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Question:
I need some help in proving that $\color{green}{\text{if $k\geq 0$, then $\sup (kS) = k\sup(S)$ and $\inf (kS) = k\inf (S)$}}$, and also that $\color{red}{\text{if $k<0$, then $\sup(kS)=k\inf(S)$ and $\inf(kS)=k\sup(S)$}}$.
Thoughts:
As far as I can tell, $S$ is a non-empty, bounded subset of $\mathbb{R}$, and $kS=\{ks:s\in S\}$, where $k\in \mathbb{R}$, so $kS\subseteq\mathbb{R}$. I suspect that we are working with the ordered set $(\mathbb{R},\leq)$, where, again, $kS\subseteq \mathbb{R}$, so if an element $\alpha\in \mathbb{R}$ is the supremum of $kS$ in $\mathbb{R}$, then $\alpha$ is an upper bound of $kS$ in $\mathbb{R}$, and $\alpha\leq \beta$ for all upper bounds $\beta$ of $kS$ in $\mathbb{R}$, and if and element $\alpha'\in\mathbb{R}$ is the infimum of $kS$ in $\mathbb{R}$, then $\alpha'$ is a lower bound of $kS$ in $\mathbb{R}$, and $\beta'\leq\alpha'$ for all lower bounds $\beta'$ of $kS$ in $\mathbb{R}$.
$$\star\star\star\star\star\star\star\star\star\star$$
Here, an upper bound is an element $\gamma\in\mathbb{R}$ such that for all $\zeta\in kS$ we have $\zeta\leq\gamma$. Further, a lower bound in this case is an element $\gamma'$ such that for all $\zeta'\in kS$ we have that $\gamma'\leq \zeta'$.
Comment:
Please let me know if I have the foundational ideas here down, so I can begin working on this. I want to get it solid.
Edit:
So it turns out that this problem was pushed to the next homework set, so I get to keep working on it. So far this is what I've got:
Consider the case where $k=0$. This means $kS=0S=\{0\}$, and so we have that
$$\sup(kS)=\sup\{0\}=0=0\sup(S)=k\sup(S).$$
Now consider the case where $k>0$. Note that $\sup(kS)$ is the element in $\mathbb{R}$ such that
$\hspace{0.25cm}$(a) $\sup(kS)$ is an upper bound for $kS$, and
$\hspace{0.25cm}$(b) if $\alpha$ is any upper bound for $kS$, then $\sup(kS)\leq \alpha$,
so if is sufficient to show that $k\sup(S)$ is the element in $\mathbb{R}$ such that
$\hspace{0.25cm}$(a') $k\sup(S)$ is an upper bound for $kS$, and
$\hspace{0.25cm}$(b') if $\alpha$ is any upper bound for $kS$, then $k\sup(S)\leq \alpha$,
namely that properties (a) and (b) hold for $\sup(kS)=k\sup(S)$. Thus, if $k\sup(S)$ is an upper bound for $kS$, then there is an $s\in S$ such that $\beta=ks$, where $\beta\in kS$. This would mean $s\leq \sup (S)$ because $\sup(S)$ is an upper bound for $S$, which means $\beta=ks\leq k\sup(S)$, and so $k\sup(S)$ is an upper bound for $kS$, as $\beta$ was arbitrary. Now in order to show that if $\alpha$ is any upper bound for $kS$, then $k\sup(S)\leq \alpha$, first suppose that $\alpha$ is an upper bound for $kS$. If this were the case, then $ks\leq \alpha$ for all $s\in S$, and so $s\leq\frac{\alpha}{k}$ for all $s\in S$, which therefore means that $\frac{\alpha}{k}$ is an upper bound for $S$; hence $\sup(S)\leq\frac{\alpha}{k}$. Multiplication by $k$ on both side of this inequality gives us that $k\sup(S)\leq \alpha$; quod erat demonstrandum. Similarly, to show $\inf(kS)=k\inf(S)$ there are two case to consider, namely $k=0$, and $k>0$. If $k=0$, then $kS=0S=\{0\}$, and so we have that
$$\inf(kS)=\inf\{0\}=0=0\inf(S)=k\inf(S).$$
Suppose now that $k>0$. In this case, just as above, is is sufficient to show that
$\hspace{0.25cm}$(a) $k\inf(S)$ is a lower bound for $kS$, and
$\hspace{0.25cm}$(b) if $\alpha$ is any lower bound for $kS$, then $\alpha\leq k\inf(S)$.
Thus if $k\inf(S)$ is a lower bound for $kS$, then there is an $s\in S$ such that $\beta=ks$, where $\beta\in kS$. Therefore $s\geq\inf(S)$ because $\inf(S)$ is a lower bound for $S$, and so $\beta=ks\leq k\inf(S)$, which means $k\inf(S)$ is a lower bound for $kS$ since $\beta$ was arbitrary. Now to show (b), suppose that $\alpha$ is a lower bound for $kS$. In this case we have that $ks\leq\alpha$ for all $s\in S$, and so $s\leq\frac{\alpha}{k}$ for all $s\in S$; hence $\frac{\alpha}{k} $ is a lower bound for $S$, and so we have that $\inf(S)\leq\frac{\alpha}{k}$. If we multiply by $k$ on both side of this inequality we get that $k\inf(S)\leq \alpha$; quod erat demonstrandum.
Comment:
$\color{green}{\text{Above}}$ is the essence of the kind of proof I'd like to exhibit for the $\color{red}{\text{second part}}$. I still haven't got the second part, so any help would be appreciated.
Note:
If you are from UCLA, or anywhere, please don't just copy and paste this as your answer.
Best Answer
This answer assumes that $S$ is bounded.
For the first part check this answer. The second part is similar.
Third part:
$\bbox[5px,border:2px solid #0000FF]{k<0\implies \sup (kS)=k\inf (S)}$
Let $k<0$.
We'll be proving $\sup(kS)\leq k\inf(S)$ and $\sup (kS)\ge k\inf (S)$.
Let $y\in kS$. There exists $x\in S$ such that $y=kx$. Obviously $kx\leq k\inf (S)\iff x\ge \inf (S)$, (because $k<0$). Since $x\ge \inf (S)$ is, by definition of $\inf$, true it follows that $y=kx\leq k\inf (S)$. Since $y$ was taken arbitrarily, it we've proved that $k\inf (S)$ is an upper bound of $kS$ and therefore it is greater than (or equal to) the smallest upper bound, i.e., $\sup (kS)\leq k\inf (S)$.
$\therefore \sup(kS)=k\inf (S)$
The fourth part is similar.