Cantor's diagonal argument and alternate representations of numbers

Question

Cantor's diagonal argument works because it is based on a certain way of representing numbers. Is it obvious that it is not possible to represent real numbers in a different way, that would make it possible to count them?

Edit 1: Let me try to be clearer. When we read Cantor's argument, we can see that he represents a real number as an infinite sequence of binary digits. Using this representation, he shows that real numbers are uncountable. An intuitive counter argument could be that maybe there is another type (perhaps incredibly strange) way of representation that would make it possible to count the real numbers. A kind of trick, like the typical one used to show the countability of rational numbers. One could thus be tempted to think that when representing real numbers as infinite sequences of binary digits, it is those representations that are uncountable, but that some other representation could be countable. It seems to me that this can be summed up like this: Is a proof of the countability of a set dependent on the representation its members in the proof?

Answer 1

You are making the common mistake to confuse between a number and its decimal representation.

An easy way to see that $\Bbb R$ is uncountable, regardless to how we can or cannot represent real numbers, is to see there is an injection from $\mathcal P(\Bbb N)$ into $\Bbb R$, defined by $\displaystyle A\mapsto\sum_{n\in A}\frac1{3^{n+1}}$.

This function depends only on the fact that this sum is a convergent sum (as it is bounded by a convergent geometric sequence), and the fact that $\mathcal P(\Bbb N)$ is uncountable depends only on its property of being the power set of $\Bbb N$.

Answer 2

You obviously do not yet fully understand Cantor's argument and its implications.

Where you are correct: Cantor's argument indeed relies on the fact that there exists a decimal representation of numbers.

Where you are wrong: It is not true, as you are implying, that Cantor's argument only works if we represent numbers in a particular way.

Cantor's argument proves that there does not exist any bijective function from $(0,1)$ to $\mathbb N$.

This statement, in itself, does not "see" the representation of numbers, so changing the representation cannot effect the truth value of the statement.

For example, say I give you the statement "there are $10$ cows in the world", and you show me a herd of $11$ cows. Then, you say "let $S_H$ be the set of all cows in this herd, and let $S_A$ be the set of all cows in the world. Then $|S_H| = 11$, and because $S_H\subseteq S_A$, we know that $|S_H|\leq |S_A|$, therefore, the set of all cows contains at least $11$ cows, and your statement is wrong."

Thus, you prove to me that my statement is incorrect. Can I now say

Well, yes, but that's just because you represented your herds of cows with sets. Maybe, there is another representation of cows in which there actually are only $10$ cows in the world.

No, of course I cannot say that. Your sets were only a tool to prove a point, and they prove the fact that there are more than $10$ cows in the world, but the fact remains true even if you use some other tool.

Answer 3

Independent of Cantor's diagonal we know all cauchy sequences (and every decimal expansion is a limit of a cauchy sequence) converge to a real number. And we know that for every real number we can find a decimal expansion converging to it. And, other than trailing nines and trailing zeros, each decimal expansions are unique.

So we know that the real numbers and the set of decimal expansions are the same.

NOW we can talk about Cantor's doagonal. Any finite set $F $ (such as the set of 10 digits), then $F^\mathbb N $ is uncountable by Cantor's diagonal. (Equivalency of trailing 9 and trailing 0 is a minor detail easily resolved.) A 1-1 corespondence between decimal expansions and the set $A =\{0...9\}^\mathbb N $/trailing 0 equiv trailing 9 is obvious.

$A \iff $ {decimal expansions} = $\mathbb R $.

So Cantor's argument is good no matter what format we do or do not represent the reals with.

Answer 4

It would not matter if we used a different representation. The reason for this is that the existence of a decimal expansion is a property of real numbers. A property of real numbers is that if I take a number $x \in [0, 1)$, then I can construct a sequence of integers that does such-and-such. The argument doesn't work because we choose to look at such-and-such representation of real numbers, but because the real numbers have a property that we can do the operations we propose. For example, the argument would have failed to establish the uncountability of $\mathbb{Q}$ because $\mathbb{Q}$ is not complete, and soo even if we constructed a "number" whose first digit was not $0, 9,$ or the first digit of the first number in the list, whose second digit was not $0, 9,$ or the second digit of the second number on the list, and so on, we would be incorrect because we cannot promise that we have constructed a rational number (in fact, there would be cases where you hadn't, as you could in fact list every rational number). But $\mathbb{R}$ is complete, so we know the Cauchy sequence that is $\left( \sum_{k = 1}^{N} a_{k} 10^{-k} \right)_{N \in \mathbb{N}}$ converges, where $a_{k} \in \{1, \ldots, 8 \}$, and moreover we know it converges to a number different from any on the list.

Now imagine a similar question: "Sure, the fundamental theorem of arithmetic says that every positive integer can be written uniquely as a product of positive primes, but what if we didn't factor them? Would such-and-such argument still hold?" The answer here is yes, because no matter how you express a natural number, you can still go back and factor it. This is because we never said that "a natural number is one that can be written uniquely as a product of prime numbers", but because rather it is a property of natural numbers that they can be written as a product of primes. Likewise, given any sequence of real numbers, no matter how you write them, we can still say, "Okay, so these are all real numbers, and so we can do this deal here and find a sequence of digits for each real number, perform the right operations on the right digits, and get a sequence that we can relate to a convergent series that converges to a real number not on our list." It's well and good to find a new way to write a real number, but it is still a real number, and you cannot re-write it so it lacks the properties of real numbers.

Answer 5

Cantor himself did not give an argument with decimal or binary representations of the reals, but a topological style argument: assuming the reals can be enumerated as $(r_n), n \in \omega$ he picks smaller and smaller nested closed intervals, the $n$-th missing $r_n$ for each $n$. He previously showed his nested intervals theorem for complete metric spaces and so he knows there is a point in the intersection of the intervals which cannot be any of the $r_n$ and we have a contradiction. So a topological argument, not one based on decimal expansions but on completeness of the reals (which is by construction of the set of reals, Cantor did that construction in an earlier paper: Dedekind used order completeness, Cantor metric completeness and Cauchy sequences to construct an isomorphic copy of the "real numbers" $\Bbb R$).

Answer 6

To prove $\mathbb{R}$ is uncountable, we assume that it isn't (i.e. that it is countable). We then make a list of all the real numbers, and state that this list is complete. However we can construct a new real from not the $1^{st}$ digit of list item $1$ (e.g. if the $1^{st}$ digit of item $1$ is $4$, pick any digit other than $4$ (and arguably not $0$ either)), and keep the rest the same ($2^{nd}$ digit from $2$, $3^{rd}$ digit from $3$, etc..), not the $2^{nd}$ from list item $2$ and the rest the same, and so on. This number is not in the list, and so $\mathbb{R}$ is uncountable.

This is Cantor's Diagonal argument.

As it is impossible to count $\mathbb{R}$, there is no way we can re-invent numbers in such a way as to make this possible.