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Sme question about irrational numbers

  1. Apr 10, 2005 #1
    Some question about irrational numbers

    Our teacher showed us Cantor's second diagonal proof.

    He said that by this proof we can show that there are more irrational numbers
    than rational numbers.

    He also said that the cardinality of natural numbers or rational numbers has a magnitude called aleph_0, where the cardinality of irrational numbers has a magnitude of 2^aleph_0.

    He said that any set that its cardinal = 2^aleph_0 has the power of the continuum, so we can conclude that the set of all irrational numbers has the power of the continuum.

    I thought about it and I have this question:


    If we have two sets of irrational numbers (let us call them S1 and S2) such that:

    S1={x: x < 1}

    S2={x: x > 1}

    then in this case the boundary point between S1 and S2 is the natural number 1.

    In this case S1 and S1 are disjoint sets, because the boundary point is a natural number.

    But then we can conclude that the irrational numbers do not have the power of the continuum, even if the magnitude of their cardinal is 2^aleph_0.

    I think that from what I have shown, the irrational numbers are uncountable (by Cantor’s
    Second diagonal method) but because S1 and S2 are disjoint sets, the set of all irrational numbers does not have the power of the continuum.

    So, my question is:

    What is the cardinality of the set of all irrational numbers, and how it is related to the continuum hypothesis?
     
    Last edited: Apr 10, 2005
  2. jcsd
  3. Apr 10, 2005 #2
    I do not see how this statement follows from your previous statements.
     
  4. Apr 10, 2005 #3

    Hurkyl

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    "Power" means cardinality. "The continuum" means the reals. "Power of the continuum" means cardinality of the reals.

    Since the reals have cardinality [itex]2^{\aleph_0}[/itex], anything with that cardinality has the power of the continuum.


    The cardinality of the irrationals follows from this fact about cardinal arithmetic:
    if α and β are infinite cardinals, then α + β = max {α, β}
     
  5. Apr 10, 2005 #4
    There is no common member to S1 and S2 because the boundrey point is not an irrational number but a natural number.

    In this case we can conclude that the set of all irrational numbers does not have the power of the continuum.
    Thank you for your better version.
    Please answer to my question:

    Is the cardinality of all irrational numbers is 2^aleph_0?
     
  6. Apr 10, 2005 #5

    matt grime

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    erm, you are very mistaken. in fact there is ablsoutely no connection between the premiss and conclusion with those two sets S1 and S2.

    just because I can write a set S, as the union of two disjoint sets, A and B, does not mean that A and B must have different cardinality from S. (infinite) cardinals do not behave like that.

    Do you know the criteria for sets having the same cardinality? One of the equivalent ones is that there is a bijection between them. So, N, the natruals is the disjoint union of the odd and even elements yet all have the same cardinality *by definition*. Prove this. And accept it is the definition.
     
  7. Apr 10, 2005 #6

    jcsd

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    Yes the set of all irrationals has the cardinality of the continuum.

    I don't see your reasoning behind your claim that the irrationals don't have the cardinality of the continuum, which is wrong anyway.
     
  8. Apr 10, 2005 #7
    In addition to the above replies, not that the set of all even natural numbers and the set of all odd natural numbers are disjoint but have the same cardinality because there exists a bijection between them.
    If you have a set of 3 distinct apples and another set of 3 distinct oranges, you don't claim that they have a different amount of elements simply because the nature of the elements are different (Ie., by claiming the set of oranges doesn't include any apples).
     
    Last edited: Apr 10, 2005
  9. Apr 10, 2005 #8
    I do not say that S1 and S2 do not have the same cardinality.

    I say that the irrational numbers are uncountable but they do not have the power of the continuum.


    Our teacher gave this example to show that rational numbers do not have the power of the continuum.

    He wrote:

    S1={x: x2 < 2}

    S2={x: x2 > 2}

    where S1 and S2 are all of R members.

    In this case the boundary point is the square root of 2, where in the left side of it we have S1 and in the right side of it we have S2, where S1 and S2 are disjoint sets of rational numbers.

    The same thing holds for:

    S1={x: x < 1}

    S2={x: x > 1}

    where S1 and S2 are all of irrational members.

    In this case the boundary point is the natural number 1, where in the left side of it we have S1 and in the right side of it we have S2, where S1 and S2 are disjoint sets of irrational numbers.

    It means that the collection of all irrational numbers are uncountable (by Cantor's second diagonal proof) but they do not have the power of the continuum, because the boundary point is the natural number 1, that cannot give us the ability to define an irrational number as a boundary point of both S1 and S2.

    Now, we can conclude that the set of all irrational numbers are uncountable but they do not have the power of the continuum.
     
    Last edited: Apr 10, 2005
  10. Apr 10, 2005 #9

    Hurkyl

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    Your teacher's example has nothing to do with cardinality. It's a lesson about connectedness -- the rationals (and the irrationals) are not a connected set. Or, loosely speaking, they do not form a continuum.

    Something doesn't have to be a continuum to have the same cardinality as R.
     
  11. Apr 10, 2005 #10
    But our teacher told us that the cardinality of all irrational numbers is 2^aleph_0 and 2^aleph_0 = the power of the continuum.

    Now you say that the set of all irrational numbers do not form a continuum.

    So please let me ask you again: what is the cardinality of the set of all irrational numbers?
     
  12. Apr 10, 2005 #11

    Hurkyl

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    I reiterate:

     
  13. Apr 10, 2005 #12
    I do not see how someone can come to this conclusion.

    If there is an unclosed hole between S1 and S2 they cannot have the cardinality 2^aleph0 = |R| = c
     
    Last edited: Apr 10, 2005
  14. Apr 10, 2005 #13

    Hurkyl

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    Cardinality merely talks about a sets. It has absolutely nothing to do with any topological or geometric ideas.
     
  15. Apr 10, 2005 #14
    Ho!??, how can you say that there is no connection beween some collection and its cardinality?


    So I think that the solution is to say that if 2^aleph_0 is the cardinality of the set of all irrational numbers, then 2^aleph0 < |R| = c
     
  16. Apr 10, 2005 #15

    Hurkyl

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    You can say it all you like, but that doesn't change the fact that [itex]|\mathbb{R}| = 2^{\aleph_0}[/itex].
     
  17. Apr 10, 2005 #16

    jcsd

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    tann your mixing up didfferent properties i.e. not all sunbstes of the rational numbers or irartional numbers bounded form above have a least upper bound (which is in the set they are subsets of) whears every subset of the real numbers which is bounded form above has a leats upper bound which is a rela number. This proeprty of the rational and irrational numbers does not dictate their cardinality.
     
  18. Apr 10, 2005 #17
    Sorry but I came here to share my thoughts with others and learn somthing from professionals.

    I do not see that you give any proof to your "facts", so please be professional and give us a proof that clearly show us that the cardinality of the set of all irrational numbers has the power of the continuum.

    And if you can, then please explain in simple words evry part of this proof.
     
    Last edited: Apr 10, 2005
  19. Apr 10, 2005 #18

    jcsd

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    Perhaps what is confusing you is that we can 'take away' elements of an iifnite set a dn still have a set with the same cardinality. This is just the dfeitnion thoguh of an inifte set i.e. a set which has proper subsets with the same cardinality as itself.
     
  20. Apr 10, 2005 #19

    Hurkyl

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    I already told you the fact that for infinite cardinals, [itex]\alpha + \beta = \max \{\alpha, \beta\}[/itex].

    Now, the cardinality of the reals equals the cardinality of the rationals plus the cardinality of the irrationals. Apply the aforementioned fact. :tongue2:
     
  21. Apr 10, 2005 #20
    I know the properties of the tarsfinite universe, but it says nothing about the cardinality of the set of all irrational numbers and the power of the continuum, and you did not give until now even a single proof that the set of irrational numbers has the power of the continuum.

    All I see here is a collection of symbols, that does not prove anything.

    So please someone give a proof?

    In N we can show that the is a bijection between N and a proper subset of N.

    Please show me how can you prove the same thing between R and the set of all irratioanal numbers?
     
    Last edited: Apr 10, 2005
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