Let A be a nonempty and bounded subset on $\mathbb{R}$ and $\lambda \in \mathbb{R}$ whit $\lambda\leqslant 0$. We difined a set $\lambda$A := { $\lambda$A : $a$ $\in$ A } Prove that sup($\lambda$A) = $\lambda$ ínf A and ínf($\lambda$A) = $\lambda$ sup A
we got a set $\lambda$A = { $\lambda$A : $a$ $\in$ A }, which is bounded because A is bounded, and how is it bounded the set got sup and inf. Because A is nonempty, A got a $a \in$ A, so $\lambda$A its nonempty Because A is bounded and got a inf, exist a $a' \in \mathbb{R}$ such that $a' \leqslant a $ si $\lambda a'\geqslant \lambda a$ for all $a \in$ A. For that reason $\lambda$A got a sup.
Let $\alpha = sup(\lambda A)$, so $\alpha$ its a upper bound of $\lambda A$ and if we do $\lambda \alpha = \lambda sup(\lambda A)$ so $\lambda \alpha$ its a lower bound of A. Besides $\beta$ its any lower bound of $\lambda A$ for $\lambda \beta \geqslant \alpha$ and $\lambda \alpha \geqslant \beta$. For that $ \lambda \alpha$ its a inf of A
***Hi I got this problem and my solution but my teacher told me that's wrong. But I don´t know why. I hope you can help me
If $\lambda=0$, $\sup{(\lambda A)}=\lambda\inf A=0$.
Now, if $\lambda<0$, define $\gamma=-\lambda$. So, $\gamma>0$. Now we find $\sup{(\lambda A)}=\sup{(-\gamma A)}=\sup\{-\gamma A: a\in A\}=-\inf\{\gamma A: a\in A\}=-\gamma\inf\{A: a\in A\}=\lambda \inf A$.
We can do a similar thing for the infimum.