Let $f : X \to Y$ be a
D18: Map such that
| (i) |
\begin{equation}
E
\subseteq F
\subseteq Y
\end{equation}
|
If $f^{-1}(E)$ is empty, then result
R7: Empty set is subset of every set establishes the claim so that we may assume that $f^{-1}(E)$ is nonempty. If
\begin{equation}
x \in f^{-1}(E) = \{ z \in X : f(z) \in E \}
\end{equation}
then $f(x) \in E$. Since $E$ is contained in $F$, then the transitivity of the subset relation guarantees that $f(x) \in F$ and therefore $x \in f^{-1}(E)$. Since $x \in f^{-1}(E)$ was arbitrary, we have the inclusion $f^{-1}(E) \subseteq f^{-1}(F)$,. $\square$
