Negative of Infimum is Supremum of Negatives

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

Let $$T$$ be bounded below.

Then:
 * $$\left\{{x \in \R: -x \in T}\right\}$$ is bounded above;
 * $$-\inf_{x \in T} x = \sup_{x \in T} \left({-x}\right)$$.

Proof
Let $$B = \inf T$$.

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

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

Hence $$-B$$ is an upper bound for $$S$$, and so $$\left\{{x \in \R: -x \in T}\right\}$$ is bounded above.

If $$C$$ is the smallest upper bound for $$S$$, it follows that $$C \le -B$$.

On the other hand, $$\forall y \in S: y \le C$$.

Therefore $$\forall y \in S: -y \ge -C$$.

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

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

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