Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I know that uncountably product of $\{0,2\}$ is not-metrizable, separable, compact, Hausdorff and not second countable. But what we can say about Lindelöf property? I think it is not Lindelöf, but how can I show it? Could you give me any idea?

share|improve this question
    
If by lindelof property, you mean wether it is a lindelof space, then the answer is trivially yes. Compactness implies lindelof. –  Alex Becker Dec 23 '12 at 5:49
    
oh, yes, I have to sleep, so sorry. –  ege Dec 23 '12 at 5:52
    
Not all uncountable products of a 2-point space are separable (you need at most continuum many copies). All such spaces are ccc though. –  Henno Brandsma Dec 23 '12 at 8:20
add comment

1 Answer 1

It is obviously Lindelöf (every open cover has a countable subcover) as it is already compact (every open cover has a finite subcover). It is not hereditarily Lindelöf, nor hereditarily separable, as the subspace of all points with exactly one coordinate equal to 2 (and all others 0) is discrete and uncountable, and so not Lindelöf and not separable.

share|improve this answer
add comment

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.