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It was established in Baby Rudin that

If x is a positive real and y is a real, then there exists a natural number n such that nx>y.

It is then widely referenced in subsequent theorems without further explanation. I would like to verify whether this is correct or not.

In proving the existence of decimals in positive real, it was said that

Let x be a positive real and n be the greatest integer such that n is less than or equal to x.

The statement is immediately followed by "The existence of such n depends on the Archimedean property", However, this is less than or equal to instead of the greater than, which means it is not a direct application. So I wonder, is the following proof right, or can it be simplified?

  1. Denote $A=\{n\in\mathbb{Z^+}:n\geq x\}$
  2. By the Archimedean property, there exists $n_0\in\mathbb{Z^+}$ such that $n_0>x$
  3. The set is not empty
  4. Now consider the set $B=\{n-x:n\in\mathbb{Z^+}\land n\leq n_0\land n-x\geq 0\}$
  5. We know $n_0$ is finite so the set has finite members. As a result, the infimum equals minimum
  6. $x\in[\inf B-1+x,\inf B+x]$
  7. $n_0=\inf B-1+x\leq x$

Here I show there is an integer. Am I correct? The problem is not about the proof. The matter is about the degree which was ignored by Rudin. The proof applies the concept of infimum and order set. Archimedean property is just part of it.

So, is there better proof that uses the archimedean property?

DanielWainfleet
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Andes Lam
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    The archimedean property insures the existence of some $n\ge 1$ with $x\in(0,n]$. Consider now the finite set ${1,2,\dots,n}\cup{x}$, ordered by the usual order of the real numbers. Pick the subset of all elements $\le x$ of it, and take the maximum. – dan_fulea Jul 26 '21 at 11:00
  • @dan_fulea Thanks, this makes a lot of sense. I believe the "finite" can hit home earlier had i written in your way in my post. – Andes Lam Jul 26 '21 at 11:02
  • @dan_fulea Excuse the naive question, but isn't proving the existence of decimals numbers in the reals extremely trivial; just take $1$, divide it by $10$ - which you're allowed to do by construction of the reals, and now you have $0.1$... – FShrike Jul 26 '21 at 11:02
  • @AndesLam Maybe you could explain? Sorry for distracting from your question, but I do not see why all the set theory is necessary – FShrike Jul 26 '21 at 11:06
  • @IdioticShrike I think you catch the drift of intuition but not rigor. Division, insofar as baby rudin is concerned at this point, which is chapter 1, is merely "defined" as the inverse of a non-zero real. It's not even defined tbh. What I am doing here is mimicking the behavior of division, which is yet defined, in a most rigorous, or primitive, fashion. – Andes Lam Jul 26 '21 at 11:06
  • @AndesLam Ah - but I thought the set of reals was constructed rigorously by saying: "Let all natural numbers be a real number. Now take any two different real numbers. Now find their midpoint, and add this to the set of reals. Repeat ad infinitum" and that midpoint-finding process entails a division by two, right? – FShrike Jul 26 '21 at 11:08
  • @IdioticShrike Short after constructing "the" real numbers, you have something else as real numbers than you may expect from the school, where "real numbers" are introduced by writing them. Try to understand some construction of them, this is the core of analysis. Then try to use this construction, as things are constructed from axioms and thin air, to define "decimals" of some / any number. To understand the point, let us consider an example, assume i give you the number defined by the limit of the "simple" recursive sequence $a_0=1$, $$a_{n+1}=\left(a_n+\frac 7{a_n}\right)\ .$$ Decimals? – dan_fulea Jul 26 '21 at 11:11
  • @IdioticShrike I think here's where we part ways. In the second international edition of baby rudin, which is the copy in my possession, the set of reals is constructed through cuts. The book, till the proof and application of archimedean rn, has only defined the axioms of field and order set. Just curious, which book are you using? Also, the definition of reals you mentioned sound more like nested interval property than construction of reals. Just saying. – Andes Lam Jul 26 '21 at 11:11
  • There was no need of $B$. Consider $\inf A$ and show $\inf A\in A$. If $x=\inf A$ then we can take $n=\inf A$ otherwise take $n=\inf A-1$. – Paramanand Singh Jul 26 '21 at 12:38

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Since an answer is still missing, I will try to fill this up:

Proof:

Let $x \in \mathbf{R}^+$. By the Archimedean Property, there exists $n \in \mathbf{Z}^+$ big enough such that $n > x$. Define the nonempty (why?) set $$ A := \{ m \in \mathbf{Z}: 0 \leq m \leq n \}. $$ Note that $A$ is finite as $n$ is. Now consider the nonempty (why?) set $$ B := \{ m \in \mathbf{Z}: 0 \leq m \leq x \}. $$ Then $B \subseteq A$ since $n > x$, so $B$ is finite and bounded above. Thus we can take $n_0 = \sup(B) \equiv \max(B)$ as desired. We leave why $\sup(B) \equiv \max(B)$ to the exercise below.

Exercise:

For any finite set $A \subseteq F$ on an ordered field $F$ (with the least upper bound principle, if we wish), we have $$ \sup(A) = \max(A). $$

Note:

Taking supremum and the exercise is not necessary if one knows any finite set on a totally ordered field has a maximum and a minimum. See for example, here: A finite set always has a maximum and a minimum.

Mathematics_Beginner
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