# compact set in lower limit topology

Let R be the set of real numbers with the topology generated by the basis {[a, b) : a < b, a, b ∈ R}. If X is the subset [0, 1] of R, which of the following must be true?
(I) X is compact
(II) X is Hausdorff
(III) X is connected

II and III must be true.
I want to show that $[0,1]$ is not compact in $R_l$, I need to find an open covering that doesnt have finite subcover.
So I'm consider something like $\bigcup[1/n,2), n\in N$ .
It's open in $R_l$ and covers $[0,1]$ and doesn't have finite subcover.
Is this ok ?

• Note: $\{1\}$ is relatively open in $[0,1]$ in the lower limit topology. Your family doesn't cover $0$. – Daniel Fischer Oct 4 '15 at 13:01
• so [1-1/n) $\cup$ {1}, n $\in$ N should do the job. Thanks. – Geralt of Rivia Oct 4 '15 at 14:01

Hint: Consider the set $B : [0,k)\cup [1,2)$ where k ranges from $(0,1)$. This is a cover for your set but as you can check it does not have a finite sub cover.
On the other hand your set did not cover $0$