Is the set $A=\{x\in l_p :x_n \in \mathbb Q\}$ countable?

Question

Is the set $A=\{x\in \it l_p :x_n \in \mathbb Q\}$ countable?If yes,then I can show that $l_p$ space is separable with resepect to the metric $d(x,y)=(\sum _nx_n^p)^{1/p}$.

Answer 1

No, the set is not countable: finite products of countable are countable, countable such products are not. Even $2^\mathbb{N}$ is not (a countable product of two point sets). But take those sequences (rational ones) that are $0$ after some index and try to show that set is countable and dense.

Answer 2

The answer is no.

Pick your favourite element $x \in l_p$ with $x_n \in \mathbb Q$ and $x_n \neq 0$ for all $n$.

For each subset $B \subset \mathbb N$ define $$y_B(n)= \left\{ \begin{array}{lc} x_n & \mbox{ if } n \in B \\ 0 & \mbox{ if } n\notin B \end{array} \right.$$

Show that $B \to y_B$ defines a one to one function from $\mathcal P(\mathbb N)$ to $A$.

Answer 3

It is not. For any subset $A $ of $\mathbb N$ define $x_n= 2^{-n}$ if $n \in A$ and $0$ otherwise. This given an element of your set and the map $A \to x$ is one-to-one.