How would I go about finding lim sup $a_n$, where $\{a_n\}$ is an enumeration of the rationals in the interval $[0,1]$? Honestly, I don't even know where to start.
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For any $\epsilon\gt 0$, there is an infinite subsequence of ${a_n}$ such that each member of the subsequence is greater than $1-\epsilon$. – André Nicolas Oct 04 '15 at 23:32
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I can understand why that would be true, but how does that help me? – John Sawatzky Oct 04 '15 at 23:34
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I was trying not to give it away entirely! More concretely, there is an infinite subsequence with limit $1$, so the limsup is $1$. – André Nicolas Oct 04 '15 at 23:36
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Wouldn't the same be true for some other point on the interval, say 1/2? Sorry, I'm just doing my best to grasp what is going on. – John Sawatzky Oct 04 '15 at 23:47
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It is true for $1/2$, but the limsup is the largest possible limit of a subsequence. – André Nicolas Oct 04 '15 at 23:50
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Ah, of course. I think I'm starting to get it now, thank you! – John Sawatzky Oct 04 '15 at 23:50
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You are welcome. A fair number of the problems you are asked to do become clear once the meaning of the words has been internalized. – André Nicolas Oct 04 '15 at 23:54
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For any $\varepsilon>0$,the numbers $1-\dfrac1n$ such that $1-\varepsilon<1-\dfrac1n<1$ are an infinity, hence for any enumeration of the rationals in $[0,1]$, an infinity of them are in the enumeration with an index as large as we please. Thus $\limsup a_n=1$.
Bernard
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