Prove for $b<0$ and $bS:=\{bs:s\in S\}\Rightarrow \inf(bS)=b\cdot\sup (S)$ and $S$ is a nonempty bounded set in $\mathbb{R}$

My attempt:

Since $S$ is bounded above let $v=\sup S$ so $\forall s\in S$ we have $s\leq v$, then multiplying by $b<0$ we obtain $bs\geq bv$ so $b\cdot\sup(S)$ is a lower bound for the set $bS$ i.e. $\inf(bS)\geq b\sup(S)$. Unfortunately I struggle a lot with going the opposite direction since I never know when to start and in this case I'm not sure how to manipulate $|v-s|<\varepsilon$ to show that $inf(bS)=b\sup(S)$ since $b<0$.

  • $\begingroup$ It is easier to think of the case when $b = -1$. Showing $\inf X \leq a$ is equivalent to showing that for any $\epsilon > 0$, you can find $x \in X$ such that $x < a + \epsilon$. $\endgroup$ – An Hoa Nov 7 '15 at 23:28
  • $\begingroup$ Hmm I'm not entirely sure what you're suggesting. Is there not something I can start with to get the opposite inequality that produces $\inf(bS)\leq b\sup(S)$? $\endgroup$ – Craig Nov 7 '15 at 23:40
  • $\begingroup$ I managed to get it but thanks for the help. $\endgroup$ – Craig Nov 8 '15 at 1:01

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