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- #1

#### Alexmahone

##### Active member

- Jan 26, 2012

- 268

$\sup cA=c\sup A$ and $\inf cA=c\inf A$

**My proof:**$x\le\sup A$ for all $x\in A$.

$cx\le c\sup A$ for all $x\in A$ ie $x\le c\sup A$ for all $x\in cA$. ------ (1)

$x\le b$ for all $x\in A\implies\sup A\le b$

$cx\le cb$ for all $x\in A\implies c\sup A\le cb$

$x\le cb$ for all $x\in cA\implies c\sup A\le cb$ ------ (2)

From (1) and (2), we see that $\sup cA=c\sup A$.

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$x\ge\inf A$ for all $x\in A$.

$cx\ge c\inf A$ for all $x\in A$ ie $x\ge c\inf A$ for all $x\in cA$. ------ (3)

$x\ge b$ for all $x\in A\implies\inf A\ge b$

$cx\ge cb$ for all $x\in A\implies c\inf A\ge cb$

$x\ge cb$ for all $x\in cA\implies c\inf A\ge cb$ ------ (4)

From (3) and (4), we see that $\inf cA=c\inf A$.

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Is that ok?