# Prove that a sequence of subsequential limits contains inf and sup

by Sick0Fant
Tags: limits, prove, sequence, subsequential
 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?
 Sci Advisor P: 5,935 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.
 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?