Exercise 33.1

Proof. $\subseteq$. Suppose $x\in f^{-1}(r)$, i.e., $x\in U_r$ by definition, and $x\notin U_q$ for all $q<r$ by definition in Step 3. Then, $x\in \overline{U_r}\subseteq U_p$ for all $p>r$ by construction in Steps 1, 2.

$\supseteq$. Suppose $x\in {\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{p>r}{U}_p-{\bigcup }_{q<r}{U}_q$. This implies $x\in U_p\subseteq \overline{U_p}$ for all $p>r$, and so $f(x)\le r$ by Step $4(1)$, and also $x\notin U_q$ for all $q<r$, and so $f(x)\ge r$ by Step $4(2)$. Thus, $f(x)=r$. □