Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Recently I'm interested in this open question:

Must every star compact topological group be countably compact?

  1. star compactness ( which implies pseudocompactness ) = for every open cover $U$ of the space $X$, there exists a compact subspace $K$ such that $\cup \{u \in U: u \cap K \text{ is not empty} \} = X.$
  2. countably compact ( which implies star compactness obviously) = for every open cover $U$ of the space $X$, there exists a finite subspace $K$ such that $\cup \{u \in U: u \cap K \text{ is not empty} \} = X.$ This definition is under the $T_1$ assumed. It is equivalent to this: for every countable open cover of $X$ there is a finite subcover of $X$.

I'm not very familar with topological group. I have some questions:

Firstly, could someone complete to list the properties of star compact topological group. These of star compact topological group are I know, for example:

  1. Tychonoff = $T_0$ in every topological group,
  2. pseudocompactness from star compactness,
  3. CCC = countable chain condition, for every pseudocompact topological group has the CCC.

Secondly, if you have any idea for this open question, you could write here.

Thirdly, Is there a pseudocompact topological group but is not separable?

Thanks for any help:)

share|cite|improve this question
How does this question arise? Can you give some examples of topological groups where you can check that they are star compact without simultaneously verifying stronger compactness properties? – t.b. Jul 1 '12 at 11:19
Just to contribute something mildly constructive: take $G = (\mathbb{Z}/2\mathbb{Z})^{I}$ where $I$ is a large enough set to get an example of a compact Hausdorff group that isn't separable (a separable Hausdorff space has cardinality at most $2^{\mathfrak{c}}$ as was shown in this thread for example). – t.b. Jul 1 '12 at 14:58
Since start compactness is perhaps less familiar, I'll add a link to another question where the definition is given. – Martin Sleziak Jul 2 '12 at 10:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.