# $S$ be a non empty subset of $\mathbb{R}$ having supremum and $T=\{x\in\mathbb{R}: x-a \in S\}$, then $\forall a \in \mathbb{R}$, $\sup T=a+\sup S$.

Let $S$ be a non empty subset of $\mathbb{R}$ having supremum and $T=\{x\in\mathbb{R}: x-a \in S\}$, then $\forall a \in \mathbb{R}$ , $\sup T=a+\sup S$.

My proof:- Let $x\in T$

$\implies x-a≤ \sup S$

$\implies x≤ \sup S+a$

$\because \sup T$ act as an least upper bound of $T$

$\therefore \sup T≤a+\sup S$

We have that if $y \in S$ then $(y+a)-a \in S$
Thus $$(y+a) \in T \Rightarrow y+a=t \in T \Rightarrow y=t-a \leqslant \sup T -a$$
So $\sup S \leqslant \sup T-a \Rightarrow \sup S+a \leqslant \sup T$