1. ## sequence question !

Let S be a nonempty bounded above subset of R . prove that there is a sequence $\displaystyle x_n \subseteq S$ of elements of S that converges to sup S

2. Originally Posted by flower3
Let S be a nonempty bounded above subset of R . prove that there is a sequence $\displaystyle x_n \color{red}\in\color{black} S$ of elements of S that converges to sup S
Let $\displaystyle s=\sup S.$ Then $\displaystyle \forall\,n\in\mathbb N,$ $\displaystyle \exists\,x_n\in S$ such that $\displaystyle s\geqslant x_n>s-\frac1n$ since $\displaystyle s-\frac1n$ is not an upper bound for $\displaystyle S.$ The sequence $\displaystyle \left(x_n\right)_{n\,=\,1}^\infty$ converges to $\displaystyle s.$

3. Originally Posted by proscientia
Let $\displaystyle s=\sup S.$ Then $\displaystyle \forall\,n\in\mathbb N,$ $\displaystyle \exists\,x_n\in S$ such that $\displaystyle s\geqslant x_n>s-\frac1n$ since $\displaystyle s-\frac1n$ is not an upper bound for $\displaystyle S.$ The sequence $\displaystyle \left(x_n\right)_{n\,=\,1}^\infty$ converges to $\displaystyle s.$
it's not belongs to !!!!
both $\displaystyle x_n \ and \ S$ are sets ,so the truth subsets

4. [PHP][/PHP] You said "a sequence of elements of S". Did you mean "a sequence of subsets of S"? sup S is definitely a number. How does a sequence of sets converge to a number? I think you have misunderstood the question.