For $E \subset \mathbb{R}$ let
$E' = \{x \in \mathbb{R} : (\exists a \in E)(x=-a) \}$
and
$E'' = \{x \in \mathbb{R} : (\forall a \in E)(x=-a) \}$
For example, $\{1,2\}' = \{-1,-2\}$ because, for the number $-1 \in \mathbb{R}$ there exists the element $1 \in \{1\}$ such that $-(-1)=1 \in \{1\}$ and for the number $-2 \in \mathbb{R}$ there exists the element $2 \in \{1,2\}$ such that $-(-2)=2 \in \{1,2\}$. But $\{1,2\}''= \varnothing$, because if there where some number $x \in \{1,2\}''$, the definition would require that for all element $a \in \{1,2\}$ we should have $x=-a$. So this implies that $x=-1$ and $x=-2$. Thus $-1=-2$, an absurd.