Let $X=\mathbb{R}$ and $\tau$ the inferior semicontinuity topology, defined as: $$\tau=\{(a,+\infty) \mid a \in \mathbb{R}\}\cup\{\mathbb{R}\}$$ where $(a,+\infty)=]a,+\infty[$ is an open interval. Has $\tau$ a countable base?
While it seems to me that there aren't countable bases for $\tau$, I thought about this possible base: $$\mathcal{B}=\{(q,+\infty) \mid q \in \mathbb{Q}\}$$ Where, taken $x \in \mathbb{R} \setminus \mathbb{Q}$, we can describe the open set $(x,+\infty)$ as: $$(x,+\infty)=\bigcup (q_n,+\infty)$$ where $\{q_n\}_{n \in \mathbb{N}}$ is a sequence in $\mathbb{Q}$ such that $q_n \rightarrow x$.
Is this legitimate? I know that there are such sequences (for example the famous $\left (1+\frac{1}{n}\right)^n \rightarrow e$), but can I say that for every real number?
Yes, for every real $x$ there is a decreasing sequence of rationals $(q_n)$ such that $x < q_n$ and $q_n \downarrow x$, because the rationals are order dense in $\Bbb R$, basically by construction (of the reals). And for such a sequence we can indeed easily show $$(x, +\infty) = \bigcup_n (q_n, +\infty)$$ and indeed the set $\mathcal{B}= \{q,+\infty)\mid q \in \Bbb Q\}$ forms a countable base for this topology.