Prove that a sequence of subsequential limits contains inf and sup

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Discussion Overview

The discussion revolves around proving that the set of subsequential limits of a bounded sequence contains both its infimum and supremum. Participants explore the implications of boundedness and the behavior of subsequences in relation to their limits.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant states that since the sequence {x_n} is bounded, no subsequence can converge outside of its bounds, implying that the set E of subsequential limits is also bounded.
  • Another participant suggests generating a sequence of values z that converge to the supremum y of E by using a sequence of ε values that approach 0.
  • A further contribution questions whether the constructed sequence {z_k} actually contains elements from the original sequence {x_n} and what conclusions can be drawn from this.
  • Another participant notes that if no z can be found for any ε > 0, then y must be greater than sup(E).

Areas of Agreement / Disagreement

Participants express uncertainty about the implications of their findings and whether the constructed sequences relate back to the original sequence. There is no consensus on how to proceed or what conclusions can be definitively drawn.

Contextual Notes

Participants have not resolved the relationship between the constructed sequences and the original sequence, leaving open questions about the guarantees of inclusion in the original sequence.

Sick0Fant
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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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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.
 
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 guarantee that a {z_k} is in the original seq?
 
If you can't find a z for any e>0, then y is > sup(E)
 

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