Suppose $A$ is a nonempty set of real numbers and $r > 0$. Let $B = \left\{ra : a ∈ A\right\}$. Show that $\sup A$ exists if and only $\sup B$ exists. Furthermore, if they exist, show that $\sup B = r \sup A$.
I can't seem to figure this question out.
On
I'm not sure why I did it this way, but it was the picture I had in my head, so I wrote it down.
Let $a$ be the supremum for $A$. Define $a_n \in A$ so that $a_n \in (a-\frac{1}{n},a]$. Note that $a_n \to a$ by construction.
Now, each $a_n$ has a respective $r \cdot a_n=b_n \in B$, which is also convergent, say $b_n \to b=r\cdot a$ ( which can be deduced using continuity.) Notice that $b \leq \sup B$,
How can we show that $\sup B \leq b$?
Assume that $a_0=\sup A$. Fix $b\in B$. We have $b=ra$ for some $a\in A$. Since $a\le a_0$, we get $b=ra\le ra_0$ and $ra_0$ is an upper bound of $B$. If there were less upper bound, say $c$, then $\frac{c}{r}$ were the upper bound of $A$, which is less then $a_0=\sup A$. So, $ra_0=\sup B$. For the converse implication observe that $A=\left\{\dfrac{1}{r}B:b\in B\right\}$.