I have been asked to prove that the $l^p$ space for $1\leq p<\infty$ is separable. A space is separable when we can find a dense countable subset of that space.
The argument given in my book is the following:
Take a sequence $\{a_1,a_2,\dots\}$. There exists $n\in \Bbb{N}$ such that $\sum\limits_{i=1}^{\infty}{|a_{n+i}|^p}<\epsilon/2$. Now take the sequence $\{r_1,r_2,\dots,r_n,0,0,\dots\}$, where $r_1,r_2,\dots,r_n\in\Bbb{Q}$ and $\sum\limits_{j=1}^{n}{|a_{j}-r_j|^p}<\epsilon/2$. Then the set of all such sequences (consisting of rational numbers and $0$'s) is countable.
I wonder why such a set is countable. The $n\in\Bbb{N}$ for which $\sum\limits_{i=1}^{\infty}{|a_{n+i}|^p}<\epsilon/2$ does not have a definite upper bound. It can go on increasing. So the dense subset that we're talking about may actually be $\underbrace{\Bbb{Q}\times\Bbb{Q}\times\dots}_{\Bbb{Z}\text{ times}}$, which is uncountable.
Thanks in advance!
Best Answer
It is true that the $n\in\mathbb{N}$ for which $\sum_{i=1}^{\infty}|a_{n+i}|^p<\frac{\epsilon}{2}$ does not have a definite upper bound, and can "go on increasing". However, your inference that "the dense subset that we're talking about may actually be $\underbrace{\mathbb{Q}\times\mathbb{Q}\times\dots}_{\text{$\mathbb{Z}$ times}}$" is incorrect. $n$ may be arbitrarily large, but it is a natural number nonetheless — in particular, it remains a finite quantity.
To elucidate further, if $D$ denotes "the set of all such sequences (consisting of rational numbers and $0$'s)", i.e., the set of sequences with finite support and rational entries, then we may write $D=\{(q_1,\dots,q_n,0,\dots)\mid n\in\mathbb{N},q_i\in\mathbb{Q},1\leq i\leq n\}$. Observe then that $D\leftrightarrow\cup_{n=1}^{\infty}(\prod_{i=1}^{n}\mathbb{Q})$ - which certainly is countable, as the countable union of a finite product of countable sets.