My attempt:
It will suffice to show that every open covering of $\mathbb{R}_l$ by basic elements contains a countable subcollection covering $\mathbb{R}_l$. Let $A = \{[a_{\alpha}, b_{\alpha} ) \}_{\alpha \in J}$ be a covering of $\mathbb{R}$ by basic elements for the lower limit topology. We wish to find a countable subcollection that covers $\mathbb{R}$. For each $q \in \mathbb{Q}$, take an element $[a_{q}, b_{q}) \in A$ that covers $q$. The collection $B= \{[a_{q}, b_{q} ) \}_{q \in \mathbb{Q}} $ is a countable subcollection of $A$ that covers all rational points in $\mathbb{R}$. Consider the set $C$ of all irrational points which are not covers by $B$. Denote $[a_{x}, b_{x} )$ to be the elements in $A$ that cover $x$ and $[a_{q}, b_{q} )$ to be the element in $B$ that covers $q \in \mathbb{Q}$. There are two types of points on $C$.
- Type 1: consist of point $x \in C$ satisfying the condition: there exists an element $\{[a_{x}, b_{x} ) \} \in A $ that covers $x$ and an element $\{[a_{q}, b_{q} ) \} \in B $ such that $ [a_{q}, b_{q} ) \subset [a_{x}, b_{x} ) $ (this implies $q \in [a_{x}, b_{x} )$). Then we can replace $[a_{q}, b_{q} ) $ by $[a_{x}, b_{x} )$ in $B$, the new collection will remain countable and cover all rational points together with point $x$. In general, apply this procudre we can find a countable subcollection that cover all rational points and all points in this type.
- Type 2: consists of all points which does not satisfy the condition in Type 1, we claim that type 2 only contains a countable points in $C$. Since it does not statisfy the condidtion in type 1, for each point $x$ in this type, any element $[a_{x}, b_{x} ) \in A$ that covers $x$ and any rational point $q \in [a_{x}, b_{x} )$ and its corresponding covering element $[a_{q}, b_{q} )$ in $B$, we have $b_{q} > b_{x}$. Because $b_{q} > b_{x}$ for all $q \in [a_{x}, b_{x} )$, it is clear that $[a_{x}, b_{x} )$ contains only one point in $C$, namely $x$ itself. So we can define a map $f$: All points in type 2 $\rightarrow \mathbb{Q}$ which maps $x$ to any rational point in $[a_{x}, b_{x} )$. If $x < y$, then $x < f(x) < y < f(y)$, hence this map is injective, therefore, type 2 contains only countable points of $C$. So we can pick a countable subcollection $B_1$ of $A$ to cover all points in type 2, and $B \cup B_1$ forms a countable subcollection of $A$ that cover $\mathbb{R}$ as desired.
Is the proof above right to show that $\mathbb{R}_l$ is Lindelöf?
Best Answer
It's right! It's basically the same idea as Munkres did in his book!