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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?

user193319
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1 Answers1

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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!)

Eric Wofsey
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    Ah, of course! This is embarrassing...I already proved that the lower limit topology was finer. So obviously every open set in the standard topology is also open in the lower limit topology...By the way, thanks!! – user193319 Sep 24 '16 at 20:41