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

Conversely, suppose $-\mathrm{\ell }$ is an upper bound of $-S$. For $s\in S$ we have $-s\in -S$, so $-s\le -\mathrm{\ell }$, and applying 2.1 gives $\mathrm{\ell }\le s$; hence $\mathrm{\ell }$ 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.