Is something wrong with this counting technique of reals?

Question

I know the topic of countability of reals has been discussed a lot, but I still don't understand the proofs, including the well-kown diagonal approach. So, please forgive my dilettantism if it has place.

First of all, the main subject: is there anything wrong with this counting approach?counting reals

Here we count each node of the tree which represents all possible numbers in [0, 1). 1 could also be included, as well as the rest. Each new level - is the next digit. Here I chose binary form, but it could certainly be decimal. And I don't see any misses here. So it seems we can still count the reals.

Another question is related to the diagonal proof. But on the other hand it seems to not be a proof. At least for two reasons:

1. At any given momemnt when we try to construct a new number, we actually take into account just an already counted list, and we surely can get a number which is not yet in the list. But that seems to just mean that we have not yet counted a new number. After all, we're dealing with an infinite list, and at any moment of counting there are still more numbers wich we have not counted yet.
2. Another point which confuses me is that this same diagonal method will also prove that the set of natural numbers is also uncountable. And that certainly isn't the case. Isn't it such? Same considerations: assuming we have counted all natural numbers, we can construct another natural number which is not (yet) in the list.

Sorry for having two questions in one, but they are closely related. Moreover, the same considerations about the diagonal also confuse me in other proofs of uncountability of reals. They all seem to appeal to the same issue: when we see there is still a new number exists which is not yet in the list, we're actually taking into account an already made list. But it seems to just mean we have not yet counted our new number. Would be thankful for the clarifications.

Answer 1

Regarding question 1, every number you list will have a finite binary representation. Thus you only have enumerated a proper subset of the rational numbers. Not anywhere close to all of the reals.

Regarding question 2, your construction of a natural number not in the list has to work for any list of natural numbers, not just a specific list. If I list the natural numbers by setting the n’th entry in the list to be $n$, you cannot construct a natural number that is not in the list.

Answer 2

The numbers enumerated by your diagram are, in order, $0$, $0.0$, $0.1$, $0.00$, $0.01$, $0.10$, $0.11$, $0.000$, ... etc. Notice that all of these numbers only have at most finitely many nonzero digits, so a number like $0.01010101\ldots = 1/3$ will not be counted.

This is the main reason why the reals are unccuntable, while a set like the rationals is countable. Each rational number is specified by only a finite amount of information (i.e., an integer numerator and denominator), and for this reason they can be counted. But a single real number can have (countably) infinitely many digits, requiring an infinite amount of information to communicate.

In general, any collection in which each element requires a finite amount of information to describe (set of all computer programs, set of all polynomials with integer coefficients, etc) is countable. But a set consisting of infinitely many "infinitely sized" objects may fail to be countable.

Answer 3

The problem with trying to apply diagonalisation to the natural numbers is this:

Suppose we have a list of all natural numbers $a_1, a_2, a_3, \ldots$, and we claim to have constructed a number $N \in \mathbb{N}$ that isn't on the list by making sure that the $k$th digit of $N$ is different to the $k$th digit of $a_k$.

Since $N$ is a natural number, it's finite, and that means that it can be written in $M$ digits for some number $M$, with every "bigger" digit being fixed at zero. That means that when $k > M$, since the $k$th digit of $N$ is zero, the $k$th digit of $a_k$ must be non-zero by the construction of $N$.

But that's a problem, because there are $10^M$ numbers with fewer than $M$ digits, and $10^M > M$, meaning that there must be a number $a_k$ such that $k > M$ but $a_k$ has fewer than $M$ digits, which means that $a_k$ matches $N$ in the $k$th digit.

That contradicts our claim that we constructed $N$ to not match any digit on the diagonal, which means that either (a) $N$ is actually on the list, or (b) the list doesn't actually contain all of the natural numbers, either way we haven't made the natural numbers uncountable.

We're fine doing this with the real numbers because a finite real number can have infinitely many non-zero digits in its decimal expansion, something that doesn't work when you're looking at digits to the left of the decimal point.