Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
5  
You should trust yourself a little more. Your cover perfectly proves it's not compact. –  Daniel Fischer Dec 11 '13 at 19:56
add comment

1 Answer 1

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.