Suppose that $S$ is a nonempty subset of $\mathbb{R}$ and $S$ is bounded above. Also suppose that $T=\{x\in\mathbb{R}:x$ is an upper bound of $S\}$ Show that $\sup(S)=\inf(T)$.
My proof: Since $S$ is nonempty bounded above subset of $\mathbb{R}$, $\sup(S)$ exists. Let $\sup(S)=\alpha$. It is clear that $\alpha\in T$ because $\alpha$ is an upper bound of $S$. Also, $\alpha\leq x$ for all $x\in T$ because $\alpha$ is a supremum. Since $\alpha\leq x$ and $\alpha\in T$, we can conclude that $\inf(T)=\alpha=\sup(S)$ as required.
Is this proof ok?