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Showing a sequence converges to its supremum

  • #1
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Homework Statement

: [/B]Let a = sup S. Show that there is a sequence x1, x2, ... ∈ S such that xn converges to a.


Homework Equations

: [/B]I know the definition of a supremum and convergence but how do I utilize these together?


The Attempt at a Solution

:[/B] Given a = sup S. We know that a = sup S if: 1) a ∈ S and a is called an upper bound, and 2) if b is also an upper bound, then b ≥ a. Since a = sup S, given ε>0 and the xn ∈ S, we know that a - ε < xn ≤ a since a is a least upper bound. This means that since xn ∈ S and a = sup S, that xn can never exceed the value of a, given as sup S.


** I am stuck, any help is beneficial.
 

Answers and Replies

  • #2
andrewkirk
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Try proof by contradiction. First write down the definition of convergence. Then think about what it means if no sequence converges to a. In that case every sequence will have a certain property, and that shared property will challenge the notion that a is a supremum.
 
  • #3
Ray Vickson
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Homework Statement

: [/B]Let a = sup S. Show that there is a sequence x1, x2, ... ∈ S such that xn converges to a.


Homework Equations

: [/B]I know the definition of a supremum and convergence but how do I utilize these together?


The Attempt at a Solution

:[/B] Given a = sup S. We know that a = sup S if: 1) a ∈ S and a is called an upper bound, and 2) if b is also an upper bound, then b ≥ a. Since a = sup S, given ε>0 and the xn ∈ S, we know that a - ε < xn ≤ a since a is a least upper bound. This means that since xn ∈ S and a = sup S, that xn can never exceed the value of a, given as sup S.


** I am stuck, any help is beneficial.
You will have trouble proving this, because it is false. Here is a simple counterexample:
[tex] S = \{1, 1/2, 1/4, 1/8, 1/16, \ldots \} [/tex]
We have ##a = \sup S = 1##, but there is no subsequence of ##S## that converges to 1.

Now, if you had been speaking of ##\limsup S## instead of ##\sup S## it would have been a different story.
 
  • #4
andrewkirk
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You will have trouble proving this, because it is false. Here is a simple counterexample:
[tex] S = \{1, 1/2, 1/4, 1/8, 1/16, \ldots \} [/tex]
We have ##a = \sup S = 1##, but there is no subsequence of ##S## that converges to 1.
My interpretation of the OP was that S is a set (ie unordered) not a sequence. There is a sequence in the set [tex] S = \{1, 1/2, 1/4, 1/8, 1/16, \ldots \} [/tex] that converges to 1, which is the sequence ##x_n=1\forall n##..
 
  • #5
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You can consider the case ##a\in S##, in which case the constant sequence works.
For the case ##a\notin S##, you should argue that there is always something in ##S## between ##a-\frac{1}{n}## and ##a##, for ##n## big enough
 
  • #6
Ray Vickson
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My interpretation of the OP was that S is a set (ie unordered) not a sequence. There is a sequence in the set [tex] S = \{1, 1/2, 1/4, 1/8, 1/16, \ldots \} [/tex] that converges to 1, which is the sequence ##x_n=1\forall n##..
What I wrote was a set, not a sequence, although it might look like one.

Of course the sequence ##x_n = 1 \; \forall \; n## converges to the sup, but I suspect that is not what the questioner had in mind; after all, that would make every point of every set a limit point, and that would more-or-less throw out any or all useful concepts in point-set Topology. Although, to be fair---who knows what the questioner really wanted, or even if the OP stated the problem accurately?
 

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