# Does there Always Exist a Finer/Coarser Topology such that the space is compact?

Sorry for the somewhat vague title. My questions are as follows:

Suppose $$T$$ is a quasicompact space. Does there exist a finer topology on $$T$$ such that it becomes compact?

Suppose $$T$$ is a Hausdorff space. Does there exist a coarser topology on $$T$$ such that it becomes compact?

The way I see this is that quasicompact/Hausdorff is a upper/lower bound on how fine the topology can be. This bound is also tight, in the sense that if two topologies $$\tau_1 \subset \tau_2$$ are both compact, then $$\tau_1 = \tau_2$$. So the questions ask if we can always get inside this bound from some starting point $$T$$.

I am not sure what the answer is. I have tried using Zorn's lemma, but quasicompactness and Haustorff properties aren't preserved at the obvious upper/lower bound of chains of topologies on a space X (made from taking union/intersection of all topologies in the chain).

I have also tried to construct counterexamples, but they havn't really worked either. I suspect that my examples are all too "nice", but I feel there are few pointers as to what a counterexample would look like.

Thanks for taking the time to read this. I would greatly appreciate any insight into this question.

• Do you need $T$ to remain Hausdorff for the second question? Aug 13 '20 at 14:25
• @bitesizebo The use of quasicompact indicates that "compact = quasicompact and Hausdorff". Aug 13 '20 at 14:26
• $\{\emptyset, T\}$ is compact but probably not what you are looking for.. This indeed suggests that you mean compact = open cover condition + Hausdorff. Aug 13 '20 at 15:10
• I think this is the Bourbaki usage of "compact". Aug 13 '20 at 15:20