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I just had a chat with my real analysis teacher today, we talked about how there is no $\sigma$-algebra such that it has countably many elements, but finitely many.

In the end, he said that I can visualize it by drawing out the $\sigma$-algebra on (0,1) using intervals. But I'm not too sure how to do that, should I draw it out on the real line? Or can I use any type of diagram?

Akaichan
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    ${\varnothing,(0,1)}$ seems to be a finite $\sigma$-algebra. – hmakholm left over Monica Sep 10 '14 at 20:02
  • You mean infinite but countable? – Ayman Hourieh Sep 10 '14 at 20:04
  • I meant there is no such $\sigma$-algebra that is countable and infinite – Akaichan Sep 10 '14 at 20:06
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    One strategy might be to first suppose there exists such ($\sigma$-)algebra, and then identify a countably infinite disjoint collection of subsets of that algebra, and then use the fact that the power set of such a countably infinite set would be uncountable, reaching a contradiction (Later edit: Huh, apparently there is one, forget about this then.). – Lord Soth Sep 10 '14 at 20:08
  • @Lord Soth: Regarding your comment (Later edit: Huh, apparently there is one, forget about this then.), I thought it was well known that any infinite $\sigma$-algebra must have cardinality $\geq c = 2^{{\aleph}_{0}}.$ For example, I'm pretty sure this is an exercise early in Halmos' Measure Theory. – Dave L. Renfro Sep 10 '14 at 21:04
  • See here http://math.stackexchange.com/questions/806778/cardinialty-infinite-sigma-field. My answer shows that every $\sigma$ algebra is either finite or uncountable. – PhoemueX Sep 10 '14 at 21:31

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