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Prove that a sequence of subsequential limits contains inf and sup 
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#1
Oct1706, 12:45 PM

P: 13

Okay. The problem I have is:
Let {x_n} be bdd and let E be the set of subsequential limits of {x_n}. Prove that E is bdd and E contains both its lowest upper bound and its greatest lower bound. So far, I have: {x_n} is bdd => no subseq of {x_n} can converge outside of {x_n}'s bounds=>E is bounded. Now, sse that y=sup(E) is not in E=> there is a z in E s.t. ye < z < y for some e > 0. Now, how would one proceed from here? 


#2
Oct1706, 03:34 PM

Sci Advisor
P: 6,068

You can generate a sequence of z's by using a sequence of e's that goes to 0. This sequence of z's the must converge to y.



#3
Oct1706, 04:36 PM

P: 13

I already had thought of that: you have y  e< z < y. Take e to be 1/k with e going to infinity, then {z_k} cgt to y, but what can we really conclude from that? Is there any guarentee that a {z_k} is in the original seq?



#4
Oct1806, 03:36 PM

Sci Advisor
P: 6,068

Prove that a sequence of subsequential limits contains inf and sup
If you can't find a z for any e>0, then y is > sup(E)



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