Why do we need the axiom of completeness? Why won't Cantor's diagonalization work without it?

Question

Okay so for my upcoming test I need to "be able to explain at least one result that would not hold if the axiom of completeness were not accepted"

My teacher suggested that I could try to explain why Cantor's diagonalization method won't work without the axiom of completeness, but I'm not really sure why it wouldn't work?

EDIT: If you can think of any easier examples I could explain for my test, that be great as well!

Answer 1

In Cantor's diagonalization argument, you assume (for a contradiction) that you can make a list $(x_1,x_2,x_3,\ldots)$ of all real numbers (let's say between $0$ and $1$ inclusive). We then construct a new number $y = .d_1 d_2 d_3 \ldots$ which differs from $x_1$ in the first digit, differs from $x_2$ in the second digit, and so on. We now say, aha, I have found a number in $[0,1]$ which is not on our list!

But wait, is $y$ really a number? Without completeness, we don't know that the infinite series \begin{equation} y = \sum_{i=1}^{\infty} \frac{d_i}{10^i} \end{equation} converges! And if that series does not converge, then we have failed to show that our list was incomplete.

Answer 2

That’s a rather strange suggestion on the part of your teacher, since there are much more straightforward examples, but it can be made to work. The version of the diagonalization that I think you have in mind argument constructs a non-terminating decimal that differs from each row from the diagonal entry in that row. In order to conclude that this decimal actually represents a real number not in your list, you must first show that it represents a real number. Let’s say that the decimal expansion looks like this: $0.d_1d_2d_3\ldots~$. We understand this to mean $\sum_{k\ge 1}\frac{d_k}{10^k}$, but that infinite series is really an abbreviation for

$$\lim_{n\to\infty}\sum_{k=1}^n\frac{d_k}{10^k}\;,\tag{1}$$

if it exists. Let $$s_n=\sum_{k=1}^n\frac{d_k}{10^k}\;,$$ the $n$-th partial sum; it’s these partial sums are all rational, and it’s easy enough to show that $s_1\le s_2\le s_3\le\ldots~$, but how do you know that the limit in $(1)$ exists? In other words, what guarantees that this $0.d_1d_2d_3\dots$, the one that disagrees with the diagonal in every row, actually represents a real number at all?