# Supremum of reals

Hi this might seem quite trivial, but I'm a little stuck so any help please, thanks. Let $a\in\mathbb{R}$ and $\alpha\in(0,1)$. I need to show the obvious fact that $\sup\{\alpha. a|\alpha\in(0,1)\}=a$, i.e. for any other upper bound $b$, $a\leq b$. So I tried by contradiction, so took an upper bound $b$ and assumed $b<a$, but seem to get nowhere. So any help please with this simple problem, am probably mixing up some inequalities but just can't get it. Thanks, any help greatly apprecited.

First of all, $\sup \{\alpha \cdot (-1) \mid \alpha \in (0,1)\} = 0$, so you need $a \geq 0$. Moreover, $\alpha a \leq a$ for any $\alpha \in (0,1)$. Hence $\sup \{\alpha a \mid \alpha \in (0,1) \} \leq a$. If $b = \sup \{\alpha a \mid \alpha \in (0,1) \} \leq a < a$, then simply remark that $$0 < \bar{\alpha} = \frac{a+b}{2a}<1$$ and $$\bar{\alpha} a > b,$$ against the assumption that $b = \sup \{\alpha a \mid \alpha \in (0,1) \} \leq a < a$.