Showing a sequence converges to its supremum

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.

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andrewkirk
Homework Helper
Gold Member
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.

Ray Vickson
Homework Helper
Dearly Missed

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:
$$S = \{1, 1/2, 1/4, 1/8, 1/16, \ldots \}$$
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.

andrewkirk
Homework Helper
Gold Member
You will have trouble proving this, because it is false. Here is a simple counterexample:
$$S = \{1, 1/2, 1/4, 1/8, 1/16, \ldots \}$$
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 $$S = \{1, 1/2, 1/4, 1/8, 1/16, \ldots \}$$ that converges to 1, which is the sequence ##x_n=1\forall n##..

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

Ray Vickson
My interpretation of the OP was that S is a set (ie unordered) not a sequence. There is a sequence in the set $$S = \{1, 1/2, 1/4, 1/8, 1/16, \ldots \}$$ that converges to 1, which is the sequence ##x_n=1\forall n##..