Let us assume that construction of a number, $z$, which cannot be indexed (or counted) in $\mathbb{N}$, proves by contradiction that the cardinality of the set containing $z$ is greater than $|\mathbb{N}|$ . Then for any set $S$ that contains $z$, for which there is no bijection with $\mathbb{N}$, $|S| \gt |\mathbb{N}|$.

→ (This is the basis for Cantor's Diagonal argument, right?)

Now let’s examine a subset of $\mathbb{N}$ in which numbers start with 1, and are non-repeating infinite sequences of integers:

$S =$ { non-repeating infinite integer sequences beginning with number 1 }

→ (the reason i'm starting with 1 is to prevent cases where a majority of the leading digits are 0, and to create a similar situation where we start with 0.xxx)

There are infinitely many non-repeating infinite integer sequences that start with 1, just as there are infinitely many non-repeating infinite integer numbers that start with 2, 3, 4 etc.

→ (isn’t this true?)

Elements in this set $S$ can be represented as $1d_1d_2d_3d_4$…. where the sequence of digits $d_i$ is infinite and non-repeating.

An infinite subset of $\mathbb{N}$, here called $S$, and the parent set $\mathbb{N}$ have the same cardinality (e.g. odd numbers and even numbers), so there should exist a bijection between $S$ and $\mathbb{N}$.

→ (isn’t this true?)

The elements of $S$ can be listed as $s_1$, $s_2$, $s_3$… and while we cannot tell exactly which number from $S$ got to be the first one, or the second... we do know that every element in $S$ must get an index from $\mathbb{N}$

Let’s construct the number $z$ that will be in $S$, but will differ from each and every number that has received an index. The number $z$ will have the form $1e_1e_2e_3e_4$… where the $i$th digit past the leading 1 in $z$ will differ from the $i$th digit of $s_i$. We can do this by setting $e_i = s_{ii}+1 (mod 10)$ where $s_{ii}$ is the $i$th digit of $s_i$.

Now, $z$ is in $S$, but it will not receive an index in $\mathbb{N}$. If it did get one, say $k$, then we have $z = s_k$, but the $k$th digit of $z$ is, by construction, different from the $k$th digit of $s_k$, which is a contradiction. Hence our assumption is incorrect. (here's the core of the Diagonal argument)

This means that either…

a) Cardinality of a subset of $\mathbb{N}$ (here $S$) is somehow larger than $|\mathbb{N}|$?

b) The initial assumption that construction of $z$ proves $S$ has a greater cardinality than $\mathbb{N}$ is incorrect?

c) I’m missing something else…

  • $\begingroup$ math.stackexchange.com/questions/263677/… this should answer you question $\endgroup$ – B.Swan Feb 16 '17 at 16:29
  • $\begingroup$ I'm not sure how that answers the question. It does confirm my suspicion that S and N should have the same cardinality, but I'm not sure what else I'm doing wrong here. $\endgroup$ – tmsimont Feb 16 '17 at 16:39
  • $\begingroup$ Why do you say your set $S$ is a subset of $\mathbb{N}$? The elements of $S$ are not natural numbers... $\endgroup$ – Eric Wofsey Feb 16 '17 at 17:00
  • $\begingroup$ @EricWofsey That's where I'm thinking my flaw is here, but I'm not understanding why my set $S$ isn't a subset of $\mathbb{N}$ -- can't you construct an element of $\mathbb{N}$ that is an infinite non-repeating sequence of digits? Or, if you were to describe such a sequence, would it not be an element of $\mathbb{N}$? $\endgroup$ – tmsimont Feb 16 '17 at 17:02
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    $\begingroup$ "can't you construct an element of N that is an infinite non-repeating sequence of digits?" No. Elements of N are not infinite non-repeating sequences of digits. $\endgroup$ – Did Feb 16 '17 at 19:46

Your argument is wrong at many levels :

Your first conclusion that any set $S$ containing $z$ has $|S| > |\mathbb{N}|$ is false. What Cantor's diagonal argument shows is that if I fix any function from $\mathbb{N}\to \mathbb{R}$, we cand exhibit $z$ that's not in the image of the function, i.e. said function isn't injective. But by no means does that imply what you said.

Moreover, the set $S$ you construct afterwards is not a subset of $\mathbb{N}$, why would it be ? "Integers are finite"...

  • $\begingroup$ Wait are you saying integers are finite? I'm not sure I understand what you're saying since you've got quotes around that part of your response $\endgroup$ – tmsimont Feb 16 '17 at 16:33
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    $\begingroup$ What he means is that every integer has a finite number of digits, not that the set of all integers is finite. $\endgroup$ – user247327 Feb 16 '17 at 16:42
  • $\begingroup$ So that means the core of my problem here is that there is no way to have an integer that is an infinite non-repeating sequence of digits? If the integer set is infinite, why is that the case? $\endgroup$ – tmsimont Feb 16 '17 at 16:45
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    $\begingroup$ An integer is a "counting number", it's $0, 1, 2,...$ and their friends. I don't see how it could be counterintuitive that integers are "finite". You can design a new set of infinite sequences of integers starting with a $1$, and call it $S$, but it will not be a set of integers. There is an infinite set of integers starting with a $1$ (there's $1, 10, 100, 1000, 10000,...$ and so on and many more), but they're all finite. Here I'm just trying to make you "feel" why, but that's not how maths is done. In maths, you prove that integers have a finite number of digits $\endgroup$ – Max Feb 16 '17 at 18:00
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    $\begingroup$ To make it clear : integers aren't just some digits that follow one another : $54$ isn't just a $5$ and a $4$. Usually, integers are defined as finite ordinals, and then you prove their properties (such as $x+y = y+x$). One of these properties (that you can prove and not guess or intuite ) is that any integer $n$ can be uniquely written as $\displaystyle\sum_{i=0}^d a_i 10^i$ with $a_i\in \{0,...,9\}$. This writing is unique and finite. So anything that you can conceive of that is an infinite string of numbers is not an integer $\endgroup$ – Max Feb 16 '17 at 18:03

Your first paragraph is a misinterpretation (I may edit this answer to point out later errors as I look over the question again).

The basis of Cantor's Diagonal argument is that if given an arbitrary indexing (which is really a map $\mathbb{N}\to S$) you can always construct $z\in S$ not indexed (that is the map is not surjective) then $|S|>|\mathbb{N}|$. So it's not that any set containing $z$ has larger cardinality than $\mathbb{N}$, rather if for any indexing you can find $z\in S$ not indexed then $|S|>|\mathbb{N}|$.

  • $\begingroup$ I would like to point out that my answer is a rephrasing of the answer by @Max, who got there first. $\endgroup$ – Robert Chamberlain Feb 16 '17 at 16:54
  • $\begingroup$ Yes I see now that my first paragraph is a mangling and over simplification of Cantor's Diagonal argument, but the rest of my question doesn't directly relate to that and is instead just a re-construction of the diagonal argument on a subset of $\mathbb{N}$ -- I think the real problem I'm seeing now is that my $S$ may not be an actual subset of $\mathbb{N}$, but I'm not sure why. $\endgroup$ – tmsimont Feb 16 '17 at 17:05
  • $\begingroup$ the digits of $\pi$ for example don't repeat $\endgroup$ – Robert Chamberlain Feb 16 '17 at 17:19
  • $\begingroup$ Right -- it just seems odd to me than in an infinite set of integers, meaning there is no bound, no end... that we'd not be able to have all of the digits of $pi$ represented as an integer, without a decimal point. If we're saying that's not possible, we're basically putting a limit on the so-called "infinite" set of integers, which seems contradictory to me. $\endgroup$ – tmsimont Feb 16 '17 at 17:33
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    $\begingroup$ I guess that's the point, the set of real numbers are 'more infinite' than the set of natural numbers. It can be very counter intuitive! $\endgroup$ – Robert Chamberlain Feb 16 '17 at 17:55

There is no element "which cannot be indexed" in the absolute.

The Cantor argument shows that for a given indexing, there are always missed elements, but those that are missed depend on the particular indexing.

On the opposite, having an element which is "distinguished" allows you to count it in addition to all the others (by shifting all indexes).


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