Bruce Harrisface
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If n is in the set of natural numbers, and An = [ -1/n , 2 ) when n is odd, and [ -3 , 1/n ) when n is even, how do I find the limsup and liminf of An?
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TeeEm
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I do not think there are any purists left here ...
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TheBBQ
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Try setting n = 2m+1 for odd and n = 2m where m is a natural number. Play around and see what you can deduce in each case.
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morgan8002
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They don't exist.
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Gregorius
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(Original post by Bruce Harrisface)
If n is in the set of natural numbers, and An = [ -1/n , 2 ) when n is odd, and [ -3 , 1/n ) when n is even, how do I find the limsup and liminf of An?
Let's take a look at the definition of liminf (and I'll leave limsup to you).

\displaystyle \lim \inf_{n \rightarrow \infty} A_n = \lim_{n \rightarrow \infty}\left(\inf_{m \ge n} A_m\right)

So, the first step is to work out what you get from

\displaystyle \inf_{m \ge n} A_m

Then let  n \rightarrow \infty
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atsruser
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(Original post by Gregorius)
Let's take a look at the definition of liminf (and I'll leave limsup to you).

\displaystyle \lim \inf_{n \rightarrow \infty} A_n = \lim_{n \rightarrow \infty}\left(\inf_{m \ge n} A_m\right)

So, the first step is to work out what you get from

\displaystyle \inf_{m \ge n} A_m

Then let  n \rightarrow \infty
Far be it from me to criticise you but I'm confused: he's dealing with sets and your definition is for sequences, no? Shouldn't there be some unions and intersections floating around?
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Gregorius
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(Original post by atsruser)
Far be it from me to criticise you but I'm confused: he's dealing with sets and your definition is for sequences, no? Shouldn't there be some unions and intersections floating around?
Goodness, I'm dozy this morning! Apologies OP, atsruser is quite right. For liminf

 \displaystyle \lim \inf_{n \rightarrow \infty} = \cup_{N=1}^{\infty} \cap_{m \ge N} A_m

then apply my previous remarks to the intersection on the right hand side.
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atsruser
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(Original post by Gregorius)
Goodness, I'm dozy this morning! Apologies OP, atsruser is quite right. For liminf

 \displaystyle \lim \inf_{n \rightarrow \infty} = \cup_{N=1}^{\infty} \cap_{m \ge N} A_m

then apply my previous remarks to the intersection on the right hand side.
Phew. I thought that there'd been a breakthrough at York that had shaken mathematics to its very foundations for a bit.

Still, yesterday, I wasted 15 minutes on one of those Australian A level questions by using the wrong value for \cos \frac{\pi}{4} - maybe this website is jinxed. I mean, it can't be *us*, can it?
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Gregorius
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(Original post by atsruser)
Maybe this website is jinxed. I mean, it can't be *us*, can it?
No, definitely not. Something is wrong with the balance of the universe this morning.
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