How to understand Cantor's diagonalization method in proving the uncountability of the real numbers?

Question

Cantor's diagonalization method prove that the real numbers between $0$ and $1$ are uncountable. I can not understand it.

1. About the statement. I can 'prove' the real numbers between $0$ and $1$ is countable (I know my proof should be wrong, but I dont know where is the wrong).
   • Proof: The real numbers between $0$ and $1$ can always be written as "$0.$xxxx...". If we remove the '$0.$', the number after it "xxxx..." is always an integer. In addition, in order to has the well-defined addition of infinite long integers, we also ask the reflection operation, i.e. the corresponding of "$\sqrt{2}-1=0.4142...$" is "$...2414$". Through this way, we may build one to one correspondence from real numbers in $(0,1)$ onto the natural numbers. Thus the real numbers in $(0,1)$ is countable.
   • Remark: As far as I understanding, both rational numbers ( $1/3=0.333...$) and irrational numbers ($\sqrt{2}-1=0.4142...$) in the region $(0,1)$ are corresponding to natural numbers in the ways of removing the '$0.$' and reflection. Here '$0.333...$' and '$0.4142...$' are corresponding to infinite long integers '$...333$' and '$...4142$'.
   • EDIT： The infinite long integer has the right end is not an infinite number, because for any infinite long integer has the right end (suppose it is 'n'), we can well define 'n+1' (e.x. "$...2414$+1"), which is bigger than 'n', thus 'n' is not an infinity value, as the infinity value should be bigger than any natural number.
2. About Cantor's proof. Seem's that Cantor's proof can be directly used to prove that the integers are uncountably infinite by just removing "$0.$" from each real number of the list (though we know integers are in fact countably infinite).
   • Remark: There are answers in Why doesn't Cantor's diagonalization work on integers? and Why Doesn't Cantor's Diagonal Argument Also Apply to Natural Numbers? It is said the main reason is integers cannot have an infinite number of digits. I can not understand this. As I show above, by remove '0.' from '$\sqrt{2}-1=0.4142...$', we can get an integer, which is infinite long. And also, integers are in fact countably infinite also means we can have infinite big integer.

Answer 1

I think the confusion is over the definition of an integer. An integer, as usually defined, has only a finite number of digits. An infinite collection of digits, like "4142...", is not an integer. In your last remark, you declare "4142..." to be an infinitely long integer, but math doesn't work that way. You can similarly prove that there are uncountably many integers by declaring every real number to be an integer, but that won't have demonstrated anything other than linguistic creativity.

Answer 2

Yes, the issue with your proof is still that there is no such thing as an "infinitely long integer." An integer by definition is a finite number with finitely many digits; after all, if you have infinitely many digits, how do you know what place value each one has? That is, which one is in the ones place, which in the tens place, and so on.

The point of being countably infinite is that it doesn't apply to any one integer. It only applies to them as a group. That is, when you are counting, at any step of the way you have gone finitely many steps. You can get to any specific positive integer you want, by starting from 0 and counting up; or to any integer by counting 0, 1, -1, 2, -2, etc. But if you want to get to all the integers, then you can never stop counting.

I'm afraid it may be more confusing to say this, but: what is this "integer" 3333... that you propose? What are its prime factors, since it is not 1, 0, or -1? What is this integer minus 200? The point being, for any integer, you should always be able to subtract, add, or multiply by any other integer. If you allow "infinitely many digits" per integer, then you cannot actually do integer arithmetic.

Answer 3

What you called integer is an infinite string of digits.

Every integer is finite and can only have finitely many digits.

An infinite string of $1111111.....$ is not even convergent as an infinite series.

Answer 4

You need to understand the difference between the following statements:

1. There are integers of arbitrary value
2. There are integers of infinite value

Only one of them is true, the first one. It means that for every fixed constant $R$ you can find an integer which is larger than $R$.
The second one means that there is an integer which is $\ge$ all other integers. But this is clearly not the case, because adding $1$ to an integer always gives a larger integer.

Answer 5

Let us take the positive integer 1. Add 1 to it you get 2. Again add 1 you get 3. Do this again and again. Every number you get in this process will have finitely many digits. These are the natural numbers. Whereas the number obtained from real decimal number by "omitting" the decimal point is not a natural number (i.e., not obtainable by the procedure I outlined above).

Hope this clarifies.

Answer 6

Due to the discussion in my previous answer I hopefully figured out what the problem is. ( I write this as a new answer because it is quite different from the first one and I don't want to add another comment to it).

The claim is, that the "numbers" with infinitely many digits are integers.

By the Peano axioms such a number $x$ is a natural number if and only if its predecessor is or if $x=0$.
Clearly $x\neq 0$, so we need to show that the predecessor of $x$ is a natural number. But the predecessor still has infinitely many digits. Repeating this process again and again, we still get "numbers" with infinitely many digits. However, if $x\in \mathbb N$, then at some point, we would arrive at $0$.

On a more informal note: Is it possible to count (starting at 0) to a number with infinitely many digits? No. So the "number" cannot be a natural.

Answer 7

Here's a perhaps more fathomable way to phrase what everyone has already said: even if you were to include $\infty$ as an integer, it would be just one integer. On the other hand, counting 4142... and 1088... separately means you're adding a much larger number of infinities to your set.

How many numbers exactly? There are ten choices for each digit, there an infinite number of digits so indeed you're adding $10^{\aleph_0}=2^{\aleph_0}$ numbers to your integers, which is precisely the cardinality of the reals.

Answer 8

Cantor's proof is applicable only to real numbers because the decimal part of a real number can be expressed as any infinite series of digits without invalidating its status as a number. However, concerning integers, if you claim that any infinite series of digits is an integer, it implies considering infinity itself as an integer, which is mathematically unacceptable. Therefore, the criterion that makes an integer an integer involves ensuring that the series of digits representing it is not infinite. This renders Cantor's proof unacceptable in the context of integers.