Prove that $\sup S=\sup R,$ if $S\subseteq\Bbb{R}$ is bounded from above and $R\subseteq S$.

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Please, is this correct?

Let $S$ be a subset of the real numbers bounded from above and let $R\subseteq S$ satisfy the following condition: $\forall\;x\in S,\;\exists\; r\in R\;\text{such that}\;x\leq r$.
I want to prove that \begin{align} \sup S=\sup R.\end{align}

PROOF

Let $x\in S$ be arbitrary, then $\exists\; r\in R\;\text{such that}\;x\leq r.$ The number, $r$, is an upper bound for the set $S.$ So, $\sup S\leq r,\;\text{for some} \; r\in R.$ By definition of $\sup,\;\;y\leq\sup R, \forall\; y\in R.$ In particular, $y=r,\;r\leq\sup R.$ Thus,
\begin{align}\tag{1} \sup S\leq r\leq\sup R.\end{align}
Since $R\subseteq S$, we have
\begin{align}\tag{2} \sup R\leq \sup S.\end{align}
Thus, \begin{align} \sup S=\sup R.\end{align}

Best Answer

It seems you confuse "$\color{blue}{\forall}\;x\in S,\;\color{blue}{\exists}\; r\in R\;\text{such that}\;x\leq r$" with "$\color{blue}{\exists}\; r\in R\; \color{blue}{\forall}\;x\in S,\;\text{such that}\;x\leq r$".

The sets $S = (0,1)$ and $R= S \cap \mathbb{Q}$ satisfy the condition: $\color{blue}{\forall}\;x\in S,\;\color{blue}{\exists}\; r\in R\;\text{such that}\;x\leq r$. But your conclusion $\sup S \leq r$ is not true for any $r \in R$.

To show that $\sup R = \sup S$ you may proceed as follows:

  • Show $\forall \epsilon > 0 \exists x_{\epsilon}\in R: \sup S - \epsilon < x_{\epsilon}$ $$\Rightarrow \forall \epsilon >0: \sup S - \epsilon < \sup R \Rightarrow \sup S \leq \sup R$$ That's exactly what was to be shown.
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