# How can I show that the sup(S)=lim{Xn} and the inf(S)=lim{Yn} as n goes to infinity

## Homework Statement

How can I show that the sup(S)=lim{Xn} and the inf(S)=lim{Yn} as n goes to infinity for both of those limits?

We are assuming S is a nonempty bounded set that is a subset of the Real numbers. Also, {Xn} and {Yn} are monotone sequences that belong to the the set S.

## The Attempt at a Solution

Here is what I think I'm thinking,

We use the fact that S is a nonempty bounded subset of real numbers. Since we can use the axiom of completeness, we know that this set has a least upper bound and greatest lower bound.

let sup(S)=a and let inf(S)=b.

Furthermore let {Xn} be a monotone sequence in S. (without loss of generality assume its monotone increasing. A similar argument can be used for a monotone decreasing sequence.)

So, we have that x1<x2<x3<...<xn<.... We know that any sequence that is bounded and monotone is convergent. Hence, Lim{Xn} exists (IF you can't use this fact, you need to prove it through a proof by contradiction) We need to show that Lim{Xn}=a.

We know that xn<=a for all n. so x1<x2....<=a. (Do you see why this is true? Monotoness is everything here, if we didnt have that fact we couldn't use this. Consider the alternating harmonic sequence. This wouldnt hold there)

Suppose now, for contradictions sake, that lim{Xn}=a' where a'=a-e for some e>0. since Xn is monotone, and a' is the limit, then we have that x1<x2<....<a'=a-e. But then a' would be an upper bound for Xn, contradicting the fact that a is the sup(S). (Obviously a+e could not be the limit because then we can make the sequence arbitrarily close to a+e, thus being able to find an s in S that is greater than a, hence another contradiction)

One way to read is $\forall S = \{x_n\} \cup \{y_n\}: \sup S = \lim \limits_{n \rightarrow \infty}x_n, \inf S = \lim \limits_{n \rightarrow \infty}y_n$, which is not true even if we say that xn is convergent monotone increasing and yn is likewise decreasing. It would be true if we added that xn > yn for all n though.
Another way is $\forall S, \exists \{x_n\} \subset S: \sup S = \lim \limits_{n \rightarrow \infty}x_n$. I'm guessing this is the real one. The easiest way to prove this is to analyse a sequence $x_n = \sup S - \frac {\sup S - \inf S}{2^n}$. Edit: although this assumes that S is an interval, which is not given.
As for your proof, I'm not sure what interpretation you have. It's something in between. You first pick xn as any monotone increasing sequence and later claim that $\sup S = \sup \{x_n\}$. There exist monotone increasing sequences in S that don't have such property.