# Proof Verification: $\mathbb{R}_l$ is Lindelöf.

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?