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Cardinality of infinite sequences of real numbers

  1. Feb 26, 2013 #1
    I have to prove that the cardinality of the set of infinite sequences of real numbers is equal to the cardinality of the set of real numbers. So:
    [tex]A := |\mathbb{R}^\mathbb{N}|=|\mathbb{R}| =: B[/tex]

    My plan was to define 2 injective maps, 1 from A to B, and 1 from B to A.
    B <= A is trivial, just map a real number x on the sequence (xxxxxxxxx...). But I can't find a injective map from A to B. Can someone help?
     
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  3. Feb 26, 2013 #2

    mfb

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    You could reduce it to an easier problem first, for example. Something like [tex]|[0,1]^\mathbb{N}|=|[0,1]|[/tex]
     
  4. Feb 26, 2013 #3
    I would start by thinking about why the cardinality of ##[0,1)^2## is equal to the cardinality ##[0,1)##. To do this you think realize that any element of ##[0,1)^2## can be written as ##(x,y)## where ##x## and ##y## have infinite decimal expansions ##x = a_1 a_2 a_3 ...## and ##y = b_1 b_2 b_3 ... ##, then you can combine these into a unique real number ##z = a_1 b_1 a_2 b_2 a_3 b_3 ...## .

    From here, you can generalize this proof to show that ##|[0,1)^\mathbb{N}| = |[0,1)|## by recalling the proof that the rational and natural numbers have the same cardinality. At this point you should be almost home.

    Good Luck!
     
    Last edited: Feb 26, 2013
  5. Feb 26, 2013 #4

    jbunniii

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    You need to be careful here because a real number can have more than one decimal expansion (for example, 0.5000... = 0.4999...) so the mapping as written is not well defined. I believe this problem can be avoided by always choosing the expansion that does not terminate. This of course necessitates a bit of work to show that every real number has exactly one nonterminating decimal expansion.
     
  6. Feb 26, 2013 #5
    Great point, jbunniii, we should definitely make sure to deal with multiple expansions.

    (By the way, I'm sorry that I have edited and updated my post multiple times. I'm still trying to figure out how to use the Tex features properly.)
     
  7. Feb 26, 2013 #6
    You mean countably infinite, right?
     
  8. Feb 26, 2013 #7

    jbunniii

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    The real numbers are not countable.
     
  9. Feb 27, 2013 #8

    mfb

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    Sure, but the number of elements in the sequences is.

    "the cardinality of the set of ([countable] infinite) sequences of real numbers"
    I added "()" to clarify the structure.
     
  10. Feb 27, 2013 #9

    jbunniii

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    I hadn't thought to parse it that way. It makes more sense, but doesn't "infinite sequence" always mean "countably infinite sequence"?
     
  11. Feb 27, 2013 #10

    WWGD

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    How about this:



    If you accept that the Reals are uncountable and the rationals are countable, and that a
    number is rational iff it has an eventually-periodic exoansion:

    First show that the set , say S , of sequences in ℝN that are eventually-periodic are countable, and then set up a bijection between
    N\S and the irrationals, sending a sequence to its "natural" decimal expansion.
     
  12. Feb 28, 2013 #11

    pwsnafu

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    By definition, a sequence is a function whose domain is a countable totally ordered set.
     
  13. Feb 28, 2013 #12
    Mathematicians do occasionally discuss "transfinite sequences" or sequences with an arbitrary index that may the uncountable. If the sequence is of real numbers and an uncountable number of the terms are nonzero, then the sum of the sequence necessarily diverges.
     
  14. Feb 28, 2013 #13

    pwsnafu

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    The proper concept for that is to use the theory of nets.

    Edit: just because "transfinite sequence" has the word sequence in it, does not mean it is in fact a "sequence" in the technical sense.
     
    Last edited: Feb 28, 2013
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