Proving that there is a sequence in S, such that ##\lim s_n = \sup S##

[Hall]

Homework Statement

Let S be a bounded nonempty subset of R such that sup S is not in S. Prove there is a sequence (sn) of points in S such that lim sn = sup S.

Relevant Equations

sup S is the least upper bound.

Let $S=\{s_n:n∈N\}$. $\sup S$ is the least upper bound of S. For any ϵ>0, we have an m such that
$\sup S−\epsilon \lt s_m$
$\sup S−s_m \lt \varepsilon$
$|\sup S−s_m| \lt \varepsilon$

I mean to say that, no matter how small ϵ is, there is always an element of S whose distance from supS is less than that of ϵ, therefore there does exist a set of points which converge to supS, but I'm unable to get a thing like $n \gt N \implies |\sup S−s_n| \lt \varepsilon$. Because, assume for a given ϵ, $s_3$ lies within ϵ of $\sup S$, but $s_4,s_5⋯s_n$ all lies beyond the coverage of $\epsilon$

 

[pasmith]

You only need to show the existence of one sequence. It would be nice if you could just set s_n = \sup S - 2^{-n} but those points might not actually be in S. But you could use them as lower bounds for s_n...

 

[PeroK]

Let $S=\{s_n:n∈N\}$. $\sup S$ is the least upper bound of S. For any ϵ>0, we have an m such that

Are you trying to prove that every sequence in $S$ converges to $\sup S$?

 

[Hall]

You only need to show the existence of one sequence. It would be nice if you could just set s_n = \sup S - 2^{-n} but those points might not actually be in S. But you could use them as lower bounds for s_n...

That’s a good hint. How about
$$
s_n = \sup S - \frac{1}{n}$$?

as n grows, 1/n will get smaller and smaller and thus converging to $\sup S$.

 

[PeroK]

That’s a good hint. How about
$$
s_n = \sup S - \frac{1}{n}$$?

How do you know that $s_n \in S$?

 

[pasmith]

That’s a good hint. How about
$$
s_n = \sup S - \frac{1}{n}$$?

as n grows, 1/n will get smaller and smaller and thus converging to $\sup S$.

You can take any strictly positive, strictly decreasing sequence (a_n) whose limit is 0, but you can never be certain that \sup S - a_n \in S for every n \in \mathbb{N}. However, since \sup S - a_n < \sup S you do know that for every n \in \mathbb{N} there exists an s \in S such that \sup S - a_n < s < \sup S. What can you do with this observation?

 

[Hall]

You can take any strictly positive, strictly decreasing sequence (a_n) whose limit is 0, but you can never be certain that \sup S - a_n \in S for every n \in \mathbb{N}. However, since \sup S - a_n < \sup S you do know that for every n \in \mathbb{N} there exists an s \in S such that \sup S - a_n < s < \sup S. What can you do with this observation?

For very $n$ there exists an $s$ such that
$$
\sup S - a_n \lt s \lt \sup S$$
(Where $a_n$ is mono tonic decreasing sequence) implies that $s \in S$, all these $s$ may form a sequence.

But there can exist many points between $\sup S -a_n$ and $\sup S$, not just one $s$.

 

[pasmith]

But there can exist many points between $\sup S -a_n$ and $\sup S$, not just one $s$.

Choose one of them.