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As shown in Cantor's diagonal argument, we make an assumption that the list of infinite sequences of binary digits is countable. After writing an enumeration, we complemented the diagonal elements to produce an element which isn't in the enumerated list. If the number $s$ is made up from complementing the digits of $s_i$ for $i=1,2,...,n$, then it is guaranteed that it is not one of the $s_i$s by construction.

My problem is how do we know that it isn't one of the $s_j$s for $i=n+1,n+2,...$. The enumeration doesn't afterall stop at n---It goes like---${s_1,s_2,s_3,...,s_n,s_{n+1},s_{n+2},...}$. So our specially constructed $s$ can(infact must) fall in the list though it doesn't fall in the first $n \hspace{0.1cm}s_i$s. Contrastingly consider the proof that the rationals are countable, we construct a similar table(by dividing the row and column number of that table), at any step we can construct a rational which is not in that table by choosing a bigger row and or or a bigger column number, so we made a number which wasn't in the list but we don't say then that the set of rationals is uncountable. (Instead it is said we can make any rational number $c/r$ in finite number of steps.)

Note that I have no problem in accepting the fact that the set of reals is uncountable (By Cantor's first argument), it is the diagonal argument which I don't understand. Also I think, this shouldn't be considered an off-topic question although it seems that multiple questions have been asked altogether but these questions are too much related and useful in drawing the (wrong)contradiction to Cantor's argument, so they have been asked. I have seen similar understanding problems with Cantor's diagonal argument in MSE but none match mine(probably) so this isn't a duplicate as well. If you think I missed some, please kindly comment down the link and I will have a look.

Asaf Karagila
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Manas Dogra
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    You're not complementing finitely many elements of the list, you are complementing all of them at once. Or what do you think is the $n+1$-th digit of $s$? Always 0? Always 1? Some random number? – Vercassivelaunos May 10 '21 at 05:40
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    So if we apply the argument to the set of reals, essentially we are making a number with infinite digits which wasn't in the list before and we are changing infinite digits right? Thereby we are producing a number which wasn't in the countable list of enumerations and thus the original set can't be countable? – Manas Dogra May 10 '21 at 06:08
  • @ManasDogra: Yes, exactly! – Vercassivelaunos May 10 '21 at 07:28

1 Answers1

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how do we know that it isn't one of the $s_j$s for $i=n+1,n+2,...$

Because $s_{n+1},\dots$ is also complemented.

at any step we can construct a rational which is not in that table by choosing a bigger row and or or a bigger column number, so we made a number which wasn't in the list but we don't say then that the set of rationals is uncountable.

I don't think you made a number that wasn't in the list; you just made a number that wasn't in a certain initial subset of the list.

Xodarap
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  • Replying to your 2nd sentence: Why can't we say the same for the reals? Namely, Why can't we say that $s$ is in the full list we are going to make but not in the certain initial subset of the list? – Manas Dogra May 10 '21 at 05:27
  • Replying to your 1st sentence: But $s_{n+1}$ when complemented will also eventually be a part of the enumeration. – Manas Dogra May 10 '21 at 05:30
  • @ManasDogra where in the full list would the number be? The point of Cantor's method is to show it can't be anywhere. – Xodarap May 12 '21 at 06:00