I wonder if $\mathbb{Q}^\mathbb{N}=\mathbb{Q} \times \mathbb{Q}\times \ldots$ is countable.
I came across this question because I was looking to see if the metric space $(l^1(\mathbb{N}),d_\infty)$ is separable. With $l^1(\mathbb{N})=\{(x_n)_n \in \mathbb{R}^\mathbb{N}|\sum_{n=0}^\infty |x_n| \ \text{converges} \}$ and $d_\infty((x_n)_n,(y_n)_n)=\sup \{|x_n-y_n| n \in \mathbb{N} \}$. So I thought if $\mathbb{Q}^\mathbb{N}$ is countable than $(l^1(\mathbb{N}),d_\infty)$ is separable. Because then $\mathbb{Q}^\mathbb{N}$ would be a countable dense subset of $\mathbb{R}^\mathbb{N}$ and so $\{(x_n)_n \in \mathbb{Q}^\mathbb{N}|\sum_{n=0}^\infty |x_n| \ \text{converges} \} $ a countable dense subset of $\{(x_n)_n \in \mathbb{Q}^\mathbb{N}|\sum_{n=0}^\infty |x_n| \ \text{converges} \} $.
Best Answer
No. $|\Bbb Q^{\Bbb N}|\ge2^{\aleph_0}\gt \aleph_0$ .