What is wrong with my proof showing the real numbers between 0<=x<1 are countable?

Question

I am wondering what is wrong with my proof showing the real numbers are countable.

1. Create a set for every real number having 1 digit (10s position) with 0<=x<10

   {0, 1, 2, ..., 9} / 10 = {0.0, 0.1, 0.2, 0.3, ..., 0.9}

   This set covers all numbers with 1 digit, but no number having 2 or more digits

2. Create a set for every real number having 2 digits (100s position) with 1 digit set and 10<=x<100

   {0.0, 0.1, 0.2, ..., 0.9} $\cup$ {10, 20, 30, ..., 90} / 1000 $\cup$ {11, 12, 13, ..., 19, 21, 22, 23, ..., 29, 30, 31, 32, 33, ..., 39, ..., 99} / 100 =

   {0, 0.01, 0.02, 0.03, ..., 0.09, 0.1, 0.11, 0.12, 0.13, ..., 0.19, 0.2, ..., 0.99} = {0, 0.01, 0.02, 0.02, ..., 0.99}

   Were the bold number are from the one digit set (I am inserting an element of the one digit set into every 10th position)

   This set contains all 2 digit numbers, but is missing all sets containing 3 or more digits.

3. Create a set for every number having 3 digits (1000 position) with the 2 digit set and 100<=x<1000

   {0.00, 0.01, 0.02, ..., 0.99} $\cup$ {100, 200, 300, ..., 900} / 100000 $\cup$ {110, 120, 130, ..., 190, 210, 220, 230, ..., 290, ..., 310, 320, 330, ..., 390, 410, ..., 990} / 10000 $\cup$ {101, 102, 103, ..., 109, 111, 112, 113, ..., 119, 121, 122, 123, ..., 129, 131, ..., 999} / 1000 =

   {0.0, 0.001, 0.002, 0.003, ..., 0.009, 0.01, 0.011, 0.012, ..., 0.019, 0.02, ..., 0.999} = {0, 0.001, 0.002, ..., 0.999}

   Were the bold number are from the one digit set (I am inserting an element of the two digit set into every 10th position)

   This set contains all 3 digit numbers, but is missing all sets containing 4 or more digits.

If I repeated this to infinity I would eventually arrive at a set containing all numbers with an infinite amount of digits after the decimal ($0.\overline{00}$ to $0.\overline{99}$). No number would be missing except for any number with an more digits than infinity, which could easily be gotten by doing one or more rounds.

The only problem I can see with this proof is if you defined the natural numbers as having a finite number of digits. However, that would contradict the idea of the natural numbers being infinite (there are only a finite number of combinations for a set of finite numbers, which is 10^( the number of digits in your limit)), which suggests that at least one natural number has an infinite number of digits.

I am very aware of cantors diagonal argument and think this proof shows a counter example. In the case of a real number, like pi mapping to a rational number I would think that since the natural numbers are infinite, I would expect there to be a number that has the exact same digits as pi. If I divide this number by 10^(n-1), were n = number of digits in pi I get pi. Again the only problem with this logic is in how you define the natural numbers. If these is the case then please tell me and point me to a source that defines this.

Again I am not a mathematician and have a limited understanding of mathematics, so I am probably wrong here.

Answer 1

The only problem I can see with this proof is if you defined the natural numbers as having a finite number of digits.

You are correct; that is the problem with this proof. The natural numbers only have a finite number of digits; as a result, you don't get real numbers with infinitely many digits on your list.

However, that would contradict the idea of the natural numbers being infinite (there are only a finite number of combinations for a set of finite numbers, which is 10^( the number of digits in your limit)), which suggests that at least one natural number has an infinite number of digits.

This would be a problem if we said "the natural numbers can have at most <INSERT LIMIT HERE> digits". But we're not saying that. No natural number has infinitely many digits, but there are natural numbers with any number of digits you choose. (For example, if you want an $n$-digit natural number, take $10^{n-1}$.)

Note that I said number of digits. "Infinitely many digits" is not a number of digits.

That's also the problem with trying to define a natural number with the digits of $\pi$. For any number $x$ with finitely many digits after the decimal, we can let $n$ be the number of digits after the decimal, and then $10^{n} \cdot x$ will be a natural number with those same digits. However, $\pi$ has infinitely many digits after the decimal. We can't multiply $\pi$ by $10^{\infty}$, because $\infty$ is not a number.

Again the only problem with this logic is in how you define the natural numbers. If these is the case then please tell me and point me to a source that defines this.

Technicalities aside, the natural numbers are defined inductively. First, $0$ is a natural number. Then, every natural number $n$ has a natural number $n+1$ that comes after it. Everything you can "count up to" by starting at $0$ is a natural number. All natural numbers are finite, because you can never reach an infinite number by counting up.

Answer 2

To show the cardinality of $(0,1)$ is that of $\mathbb{N}$, you need to construct a bijection $\mathbb{N} \rightarrow (0,1)$. Cantor's diagonal argument shows this is impossible, which raises the question: what are you showing?

Your proof shows that the set:

$$S =\{\{x : x \text{ has at most $n$ nonzero decimal digits}, x \in (0,1)\} : n \in \mathbb{N}\}$$

has the cardinality of $\mathbb{N}$. When I write it down this way, it's not hard to see that it is the case that this set is countable: it's indexed by $\mathbb{N}$! Bijections require careful reasoning, and in this case you're not matching up a natural number to a real number, you're matching up a natural number to a set of real numbers.

Written this way, it's also clear what goes wrong. Taking the union, which is likely what you intended:

$$``\mathbb{R}" = \bigcup_{s \in S} s$$

we can see that $``\mathbb{R}"$ is the countable union of countable sets, hence is countable. Why is this so? Pick some $r \in ``\mathbb{R}"$, and notice that $r$ has at most $n$ nonzero decimal digits, for some fixed $n \in \mathbb{N}$. That is to say, you don't have the transcendental numbers, so you're missing the uncountable part of $\mathbb{R}$!

This kind of curiosity is to be praised, though. I'd be lying if I said that I questioned and considered this closely every theorem I read.

---

It may be a fun exercise to show that irrational numbers that are solutions to polynomials (part of the set of algebraic numbers) are countable. This justifies where I say that the transcendental numbers are the uncountable part of the reals.

Answer 3

If I repeated this to infinity I would eventually arrive at a set containing all numbers with an infinite amount of digits after the decimal ($0.\overline{00}$ to $0.\overline{99}$).

You do not say exactly what you mean by "repeating this to infinity", but I assume that you mean that you take the union of those sets that you described. That is a countably infinite union of countable sets and hence indeed a countable set. However, it does not contain any number that is not in any of those sets, in particular it does notcontains "all numbers with an infinite amount of digits after the decimal", it only contains all numbers with any finite number of digits after the decimal, i.e. it contains only rational numbers.