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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).
X Leo
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  • Maybe it will better if you look at the number after dropping to "0." as an infinite sequence of digits (i.e ${0,1,...,9}^\mathbb{N}$), and not as an integer. In addition, what about the rest of the proof? you didn't even get to the diagonalisation part in your question – GuySa Jul 18 '18 at 20:55
  • The first part (prove (0,1) real numbers is countable) does not need diagonalization method. I just use the definition of countable sets - A set S is countable if there exists an injective function f from S to the natural numbers. The second part (prove natural numbers is uncountable) is totally same as Cantor's diagonalization method, the only difference is that I just remove "0." from each number from the list table. – X Leo Jul 18 '18 at 21:02

8 Answers8

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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.

Elchanan Solomon
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  • Is any specific reason to exclude infinitely long integer? At least, from the natural number itself, we dont need to remove them. And, I think this excluding is inconsistent with the argument integers are in fact countably infinite. As far as I understanding, the series of natural numbers is special. The number itself can be used to identify it's positions. If all the natural numbers (including the biggest one) are finite, that's means the series is finite. – X Leo Jul 18 '18 at 21:33
  • The claims you are making are not mathematical, they are philosophical. You claim that since every natural number is finite, there are only finitely many natural numbers. There are philosophical problems with that claim, but that is a separate question. The basic assumption of mathematicians is that there are infinitely many finite numbers. Our definition of integers includes only those finite numbers. If you want to challenge those philosophical positions or definitions, that is a separate discussion. – Elchanan Solomon Jul 18 '18 at 21:39
  • I dont want to do this claim. I just want to show the two mathematical statements "there are infinitely many finite numbers" and "the integers are countably infinite" are matehmatically inconsistent. And the key point of proof is that the series of natural numbers is special, the number itself can be used to identify it's positions (if we dont miss any number). If all the natural numbers (including the biggest one) are finite, that's means the total number of elements is finite. – X Leo Jul 18 '18 at 21:49
  • There is no biggest one. – Elchanan Solomon Jul 18 '18 at 22:00
  • I make the mistake about the 'biggest one'. As illustrated by klirk, the reason there is no biggest one is that for any 'n', we can find 'n+1', so we can never get the 'biggest one'. Here the conclusion is still weird for me as For finite sequence, there always exist a number, which is the same as the number of element in the sequence. If there are infinitely many finite numbers, it means that for the whole infinite sequences, we can never find a number that it's position (must be finite) is same as the number of element. @Elchanan Solomon – X Leo Jul 19 '18 at 06:24
  • Now I have the new argument (as edited in the main post) that the infinite long number (that has the right end) is not an infinite number thus does not challenge any thing. @Elchanan Solomon – X Leo Jul 19 '18 at 06:27
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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.

EGoodman
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  • I would like to ask another question. Can we do the addition $\sqrt{2}+\sqrt{3}$? From your description, this calculation means nothing as we can never fix all numbers of the both $\sqrt{2}$ and $\sqrt{3}$ – X Leo Jul 18 '18 at 21:14
  • First of all, I said integers and those are not, though they can be added (as real numbers). I think you mean the decimal expansions: when you write $\sqrt 2$ as $1.4142...$ then the digits have meaningful place values. That is, you are writing $\sqrt 2 = 1 + 4/10 + 1/100 + 4/1000 + 2/10^4 + ...$. As such, with the decimal expansion of $\sqrt 2$ and that of $\sqrt 3$, we are essentially adding the numerators of fractions in each place, and carrying as necessary. Your operation of "taking away the decimal point" removes place values, which is why "333..." is not a number let alone an integer. – EGoodman Jul 18 '18 at 21:52
  • There are two problems of the addition of two infinite long numbers: alignment of digits and carrying. We have the first problem for the infinite long integers as it has no ends in both sides so that we dont know how to make the alignment, but we dont have this problem for the real numbers in $(0,1)$, as it has end on the left. Both calculations has the problem of carrying, as the addition is always from the most right to the most left. however both the irrational reals and the infinite long integers have no ends on the right, thus we can not really do the carrying. To be continued. @EGoodman – X Leo Jul 18 '18 at 23:23
  • Based on the first problem, you show the addition can not be defined for the infinite long integers. However, this problem can be easily solved if we define infinite long integers which has the right end. Practically, from real numbers in $(0,1)$, besides remove decimal "$0.$", we also ask the reflection operation. For example, now we ask the correspondence to '$\sqrt(3)-1=0.732...$' as '$...237$'. Now, we clearly know the ones place is '7', tenth place is '3'., e.t.c. And obviously the both problems alignment and carrying are solved by this definition. @EGoodman – X Leo Jul 18 '18 at 23:24
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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.

Mohammad Riazi-Kermani
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  • That's what I can not understand. Do you mean that ∞ is excluded from natural number? Why? And, I think the series of natural numbers is special. The number itself can be used to identify it's positions. If all the natural numbers (including the biggest one) are finite, that's means the series is finite. – X Leo Jul 18 '18 at 21:29
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    @XLiu $\infty $ is not a real or complex number. It is a symbol used to indicate certain limits. There is not a biggest one in the set of natural numbers. For every natural number $n$ we have $n+1$ which is bigger than $n$ – Mohammad Riazi-Kermani Jul 18 '18 at 21:38
  • however, that implies the ' integers are not countably infinite' by my above argument. – X Leo Jul 18 '18 at 21:52
  • @XLiu No, because your function in not onto $(0,1)$ and it does not count for all real numbers in that interval. – Mohammad Riazi-Kermani Jul 18 '18 at 21:54
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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.

klirk
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  • This help to understand 'there are no integers of infinite value'. However, this is not conflict with 'there are infinite long integers'. An infinite long integers has the right end (i.e. we know exactly digits in the ones, tens place) can still be added by 1 and thus bigger, which is thus not an infinite value. – X Leo Jul 18 '18 at 23:43
  • You claim that this is an integer, but it actually is not, it's a sequence of integers.maybe another try: as you already noted, if this object you talk about was an integer, then the integers would not be countable infinite. But looking at the construction of this object, this would also work for the natural numbers. Now look at the definition of countable infinite. Clearly the natural are countable infinite. So you get a contradiction and this assumption, namely that your object is a natural number, has to be false. @XLiu – klirk Jul 19 '18 at 07:49
  • This number does not violate the Peano axioms of the naturals. Also, when we say the natural are countable infinite, it means we allow infinite times of operation "+1" from "0" to get infinite many finite numbers. In my opinion, the infinite "+1 gives infinite long natural number 'n'. However, it is still a finite number as we have 'n+1'. I dont think this makes integers uncountable infinite, as the operation "+1" itself make the series countable. To be continued. @klirk – X Leo Jul 19 '18 at 15:38
  • This argument just questions "Cantor's diagonalization method". It suppose there is a list to include all the numbers of countable infinite sets. However, we can never write such a list for any infinite set, including the countable infinite set. As for any list you written for the countable infinite set, we can get an additional number by "+1". @klirk – X Leo Jul 19 '18 at 15:45
  • I really have trouble to understand what you mean. If I understood you right, you are convinced that the integers are not countable infinite but the naturals are. If so, you really need to think about how the integers and the naturals differ. Or is it, that you think the diagonal argument is false, because it would imply (as you believe) that the integers and also the naturals are uncountable? @XLiu – klirk Jul 19 '18 at 16:02
  • Sorry for the unclear, all I discussed are naturals. I thought naturals are countable infinite. And the infinite long numbers (with the right end) are still finite, and they can be get from "0" by infinite times of "+1", thus they are still belong to the natural set. Based on this, I think there are problems with the diagonal method. @klirk – X Leo Jul 19 '18 at 16:31
  • The point why I talked about the naturals, is because by definition a set is countable infinite if there is a bijection from it to the naturals. The problem with your argument still is, that you don't know what the naturals are. There are simply no integers with infinite length. Maybe I can convince you with a different approach which does not involve cantors argument. If there were such an integer $n$ of infinite legth, then by taking the quotient $1\n$, we could get every real number between 0 and 1. But 1/n is a rational number., so this would imply that every rational number is real. – klirk Jul 19 '18 at 17:55
  • But honestly, I think you just have to digest that your image of the integers is not true. – klirk Jul 19 '18 at 18:03
  • Yes. It is true that the argument conflicts with many results we known, especially when compare to real numbers. However, I make this argument as It does not conflict with the definition of naturals - 'the Peano axioms'. You can not wipe out the infinite long numbers from natural set according to the axioms. @klirk – X Leo Jul 19 '18 at 18:31
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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.

P Vanchinathan
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  • Besides "omitting" the decimal point, I also ask the "reflection", so that the infinite long number has the right end like "...2414" , we may then define "-1" and "+1" on it. If the infinite many additions "+1" are allowed, we can get the infinite long number from a finite number. – X Leo Jul 19 '18 at 05:05
  • You are complicating a simple thing: set of natural numbers are infinite. But physical representation any SINGLE number say for the purpose of printing on a paper uses only finitely many resources like ink and paper. This is not true for real numbers. – P Vanchinathan Jul 19 '18 at 07:55
  • Well, you can say it. However, mathematics never forbid you to find a natural number of infinite long so that you can not print it. @PVanchinathan – X Leo Jul 19 '18 at 15:09
  • I differ. Mathematics never forbid to ENTERTAIN the idea of numbers with infinitely many digits. They are called real numbers. But there is a clear definition of what a NATURAL number is which does not include such numbers in its fold. To give you an analogy: Nature does not preclude living beings with tails as part of their anatomy. Such things are called animals. You are saying a human being can have tail. Well human beings are creatures who do not have tails (and many other conditions.) I have no more to say in this matter. – P Vanchinathan Jul 20 '18 at 00:38
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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.

klirk
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  • Your argument based on the assumption that total number of processes is a finite number (both finite in length and finite in the number itself), so that you can never get a finite number from a finite long number by a finite number of processes. However, this assumption is kind of 'prove itself'. In another word, if we use another assumption total number of the process can be a infinite long number (but still finite), though human can not really do the whole process, we get a different result - we can get a finite number from infinite long number by such kind of process – X Leo Jul 19 '18 at 19:29
  • No, the argument is based on the observation that substracting $1$ from a number with infinitely many digits we can never arrive at zero. Even if we repeat that infinitely many times. Just forget about the last part of the answer. As I said, its informal. – klirk Jul 19 '18 at 19:29
  • Ehm no, i substract 1, I observe that the the number has still infinitely many digits. Then i substract 1 again and make the same observation. Then I go on and on, i.e. I do this process infinitely many times. And, surprise: The situation is still the same (because it is after each step) – klirk Jul 19 '18 at 19:34
  • Where do you get this observation? the infinite many times of subtracting 1 equivalent to the subtracting a number, this number is the infinite sum of 1. You can claim this summed number is not infinite, but you can not claim that it is not infinite long. – X Leo Jul 19 '18 at 19:36
  • Do you agree that after each step the resulting number still has infinitely many digits? – klirk Jul 19 '18 at 19:40
  • That's the difference between finite step and infinite step. There are contradiction in your process. On one hand, you suppose the infinite can never be reached step by step (+1); on the other hand, you suppose you can down a process step by step and then reaches infinite steps. – X Leo Jul 19 '18 at 19:43
  • I didn't suppose that the infinite can be reached step by step. This was the assertion I proved. I also didn't ASSUME that I can down the number step by step. This follows from the peano axioms. You still didn't answer me. Do you agree that after each step the number still has infinitely many digits. And with step I mean taking the predecessor. – klirk Jul 19 '18 at 20:02
  • My answer is that of course it is still infinite many digits, however, the reason is in the way 'step by step', you always can only get finite number of step, not the "infinite number of step". – X Leo Jul 19 '18 at 20:23
  • Well, I am suddenly clear. When you says "I do this process infinitely many times", you means you reach "infinite number of step" by the way of step by step. The background is that "$S=\sum_{i=1}^{\infty}1$" is still a finite number as there is $S+1$, so that the infinite many steps by the way of step by step is still a finite step, i.e. the steps follows from the peano axioms must be a finite step. Through the finite steps from 0 we can only get finite number (both in length and in the number itself). @klirk – X Leo Jul 19 '18 at 20:25
  • This doesnt make any sense – klirk Jul 19 '18 at 20:37
  • That follows your construction. If we suppose $S=\sum_{i=1}^{\infty}1$ is an infinite number, we reach infinite by the the way of "step by step" of "+1". (The $\infty$ in $\sum_{i=1}^{\infty}$ follow your statement "do this process infinitely many times", and your construction means to do it by the way of "step by step")@klirk – X Leo Jul 19 '18 at 20:51
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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.

Abhimanyu Pallavi Sudhir
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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.

Bezina Taki
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