# What’s wrong with this “proof” that $\mathbb{R}$ is enumerable?

The fake proof:

• We know that $$\mathbb{R}$$ is uncountable, hence we cannot enumerate over it.
• But what we do know is that $$\mathbb{Q}$$, the set of rationals, is countable, and even denumerable.
• We also know that we can construct $$\mathbb{R}$$ through what are called Dedekind cuts.
• We choose to let the partition itself denote a new number and go forth to define mathematical operations on it as to be compatible with the rest of the numbers (mainly $$\mathbb{Q}$$ and our new number $$x$$)

Sidenote: I think so far this is standard, and contains nothing false. The actual argument starts below this line.

• Let us denote the set containing $$x$$ as $$S_1 := \mathbb{Q}\cup\{x\}$$. For convenience, the superscript of $$S_1$$ is how many new such numbers we have added through the cuts.

• Since $$\mathbb{Q}$$ is countable, we can enumerate over every single rational $$q\in\mathbb{Q}$$ to produce an $$r\in\mathbb{R}$$. Do this process $$n$$ times and you end up with $$S_n = \mathbb{Q}\cup{x_1}\cup{x_2}\cup\dots\cup{x_n}$$.

• But $$S_n$$ is also enumerable since it has a finite more elements than $$\mathbb{Q}$$.

• Hence – After enumerating over the entirety of $$\mathbb{Q}$$ – Start enumerating over the entirety of $$S_n\setminus\mathbb{Q}$$

• Now we will end up with even newer numbers to put in our set, which we will now call $$S_{n,k}$$ where $$n$$ represents the enumeration over $$\mathbb{Q}$$ and $$k$$ represents the enumeration over $$S_n\setminus\mathbb{Q}$$. Do this ad infinitum and you will eventually describe $$\mathbb{R}$$.

I know I went wrong somewhere, I just don’t know where.