Reflection through zero exchanges upper and lower bounds

For all $x, y \in \mathbb{R}$ the field identity $y - x = (-x) - (-y)$ holds, and $-(-x) = x$.

The map $s \mapsto -s$ sends $S$ onto $-S$ and $-S$ onto $-(-S)$, and since $-(-s) = s$ it is a bijection of $S$ with $-S$ whose inverse is itself; hence $-(-S) = S$, and $S$ is nonempty exactly when $-S$ is nonempty.

For all $x, y \in \mathbb{R}$: $x < y$ holds exactly when $y - x$ is positive, which by 1.1 is exactly when $(-x) - (-y)$ is positive, which is exactly $-y < -x$; and $x = y$ holds exactly when $-x = -y$; hence $x \le y$ if and only if $-y \le -x$.

Suppose $u$ is an upper bound of $S$. Every element of $-S$ has the form $-s$ with $s \in S$, and $s \le u$ gives $-u \le -s$; hence $-u$ is a lower bound of $-S$.

Conversely, suppose $-u$ is a lower bound of $-S$. For $s \in S$ we have $-s \in -S$, so $-u \le -s$, and applying 2.1 to this inequality gives $s \le u$; hence $u$ is an upper bound of $S$. This together with 3.1 proves claim 1.

Suppose $\ell$ is a lower bound of $S$. For $s \in S$ we have $\ell \le s$, hence $-s \le -\ell$, and every element of $-S$ is such a $-s$; hence $-\ell$ is an upper bound of $-S$.

Conversely, suppose $-\ell$ is an upper bound of $-S$. For $s \in S$ we have $-s \in -S$, so $-s \le -\ell$, and applying 2.1 gives $\ell \le s$; hence $\ell$ is a lower bound of $S$. This together with 3.3 proves claim 2.

Claim 1 says the upper bounds of $S$ are exactly the negatives of the lower bounds of $-S$, so $S$ is bounded above exactly when $-S$ is bounded below; claim 2 says likewise that $S$ is bounded below exactly when $-S$ is bounded above; combining the two, $S$ is bounded exactly when $-S$ is bounded, and $-(-S) = S$ with $S$ nonempty exactly when $-S$ is nonempty.