Is it possible to list two real numbers that lie immediately next to one another on the number line?

Question

This question stems from my general understanding of Cantor's diagonal proof, which from my understanding suggests this is impossible. At the same time it seems like such an elementary idea, to list two numbers that lie immediately next to one another, that I feel a response of some kind might exist.

Answer 1

You can't even do this for rationals, as a commenter has already pointed out. The problem is that between any two distinct rationals (or reals) there lies a third, different from both. Seen differently, there is no smallest rational (or real) greater than a fixed one, and so there's no "next" number on the line.

Some collections of numbers do have a notion of "next," most notably the natural numbers. These do have the property, in other words, that for every integer $m$ the set of natural numbers greater than $m$ has a least element, specifically, $m+1$. In general a collection with this property (slightly extended) is called well-ordered. If you know Cantor's proof you probably know that, in contrast to the reals, the rationals are countable-yet still not well ordered, unlike the natural numbers. Furthermore, there exist sets just as large as the reals which are well ordered! So this notion of next-ness or of being well ordered is not really a question of size at all, and Cantor is a red herring here.

Answer 2

You need to define the term "immediately next". With the usual ordering of $\mathbb{R}$ this is impossible (as has been noted in other answers and comments). However, if you assume that,

• $\color{red}{\le}$ is a well-ordered relation on $\mathbb{R}$.

• Every non-empty subset of $\mathbb{R}$ is also well-ordered with respect to the same ordering.

Then we can do the following (just a sketch of the main idea, though).

Let $x\in \mathbb{R}$. Define $\mathcal{S}:=\{y\in \mathbb{R}\mid x\color{red}{\le} y\}$. If $\emptyset\subset \mathcal{S}\subseteq \mathbb{R}$ then by the second assumption we can conclude that there exist $x_0\in\mathbb{R}$ such that $x_0=\min \mathcal{S}$. Now you may define this $x_0$ to be the "immediately next" real number of $x$.

Observe that here we have assumed that $x_0$ is the "immediately next" real number of $x$. What makes our use of "the" justifiable here?

In fact we can do more. By Well-Ordering Theorem we can define the "immediately next" element of an element $x\in X(\ne\emptyset)$. Can you see how?

Answer 3

The real numbers ($R$) may be defined from the rationals ($Q$) by adjoining, to $Q,$ the set of all Dedekind cuts on $Q.$

A Dedekind cut on $Q$ is an ordered pair $(A,B)$ where $A$ and $B$ are subsets of $Q$ such that (1)$A\cup B=Q,\;$ (2)$A\cap B=\phi,\;$ (3) neither $A$ nor $B$ is empty, (4) $A$ has no largest member and $B$ has no smallest member, (5) every member of $A$ is less than any member of $B.$ (Example: $B=\{x\in Q:x>0\land x^2>2\},\; A=Q\backslash B.$)

The irrational numbers are the Dedekind cuts on $Q$.

This is treated in detail (including especially the def'n of the linear order of $R$) in many books.

One consequence is that there is are rationals and irrationals between any two reals.

Another consequence is that if $S$ is any non-empty subset of $R$ and if $S$ has an upper bound (that is, there exists $x\in R$ such that no $t\in S$ is greater than $r$) then $S$ has a LEAST upper bound. In particular, this implies that there is no $x\in R$ such that $x<1$ and $x$ is greater than every member of $\{1-10^{-n}:n\in N\}.$ This can only be shown by using the def'n of $R$.