one real analysis problem to prove the existence of a sequence and limit of a sequence

Question

Question: Suppose that S is a nonempty subset of R that is bounded above and put s = sup S.Show that there is a sequence (xn) such that xn ∈ S for all n and lim as n apporaches infinity xn = s.

Hi, I am a beginner for real analysis, the question about is the one that I am trying to do on my own, but I don't know how to analyze or solve it after I write the definitions out.

What I am thinking is that,

Given: S is a non-empty subset of R that is bounded above and s equals to the sup A

Prove: <1> there is a sequence (xn) such that xn belongs to s for all n, <2> limit of xn as n approaches to infinity is s

Thoughts: since S is a nonempty subset of R that is bounded above, then there exists a number t such that t is greater or equal to s for all s belongs to S, so we know t is an upper bound of s. Since s = sup S, then s is the least upper bound of S. since t is an upper bound for S, and s is the least upper bound, then t is also greater or equal to sup S. For the limit part, according to the definition of limit, we can write that |an-s|< epsilon.

I stop here because I don't how to do the following steps, I don't know how to use the information that I have so far to explain this questions.

Any help will be super appreciated

Thanks a lot

Answer 1

Let $s= \sup S$. We know that: $$ \text{ for any $\varepsilon>0$, there is $x\in S$ such that $s-\varepsilon < x$}. $$ Thus for each $n>0$ you can choose $x_n\in S$ such that $$ s - \frac 1 n < x_n. $$ Clearly, then, $s - \frac 1 n < x_n\le s$ for all $n>0$, so $\lim_n x_n$ exists and equals $s$.

Answer 2

There is an alternating definition for $\sup$ will be useful here, $s=\sup S$ iff 1) $s$ is an upper bound of $S$ and 2) for any $\epsilon>0$, $s-\epsilon$ is NOT an upper bound of $S$. (Try prove it using your definition)

Using above statement, for all $n\in\mathbb{N}$, $s-\frac{1}{n}$ is not an upper bound of $s$, which means $\exists x_n\in S$, such that $x_n>s-\frac{1}{n}$. Also since $s$ is an upper bound of $S$, $x_n\le s$. Hence we have $$s-\frac{1}{n}<x_n\le s,\forall n\in\mathbb{N}$$

Can you continue here to justify $x_n\to s$?

Answer 3

Hint: Put $s=\sup S$. If $s\in S$, define $x_n\equiv s$ and we are done.

Otherwise $s\notin S$. For each positive integer $n$, select $x_n\in S$ with $x_n\in (s-\frac1n,s)$ (why is this possible?). Then by construction $x_n \to s$ and each $x_n\in S$.