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Cardinality of the Preimage f^{-1}(y) of f:X->Y continuous?

  1. Mar 13, 2012 #1

    Bacle2

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    Cardinality of the Preimage f^{-1}(y) of f:X-->Y continuous?

    Hi, All:

    Let X,Y be topological spaces and f:X-->Y non-constant continuous function.

    I'm curious as to whether it is possible for the fiber {f^{-1}(y)} of some y in Y

    to be uncountable, given that the fiber is discrete (this condition is intended to

    discount cases like f:ℝ-->ℝ, that are 0 in some interval, using, e.g., combinations

    of e-1/x2) . There are examples of fibers being countably-

    infinite, like in f:ℝ→ ℝ, f(x)=sinx, cosx, etc. , or the topologists sine curve on [0,1].

    I suspect it may be possible, if X is metric to use f:X-->X with f(x)=d(x,S) , i.e.,

    S is a subset of X , and d(x,S):=inf{d(x,s): s in S}. I thought S=Cantor set may work, but

    the points of C are not isolated in ℝ (in [0,1], actually). Maybe if one can define an uncountable subset of

    a metric space , all of whose points are isolated, we would be done.

    Any IDeas?
     
    Last edited: Mar 13, 2012
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  3. Mar 13, 2012 #2

    morphism

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    Re: Cardinality of the Preimage f^{-1}(y) of f:X-->Y continuous?

    Let X be any uncountable discrete space and map it any way you want onto a two-point space. This will give a continuous map whose fibre above at least one of the points in the range will necessarily be uncountable...
     
  4. Mar 13, 2012 #3

    micromass

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    Re: Cardinality of the Preimage f^{-1}(y) of f:X-->Y continuous?

    Think of the discrete metric on an uncountable number of points.

    It may be interesting to notice that if the metric space is seperable, then every set of isolated points is countable. Indeed, let S be a set of isolated points. Since the metric space is seperable, it is second countable. So S is second countable and thus seperable (not that a subspace of a seperable space is not always seperable, but I've just shown that it is in metric spaces). But every point in S is isolated, so every singleton in S is open. It is clear that S must be countable.
     
  5. Mar 13, 2012 #4

    Bacle2

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    Re: Cardinality of the Preimage f^{-1}(y) of f:X-->Y continuous?

    Thanks; sorry, I forgot to include the condition of neither of the topologies being neither discrete nor indiscrete.
     
  6. Mar 13, 2012 #5

    morphism

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    Re: Cardinality of the Preimage f^{-1}(y) of f:X-->Y continuous?

    That's still not a problem. Take my X above and let ##X' = X \sqcup \mathbb R## (disjoint union; here ##\mathbb R## is given its usual topology) and map it onto ##\{\infty\} \sqcup \mathbb R## by sending X to ##\infty## and ##\mathbb R## to ##\mathbb R## via the identity.
     
  7. Mar 14, 2012 #6

    Bacle2

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    Re: Cardinality of the Preimage f^{-1}(y) of f:X-->Y continuous?

    Just wanted to tell others that I am planning to delete my posts; I realized that I phrased my OP poorly, and that this was not the question I had in mind; I didn't want to leave you hanging.
     
  8. Mar 14, 2012 #7

    Bacle2

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    Re: Cardinality of the Preimage f^{-1}(y) of f:X-->Y continuous?

    Well, the 'edit' function seems disabled, and I can't seem to be able to delete my original post. Any idea on what I could do to delete it? It was just not the question I wanted to ask.
     
  9. Mar 14, 2012 #8

    micromass

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    Re: Cardinality of the Preimage f^{-1}(y) of f:X-->Y continuous?

    Can't you just ask the new question?? Or make a new thread??
     
  10. Mar 14, 2012 #9

    Bacle2

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    Re: Cardinality of the Preimage f^{-1}(y) of f:X-->Y continuous?

    Yeah, no problem. I just thought the question, as (poorly) posed by me, is trite and not very helpful, and maybe it is better to ask more interesting questions.
     
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