I want to prove that $\mathbb{Q}^{\infty}$ is not first countable.

Question

I don't understand the hint which is similar to the case in $\mathbb{R}^{\omega}$ in box topology. Is there a subset of $\mathbb{Q}^{\infty}$ which does not satisfy the converse of sequence lemma? Or just can find a point directly which is not countable local basis? I tried to think about these method, but it is so hard to me, unlike the case of $\mathbb{R}^{\omega}$ in box topology. I need your help.

Answer 1

The same proof and subset as for $\Bbb R^\infty$ (for the sequence lemma) work.

Let $A = \{(q_n) \in \Bbb Q^\infty: \forall n: q_n >0\}$ is a subset of $\Bbb Q^\infty$ that has $\underline{0} = (0,0,0,0,\ldots)$ as a point in its closure (a limit point even): if $\prod_n U_n$ is an open box-neighbourhood of $\underline{0}$, then each $U_n$ is open in $\Bbb Q$, containing $0$ so it contains some rational $q_n=\frac{1}{k}>0$ for $k$ large enough, say. Then $(q_n)_n \in (\prod_n U_n) \cap A $ so that $\underline{0} \in \overline{A}$.

But if $\left((q^{(m)}_n)_n\right)_m$ is a sequence in $A$, the product neighbourhood

$$\prod_n (-\frac{q^{(n)}_n}{2}, \frac{q^{(n)}_n}{2})$$ of $\underline{0}$ contains no members of that sequence so it cannot converge to $\underline{0}$.

This contradicts the sequence lemma and $\Bbb Q^\infty$ is not first countable.