Tuesday, 12 April 2016

real analysis - lim inf, lim sup, limit points



Let t and s be real numbers with $t. Suppose that as n, a2nt and a2n1s.



True or False: It must be that lim infan=t and lim supan=s and (an) has no other limit points.



Prove the statement or give a counterexample.



I think true, based on the following example, which I adapted from a wikipedia image (where s=1, t=-1 and a2n1 is the pink sequence and a2n is the light blue sequence):




enter image description here



Hence, all of what the statement said is true... but i don't know how to prove it.


Answer



I will use the following definition of lim sup and lim inf:




Let E be the set of subsequential limits of (an). Then lim supnan:=supE and lim infnan:=infE.





It suffices to show that in the setting of your question, E={t,s}.
Clearly t,sE. Now suppose (ank) is a subsequence of (an). The sequence of indices nk must either contain infinitely many even numbers or infinitely many odd numbers (or both).



Edit: This implies that (ank) contains a further subsequence that either converges to s or to t. Thus if (ank) converges, it must be to either s or t. (But it might not converge.)



So E={t,s}.


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