Rudin's Principle of Mathematical Analysis Theorem 2.14 Question

$\mathbf{Theorem 2.14:}$ Let $A$ be the set of all sequences whose elements are the digits $0$ and $1$. Then A is uncountable, meaning there does not exist a one-to-one mapping of A onto $\mathbb{Z}$.

For reference, elements of $A$ have this form $(0,1,0,1,0,0,0,1,1,1,1,1,\cdots)$

$\mathbf{Question:}$ Please bear with me. I know I'm wrong I don't know why. What is wrong with this logic?

Let $E_1$ be the set containing all the sequences with just one $1$ in the sequence. i.e. $E_1 = \left\{(1,0,0,\cdots),(0,1,0,0,\cdots),(0,0,1,0,0,\cdots),\cdots\right\}$. $E_1$ is countable.

Let $E_{2k}$ be the set containing all the sequences with only two $1$'s in the sequence where the next $1$ in the sequence is $k$ units next to the first $1$.

$E_{21} = \left\{(1,1,0,\cdots),(0,1,1,0,0,\cdots),(0,0,1,1,0,0,\cdots),\cdots \right\}$ $E_{22} = \left\{(1,0,1,0,\cdots),(0,1,0,1,0,\cdots),(0,0,1,0,1,0,\cdots),\cdots \right\}$

Let the union of the sets $E_{2k}$ for ($k=1,2,3,\cdots$) be called $E_2$. $E_2$ is countable.

Let $E_{3ij}$ be the set of sequences with only three $1$'s where the second $1$ is $i$ units away from the first $1$, and the third $1$ is $j$ units away from the second.

For example:

$E_{311} = \left\{(1,1,1,0,\cdots),(0,1,1,1,0,\cdots),(0,0,1,1,1,0,\cdots),\cdots\right\}$ $E_{312} = \left\{(1,1,0,1,\cdots),(0,1,1,0,1\cdots),(0,0,1,1,0,1,0,\cdots)\cdots\right\}$ $E_{322} = \left\{(1,0,1,0,1,\cdots),(0,1,0,1,0,1,0\cdots),(0,0,1,0,1,0,1\cdots)\cdots\right\}$ $E_{333} = \left\{(1,0,0,1,0,0,1\cdots),(0,1,0,0,1,0,0,1,\cdots),(0,0,1,0,0,1,0,0,1,\cdots)\cdots\right\}$

Let $E_3$ be the union of all the sets $E_{3ij}$ with ($i,j=1,2,3,\cdots$). Since each $E_{3ij}$ is countable the union of them is countable, so $E_3$ is countable.

The zero sequence is just one sequence, $E_1$ is countable, $E_2$ is countable, $E_3$ is countable. Now if we continue defining the sets in this manner then each set $E_k$ (where $k$ is the number of $1$'s in the sequences) will be a countable set.

So, the union of all these sets will be countable and it will represent all the sequences in A.

Thank you in advance to anyone who reads all of this and answers!

Solution 1:

Your union of $E_{ijk\cdots}$ does not contain the following element

$$(0,1,0,1,0,1,0,1, \cdots).$$

Indeed those $E$'s contain elements in $A$ with finitely many nonzero entries. You have just proved:

The subset of $A$ consisting of all sequences with finitely many $1$'s is countable.