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I've been struggling with Cantor's Diagonal Argument, and reading the questions of many others which do not quite address my concerns. My concern has to do with with the non-terminating nature of Cantor's construction. Maybe the most clear way to solicit feedback about this concern is to propose a similar argument that relies on a non-terminating construction for the natural numbers, and ask what is wrong with it (and if it's not obvious, why this is not also a problem in Cantor's construction)?

  1. Suppose you have a countably infinite list of all natural numbers.
  2. Add up all of the numbers in this list to calculate the sum x.
  3. Contradiction: x is a natural number (the sum of other natural numbers) which should be in the list, but it is bigger than every element in the list based on its construction.

Step 2 seems problematic to me in that it's just not possible to add up all of the natural numbers. But it seems like Cantor's Diagonalization argument relies on a similarly problematic construction that never ends. Otherwise, you could use his construction as a kind of generator to enumerate irrational numbers that would at least pass a specific diagonalization test... but I'm probably getting ahead of myself. I guess my primary question is about why this kind of never ending construction is OK for Cantor to use, but not OK to use in my example above.

Thanks in advance for any thoughts or references in the direction of sorting out this confusion!

Update: I do understand the deficiency in the proposal asked about here Why Doesn't Cantor's Diagonal Argument Also Apply to Natural Numbers?. My question is instead about the reasonableness of basing a construction on the contents of a infinite list.

GaryD
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    The sum of all natural numbers is not a natural number. – Angina Seng Aug 17 '17 at 15:40
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    I don't think it's a duplicate. The number was generated using a different method. – Kenny Lau Aug 17 '17 at 15:47
  • I had read this other question's answers, and do understand that rational numbers have a finite number of digits, as opposed to irrational numbers that have an infinite number of digits. My concern is really focused on this process of performing a never ending computation on a hypothetical (infinitely countable) list of elements... and when that is vs isn't reasonable. – GaryD Aug 17 '17 at 16:16
  • @GaryD Rational numbers can also have an infinite number of digits. – Kenny Lau Aug 19 '17 at 14:26

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The number $x$ that you come up with isn't really a natural number.

However, real numbers have countably infinitely many digits to the right, which makes Cantor's argument possible, since the new number that he comes up with has infinitely many digits to the right, and is a real number.

Kenny Lau
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  • Thanks for the quick response. My concern is that since irrational numbers have infinitely many digits, Cantor's construction does not terminate and therefore does not reach a number that is missing from the original hypothetical list. If cantor can imagine an irrational that is constructed in this way, how is this different from imagining a natural number that is also constructed from this kind of hypothetical list. Maybe easier than the sum, would be to propose the largest number in this list +1. – GaryD Aug 17 '17 at 16:11
  • @GaryD nobody is building the number in a step-by-step manner. (Also, rational numbers can also have infinitely many digits), so "does not terminate" is not a valid counter-argument. – Kenny Lau Aug 17 '17 at 16:26
  • thanks again. I can imagine a number being defined as Cantor's construction suggests. So maybe my problem is with understanding why this kind of construction could not be used for natural numbers. What is the difference between basing a construction like this on an infinite list of natural numbers (each with finite length) and on an infinite list of irrational numbers (each with infinite length). Isn't it equally problematic to presume that any such construction can make use of the infinite number of values in either list? Thanks again for your help! – GaryD Aug 17 '17 at 16:48
  • @GaryD The problem of using the same argument with natural numbers is that the new number you come up with would have infinitely many digits to the left, and is hence not a natural number. – Kenny Lau Aug 17 '17 at 18:04
  • thanks for your time and patience on this. I see that this sum would be infinite, but does that follow from some property of natural numbers? My thinking is that the set of natural numbers can be generated by successively adding one. And that a set of irrationals (not necessarily all of them) can be generated by repeatedly applying cantor's diagonalization. If infinitely adding one leads the sum of natural numbers to be infinite, I wonder what infinitely diagonalizing irrationals leads to... Thanks again. – GaryD Aug 17 '17 at 19:36
  • @GaryD There are infinitely many natural numbers, but every one of them is finite. You might expect the sum of any sets to be a number, but unfortunately that only works for finite sets. Cantor's argument assumes that there are countably many real numbers and then work out a contradiction, by constructing a number with countably many digits to the right that is not in the list. The newly constructed number has a countably infinitely many digits, which makes it fine. – Kenny Lau Aug 17 '17 at 20:03
  • Most objections to Diagonalization are based two misrepresentations. It isn't about real numbers, it is about infinite-length binary strings. This one is easy to get around, but (as above) people try to extend the "numberness" of the reals to the naturals, rather than recognizing that the important property is being infinite. But it is also not a proof by contradiction, so trying to find a flaw in the contradiction is just as futile. In order to a formal proof, the contradiction must follow from everything you assume, and it does not follow from assuming your list is complete. – JeffJo Aug 19 '17 at 12:53
  • Diagonalization is a proof by contraposition. If |T is the set of all binary strings, Cantor assumed only that a subset |S was countable. From the bijection, he proved that there is an element of |T that is not in |S. That is, "If |S is countable, then |S is not all of |T." Since "If A then B" is logically equivalent to "If ~B, then ~A," this proves that if |S is all of |T, then it is not countable. – JeffJo Aug 19 '17 at 13:02