Suppose S is bounded above, Prove $\inf(-x) = -\sup(x)$
I've reckon there are similar proofs here with negative signs in different ways, but I want to get some proof verification as well as some improvement if needs to be.
what i have done so far is:
Let $\sup(x) = B$ that allows us to say $B \le y$ for all $y \in S$
Thus we know that $-B \le -y$ for all $y \in -S$
Thus we know that $-B$ is a lower bound of $x$ and there exists $b \le -y$ that $y \le -b$ for all $y \in S$
hence $y \le B$, $-B \le - y$, $-B=-\sup(x) \le -y = \inf(-x)$
Any feedback is greatly appreciated!