Is that true, that every sequence of sets have coverging subsequence? We say that sequence of sets $A_1, A_2, A_3, ...$ coverging iff ${\limsup}_{n \to \infty} A_n = {\liminf}_{n \to \infty} A_n$
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can one explain the differences from the sequence of say real numbers? – Seyhmus Güngören Apr 12 '15 at 16:42
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You dont know definitions of $\limsup$ and $\liminf$ of set sequence or you ask why we cant translate proof from real numbers to sets? – Mykola Pochekai Apr 12 '15 at 16:50
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I know the definitions, the second one. Like just taking a sine function for which we have liminf=-1 but limsup=1. Every value of a sine function can be a singleton set. – Seyhmus Güngören Apr 12 '15 at 16:52
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Because proof of real numbers based on linear order and completness theorem and we dont have something like that for sets. – Mykola Pochekai Apr 12 '15 at 16:56
