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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.

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closed as off-topic by Hayden, user91500, R_D, Claude Leibovici, zhoraster Jan 16 '17 at 9:33

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  • $\begingroup$ Mate, you must show your thoughts on the problem and some partial progress you've made. Cheers $\endgroup$ – mathworker21 Jan 15 '17 at 21:36
  • $\begingroup$ My guess is that there already are questions on this site about the same problem. You could check the list of related question in the sidebar on the right. Or you can look at results of this Approach0 search. It returns, for example, this question: Constant times Supremum. $\endgroup$ – Martin Sleziak Jan 16 '17 at 6:58
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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 $A=\left\{\dfrac{1}{r}B:b\in B\right\}$.

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  • $\begingroup$ do we really need a convers implication here, if you have shown equality? $\endgroup$ – pindakaas Feb 13 '17 at 16:22
  • $\begingroup$ @pindakaas, I have shown that if $\sup A$ does exist, then $\sup B$ also exists. The converse is very simple, but formally the proof is needed. $\endgroup$ – szw1710 Feb 13 '17 at 18:40
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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 $\sup B \leq b$?

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