Infimum of $B = \{ -x \mid x \in A \}$

Question

Assume that $A$ is a non-empty set of real numbers and it's bounded above. The supremum of $A$ is $3$. Show using the definition that, $-3$ is the infimum of the set

$$B = \{ -x \mid x \in A \}$$

Answer 1

Check the following and make sure you understand it and can prove it:

$$3=\sup A\iff \forall\,\epsilon>0\;\exists\,a_\epsilon\in A\;\;s.t.\;\;3-\epsilon<a_\epsilon\le3\iff\forall\;\epsilon>0\,,\,\,\,-3+\epsilon>-a_\epsilon\ge-3\;$$

$$\text{and}\;\;-a_\epsilon\in-A=B\iff -3=\inf B$$

Answer 2

Since $\sup(A)=3\implies$ $x\leq3$ for all $x\in A$ so we have that $-3\leq-x$ for all $-x\in B$.

Thus we have that $-3$ is a lower bound for the set $B$. Hence $\inf(B)$ exists by Completeness.

Then assume for a contradiction that there exists a lower bound of $B$, call it $y$, such that $-3<y\leq-x$ for all $-x\in B$ $\iff$ $x\leq-y<3$ for all $x\in A$.

However, the latter statement is a contradiction since $\sup(A)=3$ so there cannot exist a $-y\in \mathbb{R}$ between $x$ and $3$ $\iff$ there does not exist a $y\in\mathbb{R}$ such that $-3<y\leq -x$ for $-x\in B$, hence $\inf(B)=-3.$

In general, it can be shown that if $S$ is bounded above, then $-S:=\{-s:s\in S\}$ is bounded below. So $\sup(S)\in\mathbb{R}$ and $\inf(-S)\in\mathbb{R}$. Furthermore, we have the relation: $$\sup(S)=-\inf(-S)$$

Answer 3

Hint: There are two parts to the definition of the supremum, as you have presented it. If $p$ is the supremum of $A$, then

1. $p$ is an upper bound of $A$. That is, if $a \in A$, then $a \leq p$.
2. $p$ is the smallest upper bound of $A$. That is, if $q$ is another upper bound of $A$ (which is to say that $a \leq q$ for every $a \in A$), then it must be the case that $p \leq q$.

Similarly, there are two parts to the definition of the infimum. Now, using the fact that $3$ satisfies the definition of "supremum" for $A$, show that $-3$ must satisfy the definition of "infimum" for $B$.

Answer 4

$\sup A =3$;

1) Then $a \le \sup A$ , $a \in A$, recall : $\sup A$ is the least upper bound of $A$.

It follows:

$-a \ge -\sup A$, $a \in A$;

$-\sup A$ is a lower bound of $B:= ${$-a |a \in A$}.

2) Need to show that $-\sup A$ is the greatest lower bound of $B$.

Assume there is a $L > -\sup A$ s.t.

$-\sup A < L \le -a$ , $a \in A$.

Then

$\sup A >-L \ge a$, $a \in A$ .

We have found a lesser upper bound of $A$, a contradiction.

Hence $L=-\sup A =-3 =\inf B$.