# Proving uncountability of $\mathbb R$ only using the complete ordered field axioms

If we define the real numbers abstractly as a complete ordered field (like described in the Wikipedia page), how can we prove that they are uncountable? In other words, using just the axioms of a complete ordered field, how can we show that any model satisfying those axioms is not isomorphic to N?

The usual proof I've seen of uncountability of $\mathbb R$ is through a diagonalization argument, but that involves a specific representation of $\mathbb R$ - namely infinite decimal sequences. How can we do it in the abstract, without that (or any other) specific representation?

• Do you even need it to be a field? I thought that any densely ordered (i.e. between any two reals is another real), complete set must be uncountable. – Akiva Weinberger Apr 26 '15 at 19:26

There are many proofs of the uncountablity of the reals. Firstly though, note that the Cantor proof you mention only uses the axioms of complete ordered field, since the fact that decimal expansions exist is a consequence of those axioms.

You are perhaps looking for a proof that is more directly related to analytical properties of the reals, rather than any typographic representation of the reals. Well, Baire's theorem states that a non-empty complete metric space is not of first category. If the reals were countable, then they would be of first category. As some real exists, the reals are not countable.

• The first proof, using Cantor's lemma, is essentially a proof using the completeness of the field. – Asaf Karagila Apr 26 '15 at 19:19

This is essentially Cantor's original proof.

Given a sequence $x_1,x_2,\ldots$ of real numbers in $[0,1]$ we construct a decreasing sequence of intervals, $[a_n,b_n]$ such that for each $k$ there is some $n$ for which $x_k\notin [a_n,b_n]$.

Now the set $\{a_n\mid n\in\Bbb N\}$ is bounded by all the $b_n$'s, and the completeness axiom tells us that there is some $a$ such that $a_n\leq a$ for all $a$, and $a\leq b_n$ for all $n$.

Therefore $a\neq x_k$ for all $k$, since each $x_k$ gets omitted at some interval, and $a$ belongs to all the intervals.

Take a sequence $u_n \in \mathbb{R}$. Split $[0,1]$ into 3 equally. Take the partition furthest from $u_1$, split it into three and repeat with $u_2$. Etc. The sequence of intervals will converge to a real number not in the sequence $u_n$.

I think this is similar to Asaf Karagila's answer but without the symbols.