# Is [0,1] compact in $T_{up}$?

I want to know:

whether [0,1] is compact in the upper topology .

My guess is that it is compact, because $T_{st}$ is compact on [0,1] and $T_{up} \subset T_{st}$, but does this directly give us [0,1] is compact in $T_{up}$ as well? Since the definition of compactness is the open covering admits finite subcovering of [0,1]. Since we can find such subcovering in $T_{st}$, then we know such subcoverin also exists in $T_{up}$, is this correct?

Please give a formal prove, and thank you in advance for the help!

Upper topology:= {∅, R}∪{(a, ∞) | a ∈ R} defines a topology on R,

Let ${\cal C}\subseteq\tau_{up}$ be an open covering of $[0,1]$. Since $\tau_{up}\subseteq\tau_{st}$, we have that ${\cal C}\subseteq\tau_{st}$. Now, as $[0,1]$ is compact in $\tau_{st}$, there exists a finite ${\cal C'}\subseteq{\cal C}$ such that $[0,1]\subseteq\bigcup {\cal C'}$. Hence, $[0,1]$ is compact in $\tau_{up}$.
Consider $A=[0,1]$ and assume $\{U_\alpha\}$ be an open cover of $A$ with $U_\alpha\in T_{up}$, for all $\alpha$. Since $T_{up}\subset T_{st}$, so $U_\alpha\in T_{st}$ (i.e. $U_\alpha$ is open in standard topology). Therefore there is a finite subcover of $\{U_\alpha\}$ which covers $A$ in $(\mathbb{R},T_{st})$ because $A=[0,1]$ is compact in standard topology. Finally, the same subcover, is also covers $A$ in $T_{up}$ topology. Thus $A$ is compact in $(\mathbb{R},T_{up})$.