Two questions re. Cantor's diagonalization argument

Question

I was watching a YouTube video on Banach-Tarski, which has a preamble section about Cantor's diagonalization argument and Hilbert's Hotel. My question is about this preamble material.

At c. 04:30 ff., the author presents Cantor's argument as follows. Consider numbering off the natural numbers with real numbers in $\left(0,1\right)$, e.g.

$$ \begin{array}{c|lcr} n \\ \hline 1 & 0.\color{red}50321642239817 \ldots \\ 2 & 0.0\color{red}7829136011205 \ldots \\ 3 & 0.31\color{red}11370055629 \ldots \\ 4 & 0.999\color{red}9261457682 \ldots \\ 5 & 0.0001\color{red}042507334 \ldots \\ \vdots & \vdots \end{array} $$

Then you could form a new real in $\left(0,1\right)$ not already in the list, e.g. $0.\color{red}{68281} \ldots$. Hence there are more reals than naturals.

I have two questions about this:

1. Couldn't you run exactly the same argument (erroneously) for rational numbers in $\left(0,1\right)$? E.g. say I choose powers of $\frac{1}{2}$, giving:

$$ \begin{array}{c|lcr} n \\ \hline 1 & 0.\color{red}4999999999999 \ldots \\ 2 & 0.2\color{red}499999999999 \ldots \\ 3 & 0.12\color{red}49999999999 \ldots \\ 4 & 0.062\color{red}4999999999 \ldots \\ 5 & 0.0312\color{red}499999999 \ldots \\ \vdots & \vdots \end{array} $$

So $0.\color{red}{55555} \ldots$ is not in the list, suggesting that the cardinality of the rationals is greater than that of the naturals.

But a different argument shows that their cardinalities are the same. So there seems to be something wrong with the diagonal argument itself?

2. As a separate objection, going back to the original example, couldn't the new, diagonalized entry, $0.68281 \ldots$, be treated as a new "guest" in Hilbert's Hotel, as the author later puts it (c. 06:50 ff.), and all entries in column 2 moved down one row, creating room?
   • Admittedly, you could diagonalize this expanded list again; but then you could also move the guests down again. So the argument does not seem to show that there's any fundamental problem, i.e. that you can't continue pairing off the reals with the naturals forever?

Answer 1

For your first question, of course you can construct such a number. The only problem is that it will surely be irrational. Your example is flawed since you are enumerating some rationals. You have to enumerate all of them in order to have a contradiction. What your argument proves is that $0.5555\dots$ is not of the same form as the other numbers you chose to enumerate.

For your second question, the problem is that you assumed in the beginning that you have enumerated all the reals of $(0,1)$. Since you found a real number in $(0,1)$ that is not present in the enumeration then you have a contradiction.

Answer 2

Re: your first objection, you need to show that every (not just some) list of rationals fails to enumerate all of $\mathbb{Q}$. This means that the fact that some lists of rationals yield "antidiagonals" which are rational is irrelevant; you need to argue that every list of rationals yields a rational "antidiagonal." And you can't do this.

(This is usually ignored in the usual diagonal argument since obviously every infinite decimal expansion corresponds to a real, but it really should be stated explicitly - precisely because it will then pre-empt this particular objection.)

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Re: your second objection, consider the following dialogue. "There is no largest natural number, since for any number $n$ I can consider $n+1$." "But that $n+1$ can be taken as the new largest number. Admittedly, you could pick something bigger than that again, but then you could also move on to that bigger number again. So this argument does not seem to show that there's any fundamental problem." This has exactly the same shape as your argument, but it's easier to see that it's flawed.

When I claim "There is no $x$ with property $Y$," all I have to show is that each possible $x$ fails to have property $Y$. It doesn't matter that for every $x$ there is an $x'$ which is "more $Y$-ish than $x$" - all that matters is that no $x$ genuinely has property $Y$.

Answer 3

The crucial point of the argument (which is not explained very well by the video you watched) is that the diagonalization argument applies to any way of numbering real numbers (with natural numbers) at all. Now if the real numbers were countable, that would mean there exists some particular way of numbering them that includes all of them. Then you apply the diagonalization argument to that particular numbering and obtain a real number that is actually not on the list. This is a contradiction, since the list was supposed to contain all the real numbers.

In other words, the point is not just that some list of real numbers is incomplete, but every list of real numbers is incomplete. As you saw with your example with rational numbers, just being able to write down some infinite list that does not include all of them does not prove the rational numbers are uncountable. But if you can apply the argument to any list at all, that proves that no list can have all the real numbers, so the real numbers cannot be countable. (You might then ask, why can't you apply the same argument to show that no list of rational numbers contains all of them? The issue is that the diagonal number you're constructing that is not on the list may not be rational, since its decimal expansion may not be periodic. So the crucial property of real numbers we're using is that any infinite sequence of digits can be the decimal expansion of a real number.)

Answer 4

In your first objection you don't have a list of all rationals. You have a list of all powers of $\frac 12$. And you showed that $0.55555555 ..... = \frac 59$ is not on the list. So either 1) Your list is not a complete list of powers of $\frac 12$ or 2) The cardinalities of powers of $\frac 12$ is more than the cardinalities of the naturals, or 3) $\frac 59$ is not a power of $\frac 12$.

And as $\frac 59$ is not a power of $\frac 12$ this is not a proof that the powers of $\frac 12$ is a higher cardinality.

Or if you want to make this about the cardinality of the rationals we can conclude 1) Your list is not a complete list of the rationals or 2) the cardinalities of the rationals is more than the cardinalities of the naturals or 3) $\frac 59$ is not rational.

But as your list is not complete we still have no contradiction.

The trick to this proof needs to be that no list is complete.

Now lets say I did a fairly standard bijection from $\mathbb N$ to $\mathbb Q\cap (0,1)$ of $1\frac 12, \frac 13, \frac 23, \frac 14, \frac 34, \frac 15, \frac 25, \frac 35, \frac 45, \frac 16,.....$

So my list is

$1: 0.50000000.......$
$2: 0.33333333.......$
$3: 0.66666666.......$
$4: 0.25000000.......$
$5: 0.75000000.......$
$6: 0.20000000.......$
$7: 0.40000000.......$

And I do my change

$1: 0.\color{red}60000000.......$
$2: 0.3\color{red}4333333.......$
$3: 0.66\color{red}766666.......$
$4: 0.250\color{red}10000.......$
$5: 0.7500\color{red}1000.......$
$6: 0.20000\color{red}100.......$
$7: 0.400000\color{red}10.......$

The problem is we have no reason to assume my result $0.6471111....$ will be rational. And indeed because we know my list is complete and because we know the cardinality of the rationals is equal to the naturals, we know that our created number won't be rational.

And we can think it out. If my manufactured number were rational it would have a repeating pattern. Which means my list were made is such a way that that the $n$th digit of the $n$th term is never in the pattern. (If it were I'd have to change it. And that wouldn't result in my number.) That's a huge restriction on my list. Now admittedly it isn't obvious why such a complete list is impossible but it isn't obvious either how I would make so a list so that for no possible pattern no $n$th term will have an $n$th digit in this pattern. But apparently it is impossible.

The thing for the reals is that for any list we can makes such a decimal. And such a decimal will be a real between $0$ and $1$ (that's all we can determine about the resulting number... we don't know if it will be rational (although it probably won't be... although it could be depending on which that particular list is). So we can conclude this list is not complete.

But here's the rub. NO list can possibly be complete. And if no list can be complete a bijection from the naturals (another term of "complete list") can exist. And that means the cardinality is greater (another way of saying "no complete list can exist").

....

Admittedly, you could diagonalize this expanded list again; but then you could also move the guests down again. So the argument does not seem to show that there's any fundamental problem, i.e. that you can't continue pairing off the reals with the naturals forever?

That is a problem. That shows we will always have to expand your list and your list will never be complete. If your task is to STOP, and you fail to do it you fail to do it. To say "well, we didn't stop now, but that's okay we didn't have to stop at $k$ steps and failing to stop at $k$ steps doesn't mean we failed to stop at any step, and we can fail to fail forever, forever" is not valid. Failing to fail forever, forever is still not succeeding. End we did show you can never succeed to STOP.

Answer 5

1. If the list contains all rationals, it does contain $0.555\cdots$. If not, well you are just saying that the list of powers of $\frac12$ does not contain $0.555\cdots$.

2. Yes, you can continue forever, meaning that you'll never be done.