# Supremum and infimum of cA

#### Alexmahone

##### Active member
If $c>0$, prove that

$\sup cA=c\sup A$ 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 $\sup cA=c\sup A$.

-------------------------------------------------------------------

$x\ge\inf A$ 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 $\inf cA=c\inf A$.

-------------------------------------------------------------------

Is that ok?

#### Plato

##### Well-known member
MHB Math Helper
If $c>0$, prove that
$\sup cA=c\sup A$ 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 $\sup cA=c\sup A$.
It is not clear that you have shown $\sup cA=c\sup A$.
It is clear that $\sup cA\le c\sup A$.
What if $\sup cA < c\sup A~?$

#### Alexmahone

##### Active member
It is not clear that you have shown $\sup cA=c\sup A$.
It is clear that $\sup cA\le c\sup A$.
What if $\sup cA < c\sup A~?$
The statement $x\le cb$ for all $x\in cA\implies c\sup A\le cb$ takes care of that.

This basically means that if there is another upper bound, it will be $\ge c\sup A$. So, $c\sup A$ is the smallest upper bound of $cA$.

Last edited: