Is $(0,1]$ compact in $\mathbb{R}$?

Question

Is s $(0,1]$ compact in $\mathbb{R}$?

Since $$(0,1]=\bigcup_{n=1}^{\infty}(1/n,1]$$ and there is no finite subcover, I assume it is not compact.

Is this argument correct?

Answer 1

Yes the argument you give is perfectly fine! If you want to be absolutely rigorous, assume there is a finite subcover and invoke the archimedean principle.