# A proof from Baby Rudin using the Archimedean Property

There's a very small detail in the proof of Theorem 1.20 on page 9 that confuses me a little. I feel like I'm missing something. The Theorem proves the fact that between any two real number $x<y$, there is a rational number $p$ with $x < p < y$.

He uses the Archimedean Property (for any two real numbers $x$ and $y$, with $x>0$, there exists a positive integer $n$ or which $y < nx$) to prove that for any real number $w$, there are positive integers $m_1$ and $m_2$ for which $-m_1 < w < m_2$.

This is what confuses me: Having taken such care to prove the self-evident fact that there exist $m_i$ for which $-m_1 < w < m_2$, he concludes, with no further justification that there exists an integer $m$ with $-m_1 \le m \le m_2$ for which $m-1 \le w < m$.

Again, it seems self-evident that for any real number $w$, there exists an integer $m$ for which $m-1 \le w < m$. But how can he brush-over this when he insisted on proving $-m_1 < w < m_2$?

Perhaps the "missing piece" is that once you have such integers $-m_1$ and $m_2$, then if $m_1 + m_2 > 1$, then you can break up the interval $[-m_1,m_2]$ into two consecutive kissing subintervals of smaller integer length with integer endpoints. And, since $w \in [-m_1,m_2]$, then $w$ must be in one of these subintervals. By induction on the difference between your two integer endpoints, you can show your desired integer $m$ exists, provided $m_1,m_2$ exist (which may, in Rudin's mind, been the "hard" part to show).

• Thanks for your answer, The idea of kissing intervals is what made the fact self-evident to me. Intuitively, we can can decompose the real line into intervals of the form $$\bigsqcup_{m \in \mathbb{Z}}[m-1,m)$$ and $w$ must lie in one of them. – Fly by Night Aug 7 '15 at 18:23
• @FlybyNight yeah that is a more direct, less formal way to see it. It seems to me Rudin was being very careful to establish his $m_1$ and $m_2$ exist for an arbitrary real number. E.g, maybe there is a real number greater than any integer, if the Archimedean principle fails. But once you have your integer bounds, the rest is straightforward, e.g. basically with your argument if you don't want to do induction. – user2566092 Aug 7 '15 at 18:27
• Could you flesh out what exactly should be proved by induction? Don't solve it, just explain what I should be proving. – Fly by Night Aug 7 '15 at 18:35
• Given two finite integers $a,b$ for your interval end-points, where the interval $[a,b]$ contains your number, prove by induction there is an interval where the end-point integer endpoints differ by 1 that contains your number. To do this, if your integer end-points are $a$ and $b$ then you consider the intervals $[a,a+1]$ and $[a+1,b]$. Then either you have found your interval of length 1 that contains your number, or it's in the other interval and you apply induction using the difference $b - (a + 1)$ which is 1 less than the difference between endpoints of your original interval $[a,b]$. – user2566092 Aug 7 '15 at 18:44
• Thanks for that. It's pretty obvious when you explain it like that. – Fly by Night Aug 7 '15 at 18:51

You are right that this fact demands proof. It is safer to assume nothing self-evident.

Let $S\subseteq \mathbb N_0$ be the set of natural numbers (including $0$) larger than $w+m_1$. This set is nonempty because $m_1+m_2\in S$. Thus let $n=\min S$ (using the wellordering of $\mathbb N_0$). Then $n>0$ because $0<w+m_1$. Therefore we have $n-1\le w+m_1<n$ and from this $n-m_1-1\le w<n-m_1$, i.e., the claim with $m:=n-m_1$.

• I'm with you up to $\min S > 0$. How does that give us $n-1 \le w+m_1 < n$? It boils down to the same fact: we can fit any real number between two consecutive integers. This is what Rudin skips over too. – Fly by Night Aug 7 '15 at 18:30