Proving $\sup X=-\inf(-X)$

Question

I would appreciate help with the following question

Let $X\subseteq\mathbb{R}$ be a bounded set. Define $-X=\{x\in\mathbb{R}\mid-x\in X\}.$ Prove $\sup X=-\inf (-X)$

Answer 1

There exists $a = \sup X$. Let $Y = - X = \lbrace x\in \mathbb{R} ; -x \in X\rbrace$. We have that $x \leq a$, for every $x \in X$ then it follows that $-a \leq -x$ which means that $-a$ is a lower bound to $Y$.

Now for every $\epsilon > 0$ we have that there exists $x \in X$ such that $a - \epsilon < x \Rightarrow -a + \epsilon > - x$, so there exists $-x \in Y$ such that $-x < -a + \epsilon$, by definition of infimum we have that $-a = \inf Y$ that is $\sup X = -\inf Y= \inf (-X)$.