {a_n} of points in A with a_n->a ), which holds for all metrizable spaces (If x is a limit point of A , then B(x,1/n)
    intersects A for all n. Define a_n to be the intersection of B(x,1/n) with
    A. You can show a_n converges .) is not satisfied in it, by producing a limit point p of a set A and showing no sequence in A can converge to it.


    It also seems difficult to have 1st-countability (holds in all metric spaces;
    take B(x,1/n) , n=1,2.., as a local basis) when you have an uncountable
    product of open sets, tho there may be a way around this. (altho
    1st countability does not guarantee metrizability, but not having it shows
    non-metrizability.)


    Hope that helped
     
  14. Aug 7, 2007 #13

    WWGD

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    My previous reply

    Sorry, I did not see other replies, only the original question.
     
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