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Ordinals - set of r-v'd functions on any interval in R and cardinality

  1. Sep 28, 2007 #1
    just a cool fact I thought I'd share with anyone who's interested:

    The set of real values functions on any interval in R has cardinality at least 2^c.

    Pf: Consider characteristic functions defined on the interval, (a,b). (Note: a characteristic function is a function that can be defined on ANY domain and has range {0,1})

    Let E be a subset of (a,b), then the characteristic, g(x) function of E over (a,b) i.e.

    0 if x is not in E
    g(x) =
    1 if x is in E

    Now for each subset E of (a,b) there corresponds a unique characteristic function defined on (a,b). Hence the set of all subsets of (a,b) and the set of characteristic functions defined on (a,b) are equivalent.
     
  2. jcsd
  3. Sep 28, 2007 #2

    matt grime

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    that is clear - the set of functions from R to R has cardinality c^c, in fact, since it is, as a set, R^R.


    Yes, that is a fact everyone learns in their first meeting with sets and cardinality - it didn't need to be proved.
     
  4. Sep 28, 2007 #3
    c^c? what does that mean compared to 2^c? (2^alephnull)^(2^alephnull)?

    I've met a lot of people who come out their first analysis course without even knowing what a characteristic function is. Surely these people study sets/cardinality (naively) in these courses? Outside of set theory classes where are cardinal #'s greater than c discussed?
     
  5. Sep 28, 2007 #4
    double-post sorry.
     
  6. Sep 28, 2007 #5

    Hurkyl

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    c = 2^N

    c^c = (2^N)^(c) = 2^(N * c) = 2^c
     
  7. Sep 28, 2007 #6
    advanced set theory... (-:
    but i guess you can meet this in logic and also in infinite combinatorics which is a new field.
    i guess that also in non standard analysis, cause there we already constrcut the line of hyperreal numbers.
    btw from what hurkyl gave you you need ofcourse to prove that for every infinite cardinality: a+a=a and a*a=a for that you need zorn's lemma, from this you can easiliy conclude from cantor bernstein theorem that for a>=b we have a+b=a and a*b=a.
     
  8. Sep 28, 2007 #7

    morphism

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    Here's an exercise for you: What's the cardinality of the set of real-valued continuous functions on R?
     
  9. Sep 28, 2007 #8

    morphism

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    In addition to loop quantum gravity's suggestions, topology (and in particular set theoretic topology) comes to mind. There are ways of using the order on an ordinal to induce a topology. Ordinals turn out to be very useful in giving examples of certain topological constructions and counterexamples to conjectures. They also pop up in curious places (thanks to the well ordering theorem and its many manifestations), e.g. one can prove that R^3 can partitioned into a union of disjoint unit discs using an ordinal-related argument.
     
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