I am being asked if the set {$1-\frac{1}{n} | n\in \mathbb N$} $\cup$ {1} is compact under Sorgenfrey topology.
I believe it is not, because we can take the open cover K = [1,2)$\cup$[0,$1- \frac{1}{1}$)$\cup$….[0,$1- \frac{1}{n}$)…
But this cover has no finite subcover, because for any $m \in \mathbb N$ we may take such that $K_m = \cup_{n = 1}^m [1,2)\cup[0, 1-\frac{1}{n})$, the element $1-\frac{1}{m}$ is in our set but not in our subcover.
Is this correct? I don't understand compactness very well.
Best Answer
It looks fine to me. I would rather take$$[1,2)\cup\left[0,\frac12\right)\cup\left[\frac12,\frac23\right)\cup\left[\frac23,\frac34\right)\cup\ldots$$It is an open cover and each f its elements has one and only one element from your set. So, it is trivial that there is no open subcover.