# Showing a sequence converges to its supremum

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1. Oct 13, 2015

### Matt B.

1. The problem statement, all variables and given/known data: Let a = sup S. Show that there is a sequence x1, x2, ... ∈ S such that xn converges to a.

2. Relevant equations: I know the definition of a supremum and convergence but how do I utilize these together?

3. The attempt at a solution: 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.

2. Oct 13, 2015

### 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.

3. Oct 13, 2015

### Ray Vickson

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.

4. Oct 17, 2015

### andrewkirk

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 $x_n=1\forall n$..

5. Oct 17, 2015

### 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

6. Oct 17, 2015

### Ray Vickson

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?