Modifying proof of uncountability

Question

My professor's lecture notes offer a proof that $\mathbb{R}$ is uncountable. I think it is slightly incomplete, but I just want to be sure.

No map $f: \mathbb{N} \to \mathbb{R}$ can be surjective because if we write the decimal or binary expansion of $f(n)$, we get \begin{align*} f(0) & = a_{00}. a_{01} a_{02} a_{03} \ldots \\ f(1) & = a_{10} . a_{11} a_{12} a_{13} \ldots \\ f(2) &= a_{20} . a_{21} a_{22} a_{23} \ldots \\ & \vdots \\ \end{align*}

(I assume we can do this because the digits are in direct correspondence with $10^{-i}$ for $i \geq -1$, and the set $\{-1, 0, 1, 2, \ldots\} = \mathbb{N} \cup \{-1\}$ is countable. Please tell me if I'm not thinking of this correctly. The $a_{j0}$ digits, I assume, can just be any integer at all and might be multiple digits long.)

Now let $y = b_0 . b_1 b_2 b_3 \ldots$ where $b_j \neq a_{jj}$ for each $j$.

(I assume this step requires countable choice, so we again need the above argument on the decimal places being countable.)

So $y \neq f(j)$ since $y$ differs from $f(j)$ in the $j$th digit (calling the $a_{j0}$ term the "$0$th digit.") So $y \not \in f(\mathbb{N})$, so $f$ is not surjective.

Aside from my comments above on countable choice, which I assume is the only way we can even list the decimal places in that way, I don't see the significance of the distinction between decimal and binary expansion. I assume these are two totally separate proofs, but they require different arguments for uniqueness.

I only know about how to deal with decimals. I would just require that when I write the decimal expansion of $f(n)$, I demand that I choose the decimal expansion (it has two at most) that doesn't end in an infinite string of $9$'s. Then I require that when I choose $b_j \neq a_{jj}$, I also require $b_j \neq 9$. Since the decimal expansion, once fixed, is "unique" once I requite that it not end in $9$'s, I have $9$ remaining choices for $b_j$, so I know I can always do this.

I would like to know if my above comments are accurate as well as if anyone has insights on how to prove this with binary and, in particular, how one could guarantee that such a binary expansion is unique.

Answer 1

(I assume we can do this because the digits are in direct correspondence with $10^{-i}$ for $i \geq -1$, and the set $\{-1, 0, 1, 2, \ldots\} = \mathbb{N} \cup \{-1\}$ is countable. Please tell me if I'm not thinking of this correctly. The $a_{j0}$ digits, I assume, can just be any integer at all and might be multiple digits long.)

I believe you are asking whether it is possible to write every real as an integer followed by an infinite sequence of digits? This is basically a fact from analysis, and does not require any choice, since the digits can be constructed recursively from the real number. See for example this answer.

---

(I assume this step requires countable choice, so we again need the above argument on the decimal places being countable.)

You don't need choice, since you can once again construct the "new" real number from the list. One simply picks $b_j=0$ if $a_{jj}=1$, and $b_j=1$ otherwise, then you also avoid the problem of having repeated $9$'s (which your professor indeed seems to ignore).

---

I don't see the significance of the distinction between decimal and binary expansion. I assume these are two totally separate proofs, but they require different arguments for uniqueness.

The problem with the same proof for a binary expansion is that we cannot avoid the repeated $1$'s in the same way as we can with the decimal expansion. However, there's an easy solution: We can just add the different expansions to our list (which also works for decimal expansions).

For the real number at position $n$ of the original list, we can place it at position $2n$ on our new list, and place the same real number at position $2n+1$, but require to use the alternative expansion there, whenever that real number has an alternative expansion. Since every real number has at most two different alternative expansions, we have now included not only every real number, but also all possible expansions of real numbers in our list.

When we do the diagonalisation proof now, we create a real number that is different from every real number in our list, since it must be different from every expansions of any real number in our list.

---

An alternative way, is the following argument:

I use the notation $|X|$ to denote the cardinality of a set $X$.

1. There exists an injective function $\Bbb R\to [0,1]$ (for example something involving the arctangent), so $|\Bbb R|\leq |[0,1]|$.
2. Let $2^{\Bbb N}$ be the set of infinite sequences of $0$'s and $1$'s. Every real $x\in [0,1]$ can be written as a binary expansion $0.d_1d_2d_3d_4\cdots$, thus there exists an injective functions sending $0.d_1d_2d_3d_4\cdots$ to the sequence $(d_1,d_2,d_3,d_4,\dots)$, therefore $|[0,1]|\leq |2^{\Bbb N}|$
3. For every sequence $(d_1,d_2,d_3,d_4,\dots)\in 2^{\Bbb N}$, we can define a unique real number using the Cantor set: start with interval $I_0=[0,1]$ and divide it into three parts of equal length $[0,\frac13]$, $[\frac13,\frac23]$ and $[\frac23,1]$, if $d_1=0$ then your new interval $I_1=[0,\frac13]$, else it will be $I_1=[\frac23,1]$. Next, divide your interval $I_1$ into three parts of equal length, and if $d_1=0$ let $I_2$ be the lowest part, otherwise let $I_2$ be the highest part, and so on. The $n$-th interval has length $\frac 13^{n}$, thus diminishes, and since each interval is closed, the intersection $\bigcap_{n\in\Bbb N} I_n$ contains exactly one real number.
   If you do this construction twice, with different sequences $\vec d$ and $\vec d'$ in $2^{\Bbb N}$, you will get different intervals $I_n$ and $I_n'$ for some large enough $n$, and by construction $I_n$ and $I_n'$ are disjoint, so you will end up with different real numbers. So this shows that there exists an injection $2^{\Bbb N}\to \Bbb R$, thus $|2^{\Bbb N}|\leq |\Bbb R|$.

We now have $|\Bbb R|\leq |[0,1]|\leq |2^{\Bbb N}|\leq |\Bbb R|$, so $|2^{\Bbb N}| = |\Bbb R|$. Therefore, we can do the diagonalisation proof using just the infinite sequences of $0$'s and $1$'s, without having to worry about two different sequences representing the same real number.

As a bonus, we could even add $\Bbb N^{\Bbb N}$ into our chain of cardinalities, since each infinite sequence of natural numbers can be mapped bijectively with an irrational number (see for example this question, which has two different proofs for this fact in the answers).