Show that $y = (r_1, r_2, \dots, r_n, 0, 0, \dots)$ is countable.

Question

While proving that the space $l^p$ with $1 \le p < \infty$ is separable, the book is suggested the following subset [which is easy to prove denseness...] :

$M$ to be the set of all sequences $y$ of the form $y = (r_1, r_2, \dots, r_n, 0, 0, \dots)$ where $n$ is any positive integer and the $r_j$'s are rational.

How to prove that $M$ is countable? $\sum_{i=1}^{\infty} \mathbb{Q^i}$ 'looks' big enough to be like $2^{\mathbb{N}}$ than finitely many product of $\mathbb{Q}$ to be countable.

I am not much familiar with axiomatic set theory. Simple clear explanation would be much appreciated.

Answer 1

$M$ is countable, because

$$ \begin{align*} \left|\sum_{i\in\mathbb{N}}\mathbb{Q}^i\right| &\leq\sum_{i\in\mathbb{N}}|\mathbb{Q}|^i\\ &=\sum_{i\in\mathbb{N}}\aleph_0^i\\ &=\sum_{i\in\mathbb{N}}\aleph_0\\ &=\aleph_0^2=\aleph_0 \end{align*} $$

Or, a bit more concretely, fix a bijection $f\colon\mathbb{Q}\to\mathbb{N}$, $f(0)=0$ and inject $M\to\mathbb{N}$ by $2^{f(r_1)}\cdot 3^{f(r_2)}\cdot 5^{f(r_3)}\cdots$

Answer 2

$A_n=$

{$(r_1,r_2,..,r_n,0,0,..)|$

$ r_i \in \mathbb{Q}, 1\le i \le n$}.

Consider

$f: A_n \rightarrow \mathbb{Q}^n, $

$f(r_1,r_2,...,r_n,0,0,..) =$

$ (r_1,r_2,...r_n)$ is a bijection.

Since $\mathbb{Q}^n$ is countable $A_n$ is countable.

The countable union of countable sets

${\displaystyle \cup}_n A_n$ is countable.