# Infinum & Supremum: An Analysis on Relatedness

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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.

• :-) How funny... – Trancot Aug 26 '13 at 18:00
• For anybody curious as to what $\overline{\mathbb{R}}$ is, click here. – Trancot Aug 26 '13 at 20:04
• You changed this from a problem statement question to a solution verificatio. It's my opinion that you should ask a new question for the solution verification as they are completly different in their nature. – Git Gud Aug 28 '13 at 13:43
• +300 bounty for this question???!!! – user66733 Aug 28 '13 at 23:49
• Is this homework? – Thomas Aug 29 '13 at 12:19

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 $\sup (kS)\ge k\inf (S)$.

1. Let's first prove that $\bbox[5px,border:1px solid #FF0000]{\sup(kS)\leq k\inf(S)}$.
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., $\sup (kS)\leq k\inf (S)$.
2. Now $\bbox[5px,border:1px solid #FF0000]{\sup (kS)\ge k\inf (S)}$. Since $\sup (kS)$ is an upper bound of $kS$ if follows that for all $y\in S, \,\sup(kS)\ge ky$. Therefore $\dfrac{\sup (kS)}{k}\leq y$, for all $y\in S$. We've thus shown that $\dfrac{\sup (kS)}{k}$ is a lower bound of$S$ and hence it is smaller than (or equal to) the largest lower bound of $S$, that is, $\dfrac{\sup (kS)}{k}\leq \inf (S)$. This way we get $\dfrac{\sup (kS)}{k}\leq \inf (S)$ and $\sup (kS)\ge k\inf (S)$.

$\therefore \sup(kS)=k\inf (S)$

The fourth part is similar.

• Look at my reputation: $8888$! – Git Gud Aug 26 '13 at 19:07
• I really hope there aren't any Burmese people reading this... If so, please excuse Git Gud, he didn't know. – Trancot Aug 26 '13 at 20:00
• @Trancot Why?${}$ – Git Gud Aug 26 '13 at 20:01
• I read foreign affairs/international relations, and I just remembered the 8888 Uprising. I don't think it really will offend because most people aren't aware of things like this. – Trancot Aug 26 '13 at 20:56
• Comment, downvoter? – Git Gud Aug 28 '13 at 23:54

From the edit part I think now you are looking for some ideas regarding the second part of the problem, so I'll skip the first part.

The second part is easy once you know that for any set $A \subseteq \mathbb{R}$ which is bounded from below we have $\sup(-A) = -\inf(A)$ where $-A = \{ -a: a \in A \}$. This is a problem in baby Rudin, chapter 1, problem 5 (if my memory is working correctly, even though as far as I remember he formulates the problem for the sets that are bounded from above which doesn't matter really).

So, I'll prove that $\sup(-S) = -\inf(S)$, because once this case is settled, you can prove the second statement by realizing that when $k<0$ then $(-k)>0$ and writing:

$\sup(k(-A)) = \sup(-kA) = (-k) \sup(A) = k (-\sup(A)) = k\inf(-A)$, now if you set $-A= S$ you'll get $\sup(kS) = k \inf(S)$

To prove that $\sup(-A) = -\inf(A)$ is true for any set which is bounded from below we'll just use the definitions of $\sup$ and $\inf$.

Suppose that $A \subseteq \mathbb{R}$ is a set which is bounded from below. Therefore $\inf(A)$ exists. Let $\inf(A) = \gamma$.

That means $\forall a \in A: \gamma \leq a$ by the definition of $\inf(A)$.

$\forall a \in A: \gamma \leq a \iff \forall (-a) \in (-A): -a \leq -\gamma$

Hence, $-\gamma$ is an upper bound for $-A$. We claim that this is the least upper bound of $-A$, i.e. $\sup(-A) = -\gamma$

To show that it's the least upper bound of $-A$, assume that it's not, therefore there exists $\gamma_0$ such that $\forall (-a) \in (-A): -a \leq -\gamma_0 < -\gamma$, but this forces that:

$\forall a \in A: \gamma < \gamma_0 \leq a$ which contradicts $\inf(A) = \gamma$.

Therefore $\sup(-A) = -\gamma = - \inf(A)$. And that completes the proof.

If some parts of my answer is not clear or incomplete, don't hesitate to ask.

EDIT:

In case you are looking for reference, here you are:

5. Let A be a nonempty set of real numbers which is bounded below. Let -A be the set of all numbers -x, where $x \in A$. Prove that inf $A$ = -sup(-$A$)

p22, ch. 1, exercise #5; Walter Rudin, Principles of mathematical analysis, McGraw-Hill, Inc.

• I see... Well, thank you, and BTW, you should never stop being creative. – Trancot Aug 29 '13 at 3:03
• You're welcome. Yea, that's indeed true. But also keep in mind that most of the times the easiest way to solve a problem is just to use the most straightforward way. Moreover, this is a relatively simple problem. There are much harder and way more challenging problems in analysis that need a lot of creativity, but this one is just a simple problem to see how you can relate things you need to know with the definitions you've learned. I'm going to delete my posts since this conversation is getting a bit too lengthy. – user66733 Aug 29 '13 at 3:11