Prove that $sup S=sup R,$ if $SsubseteqBbb{R}$ is bounded from above and $Rsubseteq S$.
3
1
$begingroup$
Please, is this correct? Let $S$ be a subset of the real numbers bounded from above and let $Rsubseteq S$ satisfy the following condition: $forall;xin S,;exists; rin R;text{such that};xleq r$ . I want to prove that begin{align} sup S=sup R.end{align} PROOF Let $xin S$ be arbitrary, then $exists; rin R;text{such that};xleq r.$ The number, $r$ , is an upper bound for the set $S.$ So, $sup Sleq r,;text{for some} ; rin R.$ By definition of $sup,;;yleqsup R, forall; yin R.$ In particular, $y=r,;rleqsup R.$ Thus, begin{align}tag{1} sup Sleq rleqsup R.end{align} Since $Rsubseteq S$ , we have begin{align}tag{2} sup Rleq sup S.end{align} Thus, begin{align} sup S=sup R.end{align}
real-analysis analysis