I want to prove that $\mathbb{R}$ with the topology generated by the intervals $[a,b)$ is Lindelof.
First, it is sufficient to show that every basic sets cover has a countable subcover.
Let $\{[a_{\alpha},b_{\alpha})\}_{\alpha\in J}$ be such cover. The sets $\{(a_{\alpha},b_{\alpha})\}_{\alpha\in J}$ cover $\mathbb{R}$ and are open in the standard topology, and since $\mathbb{R}$ is separable, it is Lindelof, hence there exists a countable subcover $\{(a_n,b_n)\}_{n\in\mathbb{N}}$ such that $\mathbb{R}\subseteq\bigcup\limits_{n\in\mathbb{N}}(a_n,b_n)$. But then,
$$
\mathbb{R}\subseteq\bigcup\limits_{n\in\mathbb{N}}(a_n,b_n)\cup\{a_n\mid n\in\mathbb{N}\} = \bigcup\limits_{n\in\mathbb{N}}[a_n,b_n)
$$
Is this legit?
Best Answer
No, the $(a_\alpha, b_\alpha)$ need not cover $\mathbb{R}$ at all. That's too easy. But it does not matter because we consider $$U = \bigcup_\alpha (a_\alpha,b_\alpha)$$ which is an open subset of $\mathbb{R}$ in the Euclidean topology (so it's second countable as all subspaces of the reals are, so $U$ is Lindelöf), and the $(a_\alpha, b_\alpha)$ are by definition an open cover of $U$.
So we can find a countable set of indices (corresponding to a countable subcover) $\{\alpha_n: n \in \mathbb{N}\}$ such that $$\bigcup_n (a_{\alpha_n}, b_{\alpha_n}) = \bigcup_\alpha (a_\alpha, b_\alpha)= U$$
So what elements $x$ do we potentially miss from the union of the subcover $$\mathcal{U} = \{[a_{\alpha_n}, b_{\alpha_n}): n \in \mathbb{N}\}\text{?}$$
So consider $$P=\mathbb{R}\setminus \cup\mathcal{U}$$
We will show this set is at most countable: Any $x \in P$ must be a point of the form $x=a_\alpha$ (or it would be in some $(a_\alpha,b_\alpha)$, as it must be covered, so then $x \in U \subseteq \cup\mathcal{U}$, contradiction), and we can pick a rational number $q_x \in (a_\alpha, b_\alpha)$ for that $\alpha$ (the rationals are dense so this can be done). Note that always $(x, q_x) \subseteq U$.
Suppose $x < y$ are two distinct points of $P$. Then I claim that $q_y \le q_x$ is not possible: suppose it is, then we have that $x < y < q_y \le q_x$ and this implies that $y \in (x,q_x) \subseteq U$, which is impossible (again because then $y \notin P$).
So $q_x < q_y$ whenever $x < y$ are both in $P$. So the map $x \to q_x$ from $P$ into $\mathbb{Q}$ is injective and $P$ is indeed at most countable.
Now for each $x \in P$ add an $[a_{\alpha(x)}, b_{\alpha(x)})$ to cover it and note that
$$ \mathcal{U} \cup \{[a_{\alpha(x)}, b_{\alpha(x)}): x \in P\}$$
is a countable subcover of the original cover.