Prove that cardinal of $\mathbb{R} \neq \mathbb{N}$ without Cantor diagonal argument

Question

All proofs I know that the cardinal of $\mathbb{R} \neq \mathbb{N}$ involves the Cantor diagonal argument.

Is there any proof which not involves that? Can anyone find out an alternative proof (better elemental than using very high-level theories)?

Answer 1

How about this one?

Suppose $\mathbb R$ has the same cardinality as $\mathbb N$. Then there is a sequence $r_1, r_2, \ldots$ containing all real numbers in $[0,1]$. The collection of open intervals $(r_n - 2^{-n-1}, r_n + 2^{-n-1})$ cover all of $[0,1]$. Since $[0,1]$ is compact, some finite subcollection of these intervals cover $[0,1]$ as well.
The total length of that subcollection is less than the total length of the original collection: $\sum_{n=1}^\infty 2^{-n} = 1$. But that's impossible: an interval of length $1$ can't be covered by a finite set of intervals of total length less than $1$.