Negative of Supremum

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Theorem

Let $S$ be a subset of the real numbers $\R$.

Let $S$ be bounded above.


Then:

  • $\left\{{x \in \R: -x \in S}\right\}$ is bounded below
  • $\displaystyle -\sup_{x \in S} x = \inf_{x \in S} \left({-x}\right)$.


Proof

Let $B = \sup S$.

Let $T = \left\{{x \in \R: -x \in S}\right\}$.

Since $\forall x \in S: x \le B$ it follows that $\forall x \in S: -x \ge -B$.

Hence $-B$ is a lower bound for $T$.

Thus $\left\{{x \in \R: -x \in S}\right\}$ is bounded below.


If $C$ is the largest lower bound for $T$, it follows that $C \ge -B$.

On the other hand:

$\forall y \in T: y \ge C$

Therefore:

$\forall y \in T: -y \le -C$

Since $S = \left\{{x \in \R: -x \in T}\right\}$ it follows that $-C$ is an upper bound for $S$.

Therefore $-C \ge B$ and so $C \le -B$.

So $C \le -B$ and $C \ge -B$ and the result follows.

$\blacksquare$


Sources