Exercise 30.17

Proof. We claim ${\mathbb{Q}}^{\infty }$ is not first countable, and therefore not second countable. Suppose we have a countable basis $\{U_i\}$ at $0=(0,0,0,\dots )\in {\mathbb{Q}}^{\omega }$. Let $V_j⊊{\pi }_j(U_j)$ open in $\mathbb{Q}$ with the subspace topology induced by $ℝ$. Then, the neighborhood ${\prod <!--\; FUNCTION\; APPLICATION\; -->}_j{V}_j\ni 0$ does not contain any $U_i$, so $\{U_i\}$ is not a basis and ${\mathbb{Q}}^{\infty }$ is not first or second countable.

We now show ${\mathbb{Q}}^{\infty }$ has a countable dense subset. For, $Q^n={\mathbb{Q}}^n\times \{0\}\times \{0\}$ are countable since they are finite products of countable sets, and so their countable union ${\mathbb{Q}}^{\infty }=\bigcup <!--\; FUNCTION\; APPLICATION\; -->Q^n$ is also countable. Thus, ${\mathbb{Q}}^{\infty }$ is countable and so is a countable dense subset of itself.

We now show ${\mathbb{Q}}^{\infty }$ is Lindelöf. Suppose $V$ is an open covering of ${\mathbb{Q}}^{\infty }$. Then, since ${\mathbb{Q}}^{\infty }$ is countable, choosing for every $x\in {\mathbb{Q}}^{\infty }$ one element $V\in V$ such that $x\in V$, we get a countable subcover of ${\mathbb{Q}}^{\infty }$. □