# 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

Let $\mathbb{R}_S$ be the Sorgenfray 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.