Infinitely many elements between two Dedekind's cuts I have a problem with an informal argument I found in Pugh's book on real analysis (p.13) concerning Dedekind's cuts (notice that, in a sense, I am extrapolating from the text a formal argument).

Let $x = A|B$ and $y=C|D$ be two Dedekind's cuts, with $x < y$. Hence, $A \subset C$, with $A \neq C$.
"Proposition:" If $x < y$, then $C \setminus A$ is nonempty, and there are infinitely many elements in it.
Proof: From $x < y$ there is a $c_0 \in C \setminus A$. Since the $A$-set of a cut contains no largest element (Pugh is most probably referring to the "left" set of a cut, hence here it is $C$), there is a $c_1 \in C$ such that $c_0 < c_1$, and all the $c \in \mathbb{Q}$ such that $c_0 \leq c \leq c_1$ are in $C \setminus A$.

Thus, my problem is with the fact that here we are dealing with uncountably infinite elements (even if Pugh does not talk about different infinity magnitude), but I do not know how the uncountability pops up, considering the reference is to the rationals $c \in \mathbb{Q}$.
I do see there is another layer of infinity, namely that there is a chain $c_0 < c_1 < c_2 < \dots$ that goes on. But still, I do not find how to make the jump to the uncountability.
Any help is greatly appreciated.
Thank you for your time.
 A: If I understand correctly, you are asking how to prove that if $x<y$ are Dedekind cuts, then there are uncountably many Dedekind cuts $z$ with $x<z<y$ - the issue being that Dedekind cuts seem to come from rationals, and there are only countably many of those.
The point is that Dedekind cuts don't come from rationals, but rather sets of rationals, and there are uncountably many of these. The proof I give below is essentially Cantor's original proof of the uncountability of the reals:

Suppose there were only countably many Dedekind cuts between $x$ and $y$. List them as $$\{(E_n\vert F_n): n\in\mathbb{N}\},$$ and say that a rational interval $(p, q)$ is $n$-good if


*

*$x<p<q<y$ (so $(p, q)$ is a nonempty subset of the region we're looking at),


and


*

*$(p, q)\subseteq E_n$ or $(p, q)\subseteq F_n$ (so the $n$th Dedekind cut doesn't "split" this interval).


The key point is the following:

Exercise: For every nonempty rational interval $(p, q)$ and every $n$, there is a subinterval $(p', q')\subseteq (p, q)$ which is $n$-good.

So we can build a sequence of rational intervals $(p_n, q_n)$ such that


*

*$(p_0, q_0)\supseteq (p_1, q_1)\supseteq ... \supseteq (p_n, q_n)\supseteq (p_{n+1}, q_{n+1})\supseteq . . .$ (they're getting smaller), and

*Each $(p_n, q_n)$ is $n$-good.
Now let $G_0=\{r\in\mathbb{Q}: \exists n(r<p_n)\}$, $H_0=\{r\in\mathbb{Q}: \exists n(q_n<r)\}$. Every element of $H_0$ is bigger than every element of $G_0$, $G_0$ has no maximal element, and $H_0$ has no minimal element;so we can find a Dedekind cut $(G\vert H)$ with $G_0\subseteq G$ and $H_0\subseteq H$.
Now: Can $(G\vert H)=(E_n\vert F_n)$ for any $n$? HINT: $(G\vert H)$ lies inside the interval $(p_n, q_n)$ . . .
