I'm wondering whether or not a closed, bounded interval $[a,b]$ in $\Bbb{R}$ is compact with the discrete metric?
I tried to show it wasn't by taking the set of singletons in $[a,b]$ as an open cover, since with the discrete metric singletons are open. However I don't believe you can find a finite subcover for this as there are uncountably many points in $[a,b]$ so you can't cover it with a finite amount of singletons. So it's not compact.
Is this correct? If not, where did I go wrong?
Best Answer
Yes, you're right. In fact if $(X,\tau_{\text{dis}})$ is a topological space with $X$ infinite set and $\tau_{\text{dis}}$ the discrete topology so $(X,\tau_{\text{dis}})$ cannot be compact set.
Under the discrete topology all set they are open sets so in particular the singleton sets. Taking the open cover $\mathcal{U}=\{\{x\}: x\in [a,b]\}$ which is a cover open for the interval $[a,b]$ it has no finite subcover so $([a,b],\tau_{\text{dis}})$ is not compact space.