Let \(Y\) be an ordered set in the order topology. Let \(f, g: X \to Y\) be continuous.
Show that the set \(\ts{x \mid f(x) \leq g(x)}\) is closed in \(X\).
Let \(h: X \to Y\) be the function
\[ h(x) = \min\ts{f(x), g(x)}. \]
Show that \(h\) is continuous. [Hint: Use the pasting lemma.]
Solution
- Let \[ C=\{x\in X:f(x)\le g(x)\}. \] We show its complement is open. Suppose \(f(x)>g(x)\). If there is \(y\in Y\) with \[ g(x)<y<f(x), \] then continuity gives the open neighborhood \[ f^{-1}((y,\infty))\cap g^{-1}(( -\infty,y)) \] of \(x\), and every point in it satisfies \(f>g\).
If there is no point strictly between \(g(x)\) and \(f(x)\), then \(g(x)\) and \(f(x)\) are consecutive in the order. The rays \[ (g(x),\infty),\qquad (-\infty,f(x)) \] are disjoint open neighborhoods of \(f(x)\) and \(g(x)\) respectively. On the open neighborhood \[ f^{-1}((g(x),\infty))\cap g^{-1}((-\infty,f(x))) \] of \(x\), a point \(z\) has \(f(z)>g(x)\), hence \(f(z)\ge f(x)\), and \(g(z)<f(x)\), hence \(g(z)\le g(x)\); so \(f(z)>g(z)\). Thus \(\{f>g\}\) is open and \(C\) is closed.
- Put \[ C=\{f\le g\},\qquad D=\{g\le f\}. \] By part (a), \(C\) and \(D\) are closed and \(C\cup D=X\). On \(C\), \(h=f\); on \(D\), \(h=g\). These two definitions agree on \(C\cap D=\{f=g\}\). Since the restrictions are continuous, the pasting lemma for two closed sets (E-2CPNC) gives that \(h\) is continuous.