Is the lower limit topology finer than the standard topology on $\mathbb{R}$? Is the lower limit topology finer than the standard topology on $\mathbb{R}$? 
In Lemma 13.4 on p.82 of Munkres' Topology (2nd ed.), it is stated that the lower limit topology is (strictly) finer than the standard topology on $\mathbb{R}$. In the argument, he is using that the interval $[a,b)$ lies in the interval $(a,b)$ which is certainly not true. On the other hand, the converse is true, that is: $$(a,b) \text{  lies in the interval } [a,b).$$ 
So we can conclude that the standard topology is finer than the lower limit topology. Am I right? If not then why? I think it's an errata. I have checked some existing errata online but it's not included, though.
 A: Yes! Since one have that
$$
(a,b) = \cup_{n\ge 1} \ [a+\frac{\epsilon}{n},b)
$$
where $\epsilon < \frac{b-a}{2}$.
Note that if for topology ${\mathcal T}_1$ with basis ${\mathcal S}_1$ and topology ${\mathcal T}_2$, one have that ${\mathcal S}_1 \subseteq {\mathcal T}_2$, then ${\mathcal T}_2$ is finer than ${\mathcal T}_1$. In this case if ${\mathcal S}_2$ be a basis for topology ${\mathcal T}_2$ and ${\mathcal S}_2 \not\subseteq{\mathcal T}_1$, then ${\mathcal T}_2$ is strictly finer than ${\mathcal T}_1$.For this, it is enough to note that $[0,1)$ is not open in standard topology.
For more details you can consult this textbook: James Munkres, "Topology; A First Course".
A: Well, I don't have a copy of Munkres' book at hand but I doubt that it is said.
If $[a,b)$ is open then $(a,b)=\bigcup_{n\in\mathbb{N}}\left[ a+\frac{1}{n},b\right)$ must be open.
Conversely, $[0,1)$ is not open in the standard topology.
This means that the topology in $\mathbb{R}_l$ is finer that the topology on $\mathbb{R}.$
