Issue with Cantor's Diagonal Argument and square matrices

Question

I have an issue with Cantor's Diagonal Argument.

We suppose we have an infinite list with every real in $(0, 1)$ listed. Then we take the diagonal and change every digit in a predefined way (e.g. $+1 \pmod {10}$). And we get a new real number, not included in the list. This is supposed to prove that reals are not countable.

The problem I see is that, to apply the argument, we must have every real in a square matrix of digits (I use the matrix just as a form of representation). And we can't list every real with a square matrix. For a fixed precision, we need a matrix of $N$ columns (digits) by ($10^N$) rows (for any $N$). I understand the Cantor list would have infinite precision, but we still need $N\times(10^n)$, and we are still construing a list of $N\times N$ digits ($N$ being infinite).

So, it seems to me that the Cantor's Diagonal Argument only proves that we can't list all reals in a square matrix.

What am I missing?

Answer 1

The idea of a "list" here being a matrix is taking the diagonal argument too literally. The actual argument does not involve doing anything to a physical list of numbers (which you describe as a matrix). It instead involves showing that there cannot exist any one-to-one correspondence (or bijection) between the reals and the natural numbers. The diagonal argument is a way of visualizing the proof, but the underlying nature of the argument has nothing to do with any list of fixed, finite size. These are infinite lists (technically, infinite sequences), and the ideas of finite precision do not apply to them.

For instance, what does it even mean to have a "square" matrix, when your matrix is of infinite size? In the finite sense we understand it (if a matrix is $m\times m$ where $m$ is a natural number, it is square), but we don't have a rigorous method for talking about matrix dimensions when the matrices are infinite. Sure, one could define what that means, but it is neither necessary nor helpful in formulating the diagonal argument.

We often use the visual aide of an infinite list of reals, each real with infinite precision, but that is not necessary for the actual rigorous proof. Here is the proof without that visual aide (glossing over many details).

Suppose the reals are countable. Then there exists a one-to-one correspondence (bijection) between the reals and the natural numbers. Thus we must be able to make an infinite sequence (with a first, second, third, millionth, etc, number) of real numbers in which each real appears at least once. Consider one such sequence $\{s_n\}$. Now we are going to construct a new number which does not appear $\{s_n\}$. We take the first digit of $s_1$ and change it to $s_1+1$ modulo 10, and make that our first digit of the new number. We do the same for the second, third, and all other digits. We now have a number not appearing in the sequence $\{s_n\}$ we had before, because it differs from every term of the sequence in at least one decimal place. It must be a real number, in fact, since it is defined as and represented by an infinite decimal. Thus, our initial sequence did not actually have all the reals appear at least once, so we have a contradiction. Hence, the reals are uncountable.

Note that the argument never makes use of a matrix in any way. The fundamental idea of the proof lies in the use of infinite sequences, which we do have a rigorous grasp on. If you'd like more understanding of the details by which we construct ideas like sequences, it's turtles all the way down in that it comes back to set theory. However, I hope this gave the gist of the proof in a more understandable way, and showed that it does not in any way rely on a physical representation (such as a matrix) of the infinite sequence being dealt with.

(This got much longer than I expected, cheers!)

Answer 2

I would say that you are missing nothing. The problem is that you are adding one thing, namely a matrix. That's not part of the diagonal argument.

Answer 3

You don't need a column for each value of digit, just for each digit.

And you don't construct a list of length $\Bbb{N\times N}$. That is just the matrix itself, all the real numbers and their decimal expansions. We start with a list of length $\Bbb N$, and each one of these has a decimal expansion of length $\Bbb N$, and so each row in the matrix is actually just this decimal expansion of one of the real numbers on the list.

Next, we go over the diagonal of this matrix, i.e. nodes indexed by $(n,n)$, and use the digits that appear on that diagonal to generate a new real number which you can now prove is not any row (or column for that matter) of this matrix.

And if each row is a real number on your original list, then it means your original list is not the entire interval.

Answer 4

It is best not to think of matrices but to use the precise definition of countability.

A set $S$ is countable if there is a bijective function

$$f: \mathbb{N} \rightarrow S$$

A function is a bijection if it is 1-1 and surjective.

The proof of uncountability of the real numbers relies on the use of decimal expansions, and in general there may be more than one decimal expansion for a real number. In the following I will therefore assume that we have chosen a unique decimal representation for each real number (it does not matter which one). Suppose we had such a bijective function

$$f: \mathbb{N} \rightarrow \mathbb{R}$$

But then we could define a real number $d$ that is not in the range of $f$ by defining its decimal expansion in binary as follows. Let $x_i$ be the $i$th digit in the decimal expansion of real number $i$. Then the $i$th digit of $d$ is given by

$$d_i = \begin{cases} 0 & \text{if}\; r_i = 1 \; \text{where}\; r_i = f(i) \\ 1 & \text{if}\; r_i = 0 \; \text{where}\; r_i = f(i) \end{cases}$$

However, $d$ is by definition different from every real number for at least one decimal, so $d$ cannot be in the range of $f$. Therefore $f$ cannot be onto.

The "matrix analogy" is simply used to represent the function $f$ as an infinite table using an entry to hold each decimal for any given real number in the range of $f$; entry $(i,j)$ in the table contains the $j$th decimal of $f(r_i)$.

Answer 5

It should be pointed out that something implicit is happening in this Cantor Diagonal Construction argument (see MathTrain's answer). Recall,

A decimal representation of a non-negative real number $r$ is an expression in the form of a series, traditionally written as a sum ${\displaystyle r=\sum _{i=0}^{\infty }{\frac {a_{i}}{10^{i}}}}$ where $a_0$ is a nonnegative integer, and $a_1, a_2, ...$ are integers satisfying $0 \le a_i \le 9$, called the digits of the decimal representation.

The representation is unique up to the fact that if the ${a_i}^{'}s$ trail off in $9^{'}$s it can be simplified.

This representation is really a statement on the limit of a sequence (convergence),

$\tag 1 r=\lim _{n\to \infty }\sum _{i=0}^{n}{\frac {a_{i}}{10^{i}}}$

The Cantor argument is constructing the $a_i$ for (1) one step at a time, so that for any $N$, none of the $N + 1$ finite expansion sums

$\tag 2 s_n = \sum _{i=0}^{n}{\frac {a_{i}}{10^{i}}} \; n \le N$

can be equal to any of the first $N+1$ 'matrix rows', and that also holds true as the limits are taken.

So there are assumptions being made during the Cantor construction, the underlying facts about decimal representations of real numbers.

Finally, although it can be handled, that little detail about trailing $9^{'}$s needs to be addressed in a rigorous proof. You might want to skip that and study a Cantor proof showing that the binary sequences, $2^{\mathbb{N}}$, is uncountable (and more).

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If you look at related material in math.stackexchange, for some inexplicable reason, many people have questions about Cantor's Diagonal Argument. So, as this is the most recent instance, permit me the liberty of simply giving another proof that the real numbers are uncountable.

Recall the following property of real numbers,

Theorem 1: Let $(s_n)$ be a sequence of real numbers satisfying the following conditions,

$\quad s_0 \lt s_2 \lt s_4 \lt \cdots$
$\quad s_1 \gt s_3 \gt s_5 \gt \cdots$
$\quad \text{For every even } i \in \mathbb{N}\text{, } \; s_i \lt s_{i+1}$
$\quad \text{The alternating sequence } s_{k+1} - s_k \text{ converges to } 0$

Then $(s_n)$ converges to the limit $s= \text{LUB} \, \{s_i \, |\; i \text{ is even} \} = \text{GLB} \, \{s_i \, |\; i \text{ is odd} \}$,
and for every $n \in \mathbb{N}$, $s_n \ne s$.

Theorem 2: The set of all real numbers is an uncountable set.
Proof
To arrive at a contradiction, let $f: \mathbb{N} \to \mathbb{R}$ be a surjective mapping. We claim that we can construct a sequence $(s_n)$ satisfying the conditions of theorem 1 with a limit that can't be in the range of $f$, giving the contradiction.

Set $s_0 = f(0)$.
Let $m_1$ be the smallest integer satisfying $f(m_1) \gt s_0$, and set $s_1 = f(m_1)$.
Let $m_2$ be the smallest integer satisfying $f(m_2) \gt s_0$ and $f(m_2) \lt s_1$, and set $s_2 = f(m_2)$.
Let $m_3$ be the smallest integer satisfying $f(m_3) \lt s_1$ and $f(m_3) \gt s_2$, and set $s_3 = f(m_3)$.
etc.

Note that the increasing sequence $(m_j)$ of integers must be unbounded.

It is not difficult to show that $(s_n)$ satisfies the conditions of theorem 1; let $\alpha$ denote this limit. Observe that $f$ can't take on the value of $\alpha$, for if it did, $\alpha$ would have to appear as a term in the constructed $(s_n)$ sequence, since it is inside the alternating $(s_n$) 'clamp-down', and we keep going further out along the domain of $f$ to construct this converging sequence..
$\blacksquare$

Note: In the proof of theorem 2 we only used the fact that the range of $f$ is dense in $\mathbb{R}$, and from that constructed a number $\alpha$ that $f$ can't assume as an output value.

Answer 6

"We suppose we have an infinite list with every real in (0,1) listed. ... The problem I see is that, to apply the argument, we must have every real in a square matrix of digits." This is a common objection, that is a result of how the proof is taught incorrectly.

First, Cantor did not apply the proof to real numbers; intentionally and specifically. He applied it to infinite-length strings of the two characters "m" and "w." This is a minor point, since we can (and I will) use "0" and "1" instead. Note that such a string is just a function that maps N to the set {'0','1'}, written f:N->{'0','1'}.

Yes, every real number between 0 and 1 can be represented in binary by such a string. But some can be represented by two, which is a problem. It is easily resolved, but unnecessary to do so if we follow what Cantor actually said.

Second, he never assumed there was an infinite list of all such strings, only that there was an infinite list of some such strings. Objections like yours naturally arise when you try to imagine how to physically accomplish what Cantor claimed proved was impossible. You can't, so you quite naturally try to find the flaw.

Lastly, it isn't a proof by contradiction. To be one, you have to show the opposite of a known truth follows from all of what you assume. Not that one thing you assume contradicts something else you assume. So it is not a valid proof-by-contradiction to assume "all" and then derive "not all" from another assumption.

But it is still a valid proof. What Cantor assumed was that there is a set T comprising all of his strings, and that a subset S of it could be generated by the function s:N->S. The diagonalization method is another function d:N->{0,1}, that maps every n to the opposite of nth character of s(n). There is no square matrix here, just a definition of one of Cantor's strings as a function.

What this directly proves, is "If S is a countable subset of T, then S is not all of T." By contraposition ("If A then B" implies "If ~B then ~A"), not contradiction, this proves "If S is all of T, then S is not countable."