# Thread: naturals as a subset of the reals

1. ## 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 .

2. 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.

3. 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.

4. 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?

5. Mathematical Induction is equivalent to the Well-ordering property, so of course proving one implies the other.

6. 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 $\displaystyle \mathbb{R}$, then $\displaystyle x<y$ iff $\displaystyle x+1\le y$. 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 $\displaystyle a\in A$ such that $\displaystyle \inf A+1>a$, i.e., $\displaystyle a-1<\inf A$. So, $\displaystyle a-1<b$ for every $\displaystyle b\in A$, which implies that $\displaystyle a\le b$ for each $\displaystyle b\in A$.

7. Thanks, emakarov. One thing though. Let's refer to the subset {0,1,2,3...} of the reals as $\displaystyle \mathbf{R}_{\mathbf{N}}$. Your "lemma" that $\displaystyle x<y\ iff\ x+1\leq y$ is logically equivalent to the statement,
"for all numbers a in $\displaystyle \mathbf{R}_{\mathbf{N}}$, there are no numbers in $\displaystyle \mathbf{R}_{\mathbf{N}}$ b such that $\displaystyle a-1\ <\ b\ <\ a$. 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 $\displaystyle f[\mathbf{N}]=\mathbf{R}_{\mathbf{N}}$. I'm a very stubborn question-asker, but I try to be meticulously clear. I respect your time