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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:)

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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

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