Let $C$ be a nonempty compact set. $A$ is a family of subsets of $C$, and $A$ is a totally ordered set with respect to the inclusion $\subset$. If $A$ is countable, then the intersection of all the elements in $A$ is nonempty. But if $A$ is uncountable, does the conclusion also hold?
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If each member of $A$ is non-empty so is the intersection. Cover any member by the complements of the remaining . – Kavi Rama Murthy May 04 '22 at 10:24
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I found the answer https://math.stackexchange.com/questions/88336/intersection-of-compact-sets. – Cheburashka May 04 '22 at 10:33
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1There is absolutely no difference between the proof for the countable case and the one for the uncountable case. – Kavi Rama Murthy May 04 '22 at 11:28