Why is Cantor's diagonal argument valid?

Question

This question is about the validity of the Cantor's diagonal argument itself, not about uncountability of sets or otherwise which can probably be proved in ways I'm unfamiliar with.

It seems, to me, that the diagonal argument assumes its conclusion.

For a diagonal proof to be valid, the diagonal must be a diagonal of a square matrix.

"counterexample" if the matrix is not square:

0 1 0
0 0 0
1 0 0
1 1 1

Cantor's diagonal argument seems to assume the matrix is square, but this assumption seems not to be valid.

The diagonal argument claims construction (of non-existent sequence by flipping diagonal bits). But, at the same time, it non-constructively assumes its starting point of an (implicitly square matrix) enumeration of all infinite sequences of binary digits.

What happens when we actually attempt to construct an enumeration of all infinite sequences of binary digits?

Here is one such construction: Count from $0$ to $\infty$ in binary with least significant bit first (for example: 0, 1, 01, 11, 001, 101, 011, 111, 0001, etc). As you count, put those numbers in a list, filling in extra 0s on most significant side (on the right) as needed

First few iterations of the resulting matrix:

0
0
1
00
10
01
00
10
01
11
000
100
010
110
001
...

We immediately see the consequence of any information encoding, the width of the matrix increases $O(\log_2 n)$, while the height of the matrix increases $O(n)$. Where $n$ is the number of sequences in the matrix.

While $\lim_{n \to \infty} \log_2(n) = \infty$ and $\lim_{n \to \infty} n = \infty$, which make it seem like a square matrix. Another way to frame being a square matrix is to ensure that the ratio of ${width \over height} = 1$.

We notice that $\lim_{n \to \infty} {\log_2(n) \over n} = 0$ and $\lim_{n \to \infty} {n \over \log_2(n)} = \infty$.

Does this not mean that the ratio diverges, demonstrating the enumeration of all infinite sequences of binary digits cannot be a square matrix? Doesn't this, in turn, make the diagonal argument not valid?

sidenote: The above construction, in the limit, when we flip bits along the diagonal, would result in an infinite sequence of 1s, 1111111.... There is nothing about the construction that prevents it from generating an infinite sequence of 1s in the limit, which I think is noteworthy and takes away from what seems like diagonal argument's almost mystical generation of non-included bit sequence along the diagonal when we think only of a random arrangement of sequences instead of ordered one.

Answer 1

This post is meant to supplement Arturo Magidin's excellent post on the other thread. The point is to give you some different perspectives on the diagonal argument. This concept of "square matrix" is something you are pushing on the argument yourself. None of it is to be found within the argument itself.

If $S$ is any set, the set of all subsets of $S$ is $$\mathscr P(S) := \{A \mid A \subseteq S\}$$ This is also called the "power set" of $S$. What I will prove is:

For any set $S$, any function $f : S \to \mathscr P(S)$ is not surjective.

That is, there is always some subset $A$ of $S$ which is not in the image of $f$. In fact, we can just define $$A := \{s \in S\mid s \notin f(s)\}$$

Now for any $s \in S$, there are two possibilities. Either $s \in f(s)$ or $s \notin f(s)$. But

• if $s \in f(s)$, then $s \notin A$, and therefore $f(s) \ne A$.
• if $s \notin f(s)$, then $s \in A$, and therefore $f(s) \ne A$.

Hence $A$ cannot be in the image of $f$.

A consequence of this is that the power set of any set always has strictly greater cardinality. In particular $\mathscr P(\Bbb N)$ is not countable, because to be countable would mean there is a bijection between $\Bbb N$ and $\mathscr P(\Bbb N)$. But bijections are surjections, and by the theorem no such map is surjective.

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So there you go: a proof of uncountable sets that doesn't make use of the diagonal argument. Well... not really. That was the diagonal argument.

Given any $A \subset S$, we can define a function $\mathbf 1_A : S \to \{0,1\}$ by $$\mathbf 1_A(s) := \begin{cases}1&s \in A\\0&s \notin A\end{cases}$$ This is often called the indicator function of $A$. And given any function $g : S \to \{0,1\}$ it is quickly obvious that $g = \mathbf 1_{g^{-1}(1)}$. So you can identify the subsets of $S$ with the functions $S \to \{0,1\}$ and $\mathscr P(S)$ with the set $\{0,1\}^S := \big\{f \mid f : S \to \{0,1\}\big\}$ of all functions from $S$ into $\{0,1\}$.

If you recast my theorem above to be in terms of $\{0,1\}^S$, it becomes

For any set $S$, any function $f : S \to \{0,1\}^S$ is not surjective.

And the proof becomes: Since for each $s \in S, f(s)$ is itself a function defined on $S$, it will be more convenient to denote it by $f_s$ instead, so for $s, t \in S, f_s(t)$ is the image of $t$ under the function $f_s$.

Define $g : S \to \{0, 1\}$ by $$g(s) := \begin{cases}1&f_s(s)=0\\0&f_s(s) = 1\end{cases}$$

For any $s \in S$, either $f_s(s) = 0$, in which case $g(s) = 1$ and therefore $g \ne f_s$, or $f_s(s) = 1$, in which case $g(s) = 0$ and therefore $g \ne f_s$. Thus $g$ is not in the image of $f$, and $f$ cannot be surjective.

This is the same proof Arturo gave for $S = \Bbb N$ in the other thread, and is now recognizable as the diagonal argument.

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But note that in my original proof, there is nothing that looks like a matrix. All I did is define a single set, by applying a well-defined logical condition to the elements of $S$. And the only "squareness" comes from the fact the theorem is about functions from $S \to \mathscr P(S)$.

Further note that this theorem doesn't make any assumptions on $S$. It doesn't even have to be infinite. You can even apply it to $S = \emptyset$ (though you have to delve into technical details of what it means to be a function to fully understand it in that case).

And in addition to showing $\mathscr P(\Bbb N)$ is bigger than $\Bbb N$, it also shows that $\mathscr P(\mathscr P(\Bbb N))$ is bigger than $\mathscr P(\Bbb N)$ and so on.

Cantor's diagonal argument is really useful.