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Cantor's definition of reals

  1. Sep 16, 2007 #1
    Hi I've got a tough analysis proof. If I can do this on my own I might as well be Cantor himself.
    The first step is :
    Let C and N denote the collection of every cauchy sequence and null sequence (consisting of rationals) , prove that N is a subset of C
    Second step is :
    Prove N induces a equivalence relations on C.
     
    Last edited: Sep 16, 2007
  2. jcsd
  3. Sep 16, 2007 #2

    morphism

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    What thoughts have you had so far?
     
  4. Sep 16, 2007 #3

    Hurkyl

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    Do you understand the definitions of all the technical terms? Can you write them down?

    Anyways, you certainly have at least definition wrong. C and N, as you've defined them, are not sets1. So it doesn't make sense to ask if one is a subset of the other.


    1: At least, not in a useful sense.
     
  5. Sep 16, 2007 #4
    Sorry I meant the C and N are the collection of all cauchy sequences
    I think the step one is trivially true, since null sequence is defined as xn such that there exists N: for any n>N ,|xn|<epsilon/2, therefore for any m,n>N, |xm-xn|<|xm-0|+|0-xn|=|xm|+|xn|=epsilon by triangular inequality
    Therefore a null sequence is neccesarilly cauchy.
    I'm not sure where the significance of 'colection of sequences consisting of rationals' come into play
     
  6. Sep 16, 2007 #5

    HallsofIvy

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    What do you mean by "the cauchy sequence"? I'm pretty sure there's more than one! Do you mean the set of all cauchy sequences of rational numbers? I'm also not sure what you mean by "the null sequence (consisting of rationals)". Do you mean the set of all sequences of rational numbers converging to 0?

    Do you know the theorem that says every convergent sequence is a Cauchy sequence?

    The phrase "of rational numbers" is necessary because you haven't yet defined real numbers!
     
    Last edited: Sep 16, 2007
  7. Sep 16, 2007 #6
    thanks and you are right, I misquoted, it's the collection of all such sequences, but what property of rationals are necessary? What doesn't work if I define them using inegers.
     
  8. Sep 16, 2007 #7

    morphism

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    The exercise is attempting to construct the real numbers starting from the rational numbers. One way to do this is to take limits of Cauchy sequences of rational numbers. The problem is that not all Cauchy sequences of rational numbers converge to a rational number, so we try to 'fill in the gaps' using what we would call 'real numbers'.

    Now that you've shown that null sequences are Cauchy, you know that N sits in C. How do we usually use a subset of a set to induce an equivalence relation?

    Here's a hint. Consider the set of integers Z. The set of even integers 2Z is a subset of Z. We can let 2Z induce the following equivalence relation on Z: n~m iff n-m is even, i.e. is in 2Z. This is the familiar "mod 2" equivalence relation. Under it we do not distinguish between any two even numbers or any two odd numbers.
     
    Last edited: Sep 16, 2007
  9. Sep 16, 2007 #8
    I'm not sure about the definition of induce, why is it called induce?
    we first define n~m iff n-m is in N, then prove ~ is reflexive, transitive and symmetric, and that's it i guess?

    edited:
    Done!I've proven ~ is reflexive, transitive and symmetric
     
    Last edited: Sep 16, 2007
  10. Sep 16, 2007 #9
    I'm completely clueless on this one
    Let R := C/N, i.e., the collection of cosets of C under the equivalence
    relation induced by N. Define addition and multiplication
    on R so that R is a field with respect to these operations.
     
  11. Sep 16, 2007 #10

    morphism

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    There's no special meaning behind "induce" -- it's used in the normal English sense.

    About your definition of ~: n and m are supposed to be sequences, so what's your definition of n-m? And did you manage to prove that ~ satisfies those 3 properties?

    Do you understand what we're trying to do? Intuitively, what's going on is that we have some sequences of rational numbers that are trying to converge (because they're Cauchy) but are not finding a rational number to converge to. What we're trying to do is to artificially add a point that can be the limit of such a sequence. How would we go about doing this? Well, if two sequences are 'trying' to converge to the same 'real number', then their tails will be very close to each other, i.e. if x_n and y_n are two such sequences, then for some large enough n, |x_n - y_n| is very small (it goes to zero as n goes to infinity).
     
  12. Sep 16, 2007 #11

    morphism

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    Hopefully you can see that an element of C/N is an equivalence class of Cauchy sequences. And if you read my previous post, you'll see that each equivalence class consists of those Cauchy sequences that are trying to converge to the same limit. Essentially, we're using the entire equivalence class to be the real number the sequences are trying to converge to.

    Maybe an example will help shed some light on what's going on. Consider the three Cauchy sequences of rationals: (x_n) = (1/n), (y_n) = (1/2^n) and (z_n) = (-1/n). We know that each of them converges to 0. In particular, the tails of all three sequences are close to each other. So these three sequences lie in the same equivalence class in C/N. This equivalence class is what we would call the real number 0, and denote by [(x_n)] or [(y_n)] or [(z_n)] or [(0, 0, 0, ...)] or [(any sequence that converges to 0)].

    On the other hand, consider the sequence (w_n) defined recursively by w_1 = 1 and w_{n+1} = 1 + 1/(1 + w_n). This sequence of rational numbers is trying to converge to [itex]\sqrt2[/itex], but cannot do this because [itex]\sqrt2[/itex] is not a rational number, so it does not exist in our universe of rational numbers. In C/N, we can think of [(w_n)] as all Cauchy sequences of rationals that are trying to converge to [itex]\sqrt2[/itex]. Keeping this in mind, we can then define [itex]\sqrt2[/itex] to be this equivalence class of sequences.

    I hope everything's clearer now.
     
    Last edited: Sep 16, 2007
  13. Sep 16, 2007 #12

    thanks a lot for your help. i never thought of it that way. Stupid me, I spent 50 hours last week reading baby rudin and didn't see the whole picture.
    I've never learnd Equivalence class and couldn't find it in baby rudin either.


    I defined n-m as nx-mx where x is the index of sequence. the difference of two sequence is the difference between every element of them. I prove the 3 properties using properties of absolute value.
     
    Last edited: Sep 16, 2007
  14. Sep 16, 2007 #13
    x~y iff x-y belongs to null set N, by definition of equivalence
    therefore C-C/N=N which is trivially a subset of N
    therefore C/N~C

    Brilliant, now I know C~C/N, but how do I define a addition and multiplication operation on C/N so as to make it a field isomorphic to R?
    edit: I got it! I can define addition of two cauchy seuqnces in C/N as the addition of the the limits of these two cauchy sequences.
    similarly for multiplication.
     
    Last edited: Sep 16, 2007
  15. Sep 17, 2007 #14

    morphism

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    I had a detailed reply, but my browser lost it. Damn keyboard shortcuts.

    Anyway, what do you mean by C-C/N=N and C/N~C? C/N is a collection of equivalence classes of sequences, i.e. C/N = {[(x_n)] : (x_n) is in C}, where [(x_n)] = {(y_n) in C : (y_n) ~ (x_n)}. And because equivalence classes can have many representatives (e.g. in Z/2Z, [0]=[-2]=[2]=[2342435423554] etc.), you will have to check that the addition and multiplication operations you put on C/N are well-defined, i.e. do not depend on which representative one chooses. (For instance, if we define + on Z/2Z by [n]+[m]=[n+m], we must prove that if [n]=[n'] and [m]=[m'], then [n]+[m]=[n']+[m'].)

    Also, you said that you defined the addition of two cauchy sequences in C/N (actually, you should say "two elements in C/N") as the addition of their limits, but we do not know if these limits exist. So unless you don't mean what I think you mean, this would not be a very good definition.

    Finally, after you define + and *, make sure you check that R together with these two operations is indeed a field, i.e. check that it satisfies the axioms of a field (e.g. addition is commutative and multiplication is associative, etc.).
     
  16. Sep 17, 2007 #15
    I thought C/N is the set C without elements in N, that's my interpretation of the operator /
    Now I remember Z/2Z is all even integers.
    I see, So C/N means the collection of all equivalence classes in C(collection of all cauchy sequences): an equivalence class is defined as if A-B belongs to N, then A and B are in the same equivalence class.
    This is interesting because Z/2Z gives you one equivalence class, But C/N gives you infinitely many equivalence classes, since you have equivalence class who has one representative cauchy sequence which converges to 0, and those converging to 3 and those converging to sqrt2.

    We know that Cauchy sequences converges in R, can we somehow uses that result here?
     
    Last edited: Sep 17, 2007
  17. Sep 17, 2007 #16

    morphism

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    No, we can't use the fact that Cauchy sequences converge in R, because we don't know what R is! This is the point of the exercise: constructing R. The way it's being done will trivially give us the fact that Cauchy sequences converge in R.

    (By the way, C\N would be the set C minus N.)
     
  18. Sep 18, 2007 #17

    mathwonk

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    another satisifed baby rudin user. when are people going to stop banging their heads against that book? there are so many better ones out there. as have often been listed hereabouts. this particular construction is even in van der waerdens algebra book.
     
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