Exercise 18.8

Proof. We prove this by showing that the complement $X-C$ is open in $X$. So first let $S$ be the set of all $y\in Y$ where $y$ has an immediate successor, and denote that successor by $y+1$. Then clearly $y+1$ is well defined for all $y\in S$. Now define

for $y\in S$. As these are both rays in the order topology $Y$, they are both basis elements and therefore open. It then follows that $f^{-1}(A_{>y})$ and $g^{-1}(A_{<y})$ are both open in $X$ since $f$ and $g$ are continuous. Hence their intersection $U_y=f^{-1}(A_{>y})\cap g^{-1}(A_{<y})$ is also open in $X$.

Similarly the rays

for $y\in Y$ are also open so that the intersection $V_y=f^{-1}(B_{>y})\cap g^{-1}(B_{<y})$ is open in $X$. Then clearly the union of unions

is also open in $X$. We claim that $X-C=D$ so that the complement is open in $X$ and hence $C$ is closed as desired.

$\text{(⊂)}$ First consider any $x\in X-C$ so that clearly $f(x)>g(x)$. If $g(x)$ has an immediate successor $g(x)+1$ then $g(x)\in S$ and we have $f(x)\in A_{>g(x)}$ so that $x\in f^{-1}(A_{>g(x)})$. We also have that $g(x)\in A_{<g(x)}$ since $g(x)<g(x)+1$, and hence $x\in g^{-1}(A_{<g(x)})$. It then follows that $x\in U_{g(x)}$ and hence ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{y\in S}{U}_y$ and $x\in D$ since $g(x)\in S$. If $g(x)$ does not have an immediate successor then there must be a $y\in Y$ where $g(x)<y<f(x)$. We then have that clearly $f(x)\in B_{>y}$ and $g(x)\in B_{<y}$ so that $x\in f^{-1}(B_{>y})$ and $x\in g^{-1}(B_{<y})$. Thus $x\in V_y$ so that $x\in {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{y\in Y}{V}_y$ and $x\in D$. This shows that $X-C\subset D$ since either way $x\in D$ and $x$ was arbitrary.

$\text{(⊃)}$ Now suppose that $x\in D$. If $x\in {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{y\in S}{U}_y$ when there is a $y\in S$ where $x\in U_y$. Hence $x\in f^{-1}(A_{>y})$ and $x\in g^{-1}(A_{<y})$ so that $f(x)\in A_{>y}$ and $g(x)\in A_{<y}$. From this, it follows that $f(x)>y$ and $g(x)<y+1$. Then it has to be that $g(x)\le y$ so that $f(x)>y\ge g(x)$. If $x\in {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{y\in Y}{V}_y$ then there is a $y\in Y$ where $x\in V_y$. Hence $x\in f^{-1}(B_{>y})$ and $x\in g^{-1}(B_{<y})$ so that $f(x)\in B_{>y}$ and $g(x)\in B_{<y}$. It then clearly follows that $f(x)>y$ and $g(x)<y$ so that $f(x)>y>g(x)$. Therefore in either case we have $f(x)>g(x)$ so that $x\in X-C$. This of course shows that $X-C\supset D$ since $x$ was arbitrary. □

Proof. Let $A=\left\{x\in X\mid f(x)\le g(x)\right\}$ and $B=\left\{x\in X\mid g(x)\le f(x)\right\}$, which are clearly both closed by part (a). It is easy to see that $X=A\cup B$. First, clearly, $X\supset A\cup B$ since both $A\subset X$ and $B\subset X$. Then, for any $x\in X$, it has to be that either $f(x)\le g(x)$ or $f(x)>g(x)$ since $<$ is a total order on $Y$. In the former case, of course, $x\in A$, and in the latter $x\in B$ so either way $x\in A\cup B$. Hence $X\subset A\cup B$. It is also easy to see that $f(x)=g(x)$ for every $x\in A\cap B$. For any such $x$, we have that $x\in A$ so that $f(x)\le g(x)$, and $x\in B$ so that $g(x)\le f(x)$. From this, it clearly must be that $f(x)=g(x)$.

Since $f$ and $g$ are continuous, it then follows from the pasting lemma that the function

for $x\in X$ is continuous as well. Based on the definitions of $A$ and $B$ it is then easy to see and trivial to show that $h(x)=\min <!--\; FUNCTION\; APPLICATION\; -->\left\{f(x),g(x)\right\}$ for all $x\in X$, which of course shows the desired result. □