It applies, in the sense that you can carry it out. However, the number you obtain through the process is not a rational (it does not have a periodic decimal expansion).
To be precise, the procedure does not let you guarantee that the number you obtain has a periodic decimal expansion (that is, that it is a rational number), and so you are unable to show that the "diagonal number" is a rational that was not in the original list. In fact, if your original list is given explicitly by some bijection, then one is able to show just as explicitly that the number you obtain is not a rational.
The first thing you need to ask yourself, about finite sets, is this: When do two sets have the same cardinality?
The way mathematics works is to take a property that we know very well, and do our best to extract its abstract properties to describe some sort of general construct which applies in as many cases as possible.
So how do we compare the sizes of two finite sets? If we can write a table, in one column the set $A$ and in the other the set $B$, and each element from $A$ appears in a unique cell; and each element of $B$ appears in a unique cell. If this table have no rows in which there is only one element, then the sets $A$ and $B$ are equal. For example:
$$\begin{array}{lc}
\text{Two equal sets:} &
\begin{array}{c|c|c}A & a_1& a_2\\\hline B & b_1& b_2\end{array} \\
\text{Non-equal sets:} &
\begin{array}{c|c|c|c}A & a_1 & a_2 & a_3\\\hline
B & b_1 & b_2\end{array} &
\end{array}$$
It is clear that this methods captures exactly when two sets have the same size. We don't require one set to be the subset of another; nor we require that they share the same elements. We only require that such table can be constructed.
Well, the generalization is simply to say that there exists a function from $A$ to $B$ which is injective and surjective, namely every element of $A$ has a unique element of $B$ attached to it; and every element of $B$ has a unique element of $A$ attached to it.
It turns out, however, that this notion has a quirky little thing about infinite sets: infinite sets can have proper subsets with the same cardinalities.
Why is this happening? Well, infinity is quite the strange beast. It goes on without an end, and it allows us to "move around" and shift things in a very nice way. For example consider the following table:
$$\begin{array}{c|c|c|c|c|c}
\mathbb N&0&1&\cdots&n&\cdots\\\hline
\mathbb N\setminus\{0\}& 1&2&\cdots & n+1&\cdots
\end{array}$$
It is not hard to see that this table has no incomplete rows and that every element of the left set ($\mathbb N$) appears exactly once, and every element of the right set ($\mathbb N\setminus\{0\}$) appears exactly once!
This can get infinitely more complicated, and so on and so on.
One can ask, maybe we are thinking about it the wrong way? Well, the answer is that it is possible. We can define "size" in other ways. Cardinality is just one way. The problem is that there are certain properties we want the notion of "size" to have. We want this notion to be anti-symmetric and transitive, for example.
Namely, if $A$ is smaller or equal than $B$ and $B$ is smaller or equal in size than $A$, then $A$ and $B$ have the same size; if $B$ also has the same size as $C$ as well, then $A$ and $C$ are of the same size too. It turns out that the notion described by functions has these properties. Other notions may lack one or both. Some notions of "size" lack anti-symmetry, others may lack transitivity.
So it turns out that cardinality is quite useful and it works pretty fine. However it has a peculiarity... well, who hasn't got one these days?
To overcome this, we need to change the way we think a bit: proper subset need not have a strictly smaller cardinality. It just should have not a larger cardinality. This is the right generalization from the finite case, rather than the naive "strict subset implies strictly smaller".
To read more:
- Is there a way to define the "size" of an infinite set that takes into account "intuitive" differences between sets?
Best Answer
When you say "we're not allowed to assume that the mapping from the naturals to the reals is a bijection to begin with", what you're referencing is the nature of the proof by contradiction; we did assume that the mapping was a bijection, and we derived a contradiction by producing a number that was missed by the map. Hence, we proved that no such bijection can possibly exist. In the strictest sense, you're "allowed" to assume a bijection between the naturals and the reals; you'll just find that you can derive a contradiction from that assumption via Cantor's diagonalization argument.
Similarly, you might try and take the same approach of assuming there is a bijection between the natural numbers and the rational numbers. You could try and apply Cantor's diagonalization argument to prove that it can't be surjective, but as your quoted answer explains, this doesn't work. Moreover, a bijection between the natural numbers and rational numbers can, in fact, be constructed. This means that, try as you might, if you do everything correctly, you'll never be able to derive a contradiction from this assumption.