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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.

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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 $b = \sup \{\alpha a \mid \alpha \in (0,1) \} \leq a < a$.

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