Is the standard topology on $\mathbb{R}$ a subset of the lower limit topology $\mathbb{R_\mathcal{l}}$?
Let $\mathcal{T_l}$ denote the lower-limit topology on $\mathbb{R}$, and let $\mathcal{T_s}$ denote the standard topology on $\mathbb{R}$.
So let $\mathcal{B}$ be the basis for the lower-limit topology and $\mathcal{A}$ be the basis for the standard topology on $\mathbb{R}$. Then $\mathcal{B} = \{ \ [a, b) \ | \ a, b \in \mathbb{R} \}$ and $\mathcal{A} = \{ (a, b) \ | \ a, b \in \mathbb{R}\}$.
Now, there's a well known lemma that a topology $\mathcal{O}$ equals the collection of all unions of its basis elements.
So let $\mathcal{K}$ be an arbitrary sub-collection of $\mathcal{A}$, then $\mathcal{K} \subset \mathcal{A}$, and some $U \in \mathcal{T_s} = \bigcup_{K \in \mathcal{K}} K$, where $K = (a, b)$ for some $a, b \in \mathbb{R}$
Again let $\mathcal{J}$ be an arbitrary sub-collection of $\mathcal{B}$, then $\mathcal{J} \subset \mathcal{B}$, and some $U \in \mathcal{T_l} = \bigcup_{J \in \mathcal{J}} J$, where $J = [a, b)$ for some $a, b \in \mathbb{R}$
Lemma: Every basis element $O \in \mathcal{O}$, is an element of the topology $\mathcal{T}$ it generates, (I've put this lemma here as I use it in my argument below implicitly)
To show $\mathcal{T_s} \subset \mathcal{T_l}$, we need to show $U \in \mathcal{T_s} \implies U \in \mathcal{T_l}$.
But take $U = (a, b) \in \mathcal{T_s}$, then $U \not\in \mathcal{T_l}$ because no union of sets $[a, b) \in \mathcal{B}$ will equal $(a, b)$. It is clear that $(a, b) \subset [a, b)$, but $(a, b) \in \mathcal{T_s} \not\Rightarrow (a, b) \in \mathcal{T_l}$ (in fact $\mathcal{T_l}$ does not contain any sets of the form $(a, b)$, correct me if I'm wrong please). So how can it be true that $\mathcal{T_s} \subset \mathcal{T_l}$?