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Math Help - naturals as a subset of the reals

  1. #1
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    naturals as a subset of the reals

    How does one prove that the natural numbers are (isomorphic to) a subset of the real numbers? First of all, I take these as the definitions of these two sets:

    reals:
    Field (assoc, distrib, commun, 0, 1, negatives, multiplicative inverses
    Ordered (consists of three sets N, {0}, and P, where P is closed under +,x)
    Complete (least upper bounds of sets are again in the reals)

    naturals:
    Peano's postulates would suffice.

    Now, I'm already aware of the natural correspondence between {0,0+1,0+1+1,...}
    in the reals and the natural numbers. What my question comes down to is: how does one prove the principle of mathematical induction for {0,1,2,...}, the subset of the reals corresponding to the naturals?

    All I need in answer to this question is "vague help" or "a hint in the right direction." Or, if the answer is unbelievably complicated (which I doubt) then just let me know .
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    Hey, I just thought: could I prove that {0, 1, 2... } is well-ordered (i.e. each nonempty subset has a least element)? I think I could prove induction from that.
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  3. #3
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    What do you mean by "isomorphism"? It seems to me that a bijection f between N and any old set would do, along with the definition that S'(f(n))=f(S(n)), where S' is the successor function on the other set, in this case, the subset {0, 1, 2, ...} of the reals.

    But I haven't really studied much of this low-level (in the sense of fundamental, not easy) stuff, so I might be taking something for granted that I shouldn't be.
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  4. #4
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    An "isomorphism" is a bijection between the sets that also "preserves" the operations- f(ab)= f(a)f(b) and f(a+ b)= f(a)+ f(b).

    Mazarakham, the real numbers, being a field, has additive and multiplicative identities, 0 and 1. The natural numbers has a "first member", 0. Consider the mapping, N->R, f(n)= 0+ n(1) where "n(1)" means "add 1 to itself n times". That is precisely the mapping you suggest. You should be able to show that is a bijection and that f(sup(n))= f(n)+ 1 from which f(m+n)= f(m)+ f(n) follows. That f(nm)= f(nm) may be a little harder. Exactly how are you defining "m+ n" and "mn" for the natural numbers?
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  5. #5
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    Mathematical Induction is equivalent to the Well-ordering property, so of course proving one implies the other.
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  6. #6
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    What my question comes down to is: how does one prove the principle of mathematical induction for {0,1,2,...}, the subset of the reals corresponding to the naturals?
    Good question.

    Suppose that we can prove the following: If x and y are images of natural numbers in \mathbb{R}, then x<y iff x+1\le y. Then here is how we could prove the well-ordering property.

    Suppose A is a subset of (the images of) natural numbers. Then it is bounded from below by 0, so A has an infimum. Therefore, there exists an a\in A such that \inf A+1>a, i.e., a-1<\inf A. So, a-1<b for every b\in A, which implies that a\le b for each b\in A.
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  7. #7
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    Thanks, emakarov. One thing though. Let's refer to the subset {0,1,2,3...} of the reals as \mathbf{R}_{\mathbf{N}}. Your "lemma" that x<y\ iff\ x+1\leq y is logically equivalent to the statement,
    "for all numbers a in \mathbf{R}_{\mathbf{N}}, there are no numbers in \mathbf{R}_{\mathbf{N}} b such that a-1\ <\ b\ <\ a. Unfortunately, this result from number theory, while easily proven given the well-ordering principle, can't be proven without it. I think we'll need to keep working.

    Now, I think I should address some other posts. TinyBoss, I agree that "a bijection f between N and any old set" would mean that that "any old set" T would almost solve the problem--except for I still don't see how T is well-ordered--and in fact it won't be for many bijections. But I will give you this: that any monotone increasing sequence of real numbers is isomorphic to the Naturals. But you might as well prove it for {0,1,2,3...} for, say, visualization's sake. I too respect the simplicity of the problem and acknowledge that the answer is likely a simple one.

    HallsOfIvy: I was thinking maybe we might avoid proving the existence of the binary operations of + and x, and do a proof directly from Peano's postulates. But, setting that aside, I actually am not sure I understood your post. You lost me at "f(sup(n))." Nonetheless, suppose we had proven the existence of those binary operations. Maybe I don't understand isomorphism, but I thought that there's more to it than showing that f(ab)=f(a)f(b) & f(a+b)=f(a)+f(b). I thought you also have to show that order is maintained. Since the heart of my question is "how does one prove that {0,1,2,3...} satisfies the principle of induction," might I ask that you clarify how f(ab)=f(a)f(b) & f(a+b)=f(a)+f(b) imply the principle of induction on the image f[\mathbf{N}]=\mathbf{R}_{\mathbf{N}}. I'm a very stubborn question-asker, but I try to be meticulously clear. I respect your time
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