# Countable compactness and basic open cover

It is easily shown that in proving compactness, we can assume the arbitrary open cover is given by a basic open cover. However, the same proof does not work when proving countable compactness. The reason is that not every open set can be written as a countable union of basic open sets.

So my question is: in proving countable compactness, is there a counterexample showing that we cannot assume the cover is by basic elements? Note that such a space cannot be second countable.

The trivial example of a discrete uncountable space is not a counterexample.

• @PaulSinclair The question concerns countable compactness, not the Lindelof property. At first glance, it is not obvious (to me) that a space cannot have a basis with the property that every countable basic covering has a finite subcovering, yet not be countably compact. Commented May 9, 2023 at 23:20
• My apologies - I see I did misunderstand the property. Commented May 10, 2023 at 3:06

A discrete uncountable space $$X$$ is in fact a counterexample, even with the standard basis $$\{\{x\}:x\in X\}$$. Every countable open cover of basic open sets has a finite subcover... because such covers do not exist. But the space is not countably compact: consider any cover by any countably infinite partition of $$X$$, which has no proper subcover.
If you prefer a non-vacuous answer, consider the basis $$\{X\}\cup\{\{x\}:x\in X\}$$ for the same space. Then every countable open cover includes $$X$$, and thus includes a finite subcover.
Let $$X$$ be the disjoint union of a countably compact space $$C$$ and an uncountable discrete space $$D$$. $$\{D\} \cup \{U| \text{U open in X}\} \cup \{\{d\}| d \in D\}$$ is a basis. Any countable covering from this basis will include $$D$$ and a countable open covering of $$X$$ and so have a countable subcovering, but clearly $$X$$ is not countably compact.