Why does Cantor's diagonal argument prove uncountability?

Question

Having searched a lot about the topic, including this answer, and related to my previous question, I still do not understand, why (or how) does diagonal argument proves uncountability, and will try now to make a more formal description, why I think it does not. Since that previous question was mainly about another thing, I think it is better to create a separate question.

As a side note: I don't pretend that real numbers in particular are countable, I just don't see how the diagonal argument proves that.

Intuitive considerations

Most of the explanations, including the one I linked, appeal to the fact that we have a given list:

First, you should understand that the diagonal argument is applied to a given list. You already have all of s1, s2, s3, etc., in front of you.

But does not it already mean that we operate with a finite list? And what we really show (as I see it), is that a finite sub-set of an infinite set does not contain all the elements. Which is obvious, I think.

More formal

First, we consider we have a set $T$ of all inifinite sequences of either binary digits or elements of another finite set. We also claim there exists an enumeration for this set: $$S_1 = (a_{11}, a_{12}, a_{13}, ...)$$ $$S_2 = (a_{21}, a_{22}, a_{23}, ...)$$ $$...$$

Then we construct a sequence $s = (d_1, d_2, d_3, ...)$ such that $d_i \neq a_{ii}$. By construction, $s$ is in $T$. But what is then stated, is that $s$ differs from each $S_n$ and thus is not in the enumeration. And here comes the main point from my view. $s$ just differs from each $S_n$ in a finite enumeration, but not in the whole enumration.

In other words, though $d_i \neq a_{ii}$, it can still be equal to some $a_{i+k,i}$. Obviously, there are still elements exist in the list after $S_i$, cause otherwise, if $S_i$ was the last, our list would be finite, which would contradict to our initial consideration.

So, till now, we only proved that there exists an element of $T$ which is not in $(S_1, S_2, ..., S_i)$, and that is a finite subset of $T$. In other words, none of the finite subsets contain all the elements of an infinite set. But we still haven't come to any contradictions with our initial consideration.

Are there any flaws in these considerations?

Answer 1

I think you are missing the fact that the sequence $(S_1,S_2,S_3,...)$ is not finite, and no one said so. That's where your argument fails.

I will write what is the, in my opinion, easier version of Cantor's diagonal proof:

(*) There isn't any bijection $f:\mathbb{N} \longrightarrow \{0,1\} ^{\mathbb{N}}$

And the proof is that, if there was (let's call it $\varphi$), then $\varphi[\mathbb{N}]=\{ 0,1\} ^{\mathbb{N}}$, i.e., for every element $a=(a_n)_{n\in \mathbb{N}}\in\{ 0,1\} ^{\mathbb{N}}$ (the set of all sequences of $0$'s and $1$'s), i could find a $k\in\mathbb{N}$ such that $\varphi (k)=a=(a_n)_{n\in \mathbb{N}}$. But that's not true, cause i can construct the sequence $(b_n)_{n\in\mathbb{N}}:=(\:\overline{\varphi (n)_n}\:)_{n\in \mathbb{N}}$ where $\overline{0}=1$ and $\overline{1}=0$ [remember that $\varphi (n)$ is the $n$-th sequence of $0$'s and $1$'s and $\varphi (n)_n$ its $n$-th element]. And, as you may guessed, it differs from any of the $\varphi (k)$'s for $k\in\mathbb{N}$ in the $k$-th element of the sequence. That means that none of them is identical to the one we constructed, therefore, we found one that wasn't listed (two sequences are identical iff they are identical in every one of their components).

Why is that a proof that there isn't any bijection $f$ between $\mathbb{N}$ and $\{ 0,1 \}^\mathbb{N}$? Cause anytime you asume that such a bijection, this proves that it is in fact not surjective (it could be injective tho), cause you can find a counter-example (an element that is not in the range of $f$). And it is easy to extend this proof to any set in replacement of $\{ 0,1\} ^\mathbb{N}$.

I hope it helped you.

Answer 2

I think what you are missing is expressed when you said:

We also claim there exists an enumeration for this set ...

And then asked:

But does not it already mean that we operate with a finite list?

No, it does not. "An enumeration" means you associate an element of T with every natural number in N. That is, you have an infinite list.

There are also a couple of other subtle differences between what you said, and what Wikipedia says (which seems to be where you get your notation). "Claim there exists an enumeration" means you are dealing with a hypothetical list - which will be important in a moment - but Wikipedia's "any enumeration of elements from" means they want any (infinite) list that actually exists. And we don't assume that it does, or does not, include every element of T.

So, for example, the element for natural number n might be all "0"s, except for a "1" in position n.

The diagonal argument, by itself, does not prove that set T is uncountable. It comes close, but we need one further step. What it proves is that for any (infinite) enumeration that does actually exist, there is an element of T that is not enumerated. Note that this is not a proof-by-contradiction, which is often claimed.

The next step, however, is a proof-by-contradiction. What if a hypothetical list could enumerate every element? Then we'd have a paradox: The diagonal argument would produce an element that is not in this infinite list, but "enumerates every element" says it is in the list. It isn't this list that is a problem, it is the element of T that we could construct if the list were possible. Since that paradox can't be possible, the hypothetical list that would produce is is impossible.

Answer 3

Two sequences $(a_1,a_2,a_3,\ldots,a_n,\ldots)$ and $(b_1,b_2,b_3,\ldots,b_n,\ldots)$ are equal if and only if for every natural number $k$, $a_k=b_k$.

Two sequences $(a_1,a_2,a_3,\ldots,a_n,\ldots)$ and $(b_1,b_2,b_3,\ldots,b_n,\ldots)$ are not equal if and only if there exists an integer $k$ such that $a_k\neq b_k$.

If we have a list-of-sequences, $$S_1,S_2,S_3,\ldots,S_n,\ldots$$ with $$S_i = (s_{i1}, s_{i2},\ldots,s_{in},\ldots)$$ then a sequence $D=(d_1,d_2,\ldots,d_n,\ldots)$ is in the list if and only if there exists a natural number $i$ such that $D=S_i$. By the definition of when two sequences are equal, this happens if and only if

There exists a natural number $i$ such that for every natural number $j$ we have $d_j = s_{ij}.$

The negation, then is:

The sequence $D$ is not in the list if and only if

for every natural number $i$ there exists a natural number $j$ (which may depend on $i$) such that $d_j\neq s_{ij}$.

Given any list of sequences $S_1,S_2,\ldots, S_n,\ldots$, which we can think of as a function $f$ from the natural numbers to the set of all (binary) sequences, Cantor's Diagonal Argument constructs a list $$D_f=(d_1,d_2,d_3,\ldots,d_n,\ldots)$$ (which depends on the function $f$; that is, on the precise list given) with the highlighted property:

For every natural number $i$ there exists a natural number $j$ (namely, $j=i$) such that $d_j \neq s_{ij}$.

This proves that $D_f$ is not in the list given by $f$.

Thus, Cantor shows that:

Given any list $f\colon\mathbb{N}\to X$, where $X$ is the set of binary sequences, there is an element of $X$ that is not in the image of $f$. In particular, $f$ is not surjective.

Sure, the argument shows that $d_i\neq s_{ii}$, and this does not preclude that $d_i=s_{i+k,i}$ for some $k$... but it doesn't matter whether that happens, because $D_f$ is also not equal to $S_{i+k}$. And we know it is not equal to $S_{i+k}$ because $D_f$ differs from $S_{i+k}$ in the $i+k$-th position. And differing in a single position is sufficient for $D_f$ to not be equal to $S_{i+k}$.

Think about the sequence where $S_i=(0,0,0,0,\ldots)$ for all odd $i$, and $S_i = (1,1,1,1,\ldots)$ for all even $i$. Given any sequence $D=(d_1,d_2,\ldots)$ and any index $i$, either $d_i=s_{ii}$ or $d_i=s_{i+1,i}$ (since $d_i$ is either $0$ or $1$). Does that mean that a sequence that alternates the sequence of all $0$s and the sequence of all $1$s lists all the binary sequences? Of course not. But it would satisfy your "objection"/confusion about the diagonal sequence. So this should show you that the possibility that $d_i$ equals $s_{i+k,i}$ for some $k$ is really irrelevant. We are just checking that $D_f$ is not equal to $S_i$, and we can verify that by noting the two sequences disagree in the $i$th position. We also know $D_f$ is not equal to $S_{i+7}$ because $D_f$ differs from $S_{i+7}$ in the $(i+7)$th position. And $D_f$ is not equal to $S_{i+89179}$ because $D_f$ and $S_{i+89179}$ differ in the $(i+89179)$th position. And $D_f$ is not equal to any $S_k$, because it differs from $S_k$ in the $k$th position.