Showing a sequence converges to its supremum

[Matt B.]

Homework Statement

: [/B]Let a = sup S. Show that there is a sequence x₁, x₂, ... ∈ S such that x_n 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 x_n ∈ S, we know that a - ε < x_n ≤ a since a is a least upper bound. This means that since x_n ∈ S and a = sup S, that x_n can never exceed the value of a, given as sup S.** I am stuck, any help is beneficial.

 

[andrewkirk]

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 Statement

: [/B]Let a = sup S. Show that there is a sequence x₁, x₂, ... ∈ S such that x_n 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 x_n ∈ S, we know that a - ε < x_n ≤ a since a is a least upper bound. This means that since x_n ∈ S and a = sup S, that x_n 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]

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 $x_n=1\forall n$..

 

[geoffrey159]

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 $x_n=1\forall n$..

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?