Infinum & Supremum: An Analysis on Relatedness

Solution 1:

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.

Solution 2:

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.