# Is the subset $[0,1)$ of $\mathbb{R}$ compact in the lower limit topology?

What I have done so far is give a contradiction, namely the cover:

$\mathcal{U}=\{{[0,1-\frac{1}{n}):n\in\mathbb{N}}\}$

Because $\cup_{n\in\mathbb{N}}[0,1-\frac{1}{n})=[0,1)$, it means that there is no finite subcover that covers $[0,1)$. Is this right or am I doing something wrong$?$ For some reason I have a feeling it is compact and I am overseeing something.

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You should trust yourself a little more. Your cover perfectly proves it's not compact. – Daniel Fischer Dec 11 '13 at 19:56
yes. It's not compact and your example and method are correct.This topology on the set of reals is called the Sorgenfrey line. – user254665 Nov 21 '15 at 7:23
This also shows that $[0,1)$ is not compact in the Euclidean topology. If you knew already that it is not compact in the Euclidean topology, you could also use the fact that a non-compact set remains non-compact when we refine the topology (or that a continuous image of a compact space is compact). – Stefan Hamcke Apr 25 at 13:59

Let $\mathbb{R}_S$ be the Sorgenfrey line, i.e., $\mathbb{R}$ with the lower limit topology. As any uncountable set of real numbers contains a strictly increasing infinite sequence, if $K\subseteq\mathbb{R}_S$ is compact then $K$ is countable.