Countable compactness and basic open cover

Question

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.

Answer 1

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.

Answer 2

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.