1. ## 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.

2. 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.

3. 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.

4. 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 $\frac{1}{2}\ne\sup\text{ }S$

5. 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)

6. 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?

7. 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.

8. 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.