Two questions re. Cantor's diagonalization argument 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:

*

*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?


*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?



 A: 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.
A: 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.)

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 $Y$.
A: 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.)
A: *

*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 $0.555\cdots$.


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