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Prove that a sequence of subsequential limits contains inf and sup

by Sick0Fant
Tags: limits, prove, sequence, subsequential
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Sick0Fant
#1
Oct17-06, 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. y-e < z < y for some e > 0.

Now, how would one proceed from here?
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mathman
#2
Oct17-06, 03:34 PM
Sci Advisor
P: 6,080
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.
Sick0Fant
#3
Oct17-06, 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?

mathman
#4
Oct18-06, 03:36 PM
Sci Advisor
P: 6,080
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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