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Show that the set is countable or finite.

  1. Apr 17, 2012 #1
    Hi, can someone please help me with this problem.

    Let A be an open subset of the interval [0; 1].
    1. Show that the set W = {C(x) : x is in A} is countable or finite.

    This is what I have...

    Suppose W is an infinite subset of N. Then we have f : W-> N, which
    is one-to-one. By the fact that any infinite set contains a countable subset, then W has a countable subset E, since w is infinite. So we have a one-to-one map
    g : N ~E -> W. By Cantor-Berstein-Schroeder's Th, W~N.

    thanks
     
    Last edited: Apr 17, 2012
  2. jcsd
  3. Apr 17, 2012 #2
    I'm sorry, but what is [itex]C \left( x \right)[/itex]?
     
  4. Apr 17, 2012 #3
    Sorry, this is what I have before problem:

    Let A be an open subset of the interval [0; 1]. Our goal is to show that
    A =the union In where for all n in N, In is a possibly empty open interval and for
    all n and k in N, if n is not equal to k then In intersect Ik = empty set We begin by defi ning a relation ~ on A by the
    formula
    for all x; y in A; x ~ y if and only if every number between x and y is in A:
    For each z in A, we let C(z) = {t in A : t ~ z}


    #1 Show that the set W = {C(x) : x is in A} is countable or finite. (Hint: For this problem it's better to think of W as W = {U : there exist x in A such that U =
    C(x)}. The reason is for every element U of W will be more than one number
    x for which U = C(x).)
     
  5. Apr 17, 2012 #4
    Alright, well, I think it might be easier to define a mapping between these [itex]C \left( x \right)[/itex] and a subset of the rational numbers. Can you think of a way to assign a unique [itex]r \in \mathbb{Q}[/itex] to each [itex]U \in W[/itex]?
     
    Last edited: Apr 17, 2012
  6. Apr 18, 2012 #5
    Well, the rational numbers are countable, so any open set in R is the countable union of components. But I am not sure how to write a proof...
     
  7. Apr 18, 2012 #6
    Would this proof work?

    let r=1/n for n=1,2,3... and {In, n in N where In is the set corresponding to r=1/n, that is In is a maximal set having the property that is greater then or equal to 1/n for any a,b in In

    Let W =Un In since W is the union of countably many countable sets, W is countable.

    thanks
     
  8. Apr 19, 2012 #7
    Hmm, I'm afraid that doesn't quite work. Say [itex]A = \left( \frac{1}{2}, \frac{3}{4} \right) \cup \left( \frac{3}{4}, 1 \right)[/itex], then what would [itex]I_{2}[/itex] be?

    I think it would be better to use the fact that the rational numbers are dense within the real numbers. You should be able to show that each [itex]C \left( x \right)[/itex] is non-empty and open and thus contains at least one rational number, and since they are disjoint, these rational numbers must be unique.
     
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