# Closed sets in the lower limit topology.

Would an interval of the form $[a,b]$ be closed in the lower limit topology $\mathbb{R}_\ell$. Here is why I think it is:

Because $\mathbb{R}_\ell$ is finer than the standard topology on $\mathbb{R}$, then all the basis elements of this standard topology are in the lower limit topology; i.e., the sets $(a,b)$ are open in $\mathbb{R}_\ell$. Therefore,

$\mathbb{R} - [a,b] = (- \infty, a) \cup (b, \infty)$

$= (\bigcup_{x_1 < a} (x, a) )~ \cup ~ (\bigcup_{x_2 > b} (b,x_2)$,

which is a union of open sets. Therefore $\mathbb{R} - [a,b]$ is open and hence $[a,b]$ is closed

So, is this argument correct?

Yep, that's correct. There's no need to go through a basis: if you know that the lower limit topology is finer than the standard topology, then you can just say $R-[a,b]$ is open in the standard topology and hence also in the lower limit topology. (You might use a basis to prove that the lower limit topology is finer than the standard topology, but you don't need to repeat that argument here if you already know that fact!)