False "proof" that $\mathbb{R}$ is countable

Question

About to fall asleep, I came up with the following "proof" that $\mathbb{R}\cap[0,1]$ is countable:

For each $k \in \mathbb{N}$, let the set $S_k$ be the set containing ever positive real number less than one with decimal expansion consisting of $k$ or fewer digits. Then $$\mathbb{R} \cap [0,1] = \bigcup_{k=1}^\infty S_k.$$ But because each $S_k$ is finite, $\mathbb{R}\cap[0,1]$ is countable as it is the countable union of finite sets.

After thinking about it for a couple of minutes, I don't see what's wrong with it. Please tear it apart so I can sleep.

Answer 1

Your set $\bigcup_{k=1}^\infty S_k$ contains only rational numbers, but $\mathbb{R} \cap [0,1]$ contains also irrational numbers, so $$\bigcup_{k=1}^\infty S_k \neq \mathbb{R} \cap [0,1]$$

Answer 2

There is a very important misunderstanding here, which is the idea that "arbitrarily long" is still less long than "infinitely long". As others have pointed out, your set only contains the numbers with finitely many decimal digits. You can have as many finitely many digits as you want, true. But you cannot have infinitely many digits.

Because, okay, suppose that you had a thing with infinitely many digits. If it's really in the union, it has to live at some level of the union (in some $S_k$). But by definition, it doesn't live in any of the $S_k$. So it cannot not be in the union.

Answer 3

Suppose some irrational is in your $\bigcup_{k=1}^\infty S_k$.
Let $\dfrac{1}{\sqrt 2} \in \bigcup_{k=1}^\infty S_k$.

OK, $a\in\bigcup_{k=1}^\infty b_k$ means, that there exists at least one set $b_i$ for which it is true that $a\in b_i$.

But in your example every $S_k$ contains numbers with finite digits in their decimal expansion. So there is no such $S_k$, that $\dfrac{1}{\sqrt 2} \in S_k$.