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Cantor Set/Space and Retracts.

  1. Sep 15, 2011 #1
    Hi, All:
    I was thinking of the result that every compact metric space is the continuous image
    of the Cantor set/space C. This result is built on some results like the fact that 2nd
    countable metric spaces can be embedded in I^n (I is --I am?-- the unit interval),
    the fact that there is a continuous map between C and I, and, from what I read
    recently , the fact that every closed subset of C is a retract of C.

    How do we know that every closed subset of C is a retract of C?
     
  2. jcsd
  3. Sep 15, 2011 #2

    micromass

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    Let A be a closed subset of the Cantor set C. We know that the Cantor set is self-similar. That is, we can write

    [tex]C=C_1^1\cup C_1^2[/tex]

    where [itex]C_1^i[/itex] is homeomorphic to the Cantor set. We can also write

    [tex]C=C_2^1\cup C_2^2\cup C_2^3\cup C_2^4[/tex]

    where again [itex]C_2^i[/itex] is homeomorphic to the Cantor set. We can go on like that.

    Let

    [tex]A_n=\bigcup_{C_n^i\cap A\neq \emptyset}{C_n^i}[/tex]

    It is easy to find a retraction [itex]r_n[/itex] between [itex]A_n[/itex] and the Cantor set. Let r be the uniform limit of these retractions, then r is a retraction between A and C.
     
  4. Sep 15, 2011 #3
    What's the uniform limit? Is it some sort of inverse limit?
     
  5. Sep 15, 2011 #4

    micromass

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    It is just the pointswise limit. But the convergence turns out the be uniform. This is needed in order for the limit to be continuous.
     
  6. Sep 15, 2011 #5
    Yes, I understand the need for convergence to be uniform, e.g., {x^n} in [0,1].

    This sort of reminds me of the relation between the compactness theorem in logic

    (if every finite subset of a sentence is satisfiable, then the sentence itself is satisfiable)

    and in topology. We see compactness in terms of the finite-intersection property.

    Then the infinite product ( over I:=[0,1]) of {0,1} is, as usual, the set of possible

    functions from I into {0,1} , seen as valuations of a wff , i.e., as assignment of

    0 or 1 to each free variable in a formula. Then, by compactness/FIP, every finite

    collection has non-empty intersection, meaning that there is a valuation {1,0}

    that satisfies every finite subcollection , so that, by compactness, the sentence

    is satisfiable. Can we use this approach somehow, seeing a subset of the Cantor

    set as a valuation?
     
  7. Sep 15, 2011 #6
    BTW, I wonder what you think about this, Micromass: the infinite product of metric spaces is metrizable (product topology, of course), if the product of (X-i,d_
    i) is countable, by using,
    e.g.,

    d(x,y):=sum_i=(1,..,oo) d_i(x_i,y_i)/2^i

    ( I think we can even get a bounded metric if we choose a bounded metric for each space).

    BUT: this argument does not work for uncountable products, since then the sum

    will not converge unless all-but-uncountably-many distances are 0. Still,

    for the case of the Cantor space, we do have a countably-infinite product

    of metrizable spaces {0,1}, which is also metrizable. So, the question is: when is

    the uncountably-infinite product of metric spaces metrizable?
     
  8. Sep 15, 2011 #7

    micromass

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    Never (assuming that the spaces aren't trivial).

    This is the general theorem:

    Let [itex](X_i)_{i\in I}[/itex] be a family of spaces which are not indiscrete. The following are equivalent:
    • [itex]\prod_{i\in I} X_i[/itex] is metrizable.
    • Each [itex]X_i[/itex] is metrizable and I is at most countable.
     
  9. Sep 15, 2011 #8
    But isn't the Cantor set as an uncountably-infinite product of {0,1} metrizable?

    Also, what do you mean by indiscrete? I understand indiscrete as having only the empty

    set and the whole space being open, and this space, with more than two points, is not

    metrizable (not Hausdorff). Did you mean discrete?
     
  10. Sep 15, 2011 #9

    micromass

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    No, the Cantor set is [itex]\{0,1\}^\mathbb{N}[/itex]. The uncountable product is not metrizable.
     
  11. Sep 16, 2011 #10
    Micromass: Sorry if I'm beating a dead horse; I am trying to review my topology
    ( 2 years since I took the class, and I seem to have forgotten some) : I guess since
    the uncountable product of 1st-countable is not 1st countable, that does it.

    Also, as just a comment, the trick with the bounded metric:

    d(x,y):=Sum d_i(x_i,y_i)/2^i does not work for uncountable sums, since it

    diverges necessarily when the support is uncountable. I know this last is not a proof,

    it is just a comment.
     
  12. Sep 29, 2011 #11
    Yes, I realized what's wrong with what I was saying; by product I was using the box-product and not the standard product. Still, with the box product, a product of uncountably-many metrizable spaces can be /is metrizable, e.g., discrete spaces.
     
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