Let \(X\) be a topological space and let \(f,g:X\to\mathbb R\) be continuous functions.
Show that the set \(L=\{p\in X:f(p)\le g(p)\}\) is a closed subset of \(X\).
Show that the function \(h:X\to\mathbb R\) given by \(h(p)=\min\{f(p),g(p)\}\) is continuous.
Solution
1. The function\[ f-g:X\longrightarrow\mathbb R \]is continuous.
The map \[ (f,g):X\longrightarrow\mathbb R^2, \qquad p\longmapsto(f(p),g(p)), \] is continuous because both coordinate functions are continuous. The subtraction map \[ s:\mathbb R^2\longrightarrow\mathbb R, \qquad s(u,v)=u-v, \] is continuous, and \[ f-g=s\circ(f,g). \] Hence \(f-g\) is continuous.
2. The set\[ L=\{p\in X:f(p)\le g(p)\} \]is closed in \(X\).
1. Thus \(L\) is the inverse image of a closed set under a continuous map, so \(L\) is closed.This proves part (a).
3. For all real numbers \(u,v\),\[ \min\{u,v\} = \frac{u+v-|u-v|}{2}. \]
If \(u\le v\), then \(|u-v|=v-u\), and the right-hand side is \[ \frac{u+v-(v-u)}2=u. \] If \(v\le u\), then \(|u-v|=u-v\), and the right-hand side is \[ \frac{u+v-(u-v)}2=v. \] In either case the value is \(\min\{u,v\}\).
4. The function\[ h(p)=\min\{f(p),g(p)\} \]is continuous.
By 3, \[ h = \frac{f+g-|f-g|}{2}. \] The functions \(f\) and \(g\) are continuous, as are addition, subtraction, the absolute-value map on \(\mathbb R\), and multiplication by the constant \(1/2\). Consequently the displayed combination is continuous. This proves part (b).