# Sequence and supremum

• Mar 11th 2010, 07:35 PM
Slazenger3
Sequence and supremum
Suppose that S is a nonempty set of real numbers and that theta is an upper bound of S. Prove that the following conditions are equivalent:
(1) We have theta = Sup S.
(2) There exists a sequence (X sub n) in S such that (X sub n) approaches theta as n approaches infinity.
• Mar 11th 2010, 07:37 PM
Drexel28
Quote:

Originally Posted by Slazenger3
Suppose that S is a nonempty set of real numbers and that theta is an upper bound of S. Prove that the following conditions are equivalent:
(1) We have theta = Sup S.
(2) There exists a sequence (X sub n) in S such that (X sub n) approaches theta as n approaches infinity.

I disagree.
• Mar 11th 2010, 08:02 PM
davismj
Quote:

Originally Posted by Drexel28
I disagree.

I agree with Drexel. If you take a bounded sequence with a maximum and an infimum, then (1) does not imply (2).

In other words, if you choose a sequence such that you'll ever get sufficiently close $( < \varepsilon )$ to $\theta$ , the sequence cannot continue on until infinity. Or, if you choose a sequence that will continue on until infinity, it will always reside in the $\varepsilon$ -neighborhood of L.
• Mar 11th 2010, 08:19 PM
Drexel28
2 does not imply one. Consider $[0,1]=S$ and $x_n=\frac{1}{2}$. Then $\left\{x_n:n\in\mathbb{N}\right\}\subseteq S$ and $x_n\to\frac{1}{2}$ but surely $\frac{1}{2}\ne\sup\text{ }S$
• Mar 11th 2010, 08:35 PM
davismj
Quote:

Originally Posted by Drexel28
2 does not imply one. Consider $[0,1]=S$ and $x_n=\frac{1}{2}$. Then $\left\{x_n:n\in\mathbb{N}\right\}\subseteq S$ and $x_n\to\frac{1}{2}$ but surely $\frac{1}{2}\ne\sup\text{ }S$

There exists, not for all. This counterexample is false.

That is, there does exist a sequence in S, namely $(x_n) = 1 - \frac{1}{n}$, such that $(x_n) \to \theta$ . Now we must either prove that for any choice of S there exists a sequence like this, or that there exists an S for which no $(x_n)$ exists that implies (1)
• Mar 11th 2010, 08:37 PM
Drexel28
Quote:

Originally Posted by davismj
There exists, not for all. This counterexample is false.

What? 2 says that if I can find a sequence in $S$ whose limit is $\theta$ then $\theta=\sup\text{ }S$. How did my counterexample not work?
• Mar 11th 2010, 08:43 PM
davismj
Quote:

Originally Posted by Drexel28
What? 2 says that if I can find a sequence in $S$ whose limit is $\theta$ then $\theta=\sup\text{ }S$. How did my counterexample not work?

Yea, you're right. Somehow, I don't think that's what they meant, though.
• Mar 11th 2010, 08:49 PM
Drexel28
Quote:

Originally Posted by davismj
Yea, you're right. Somehow, I don't think that's what they meant, though.

Of course it's not what they meant. That doesn't mean we bear the burden of interpreting there obviously typo.