Let f, g : X → R be two continuous functions defined on a metric space X.
(i) Show that the set U = {x ∈ X : f(x) > g(x)} is open in X.
(ii) Show that the set F = {x ∈ X : f(x) ≥ g(x)} is closed in X.
(iii) Show that the set G = {x ∈ X : f(x) = g(x)} is closed in X.
I am trying to use the definition of continuity of a function, but I get stuck after that. Can someone help and explain the process of solving these 3 sub problems?
Best Answer
To i)
Assume that $x_0\in U,$ that is, $f(x_0)-g(x_0)=2\epsilon>0.$ Now, since $f,g$ are continuous at $x_0$ there exists
$$\delta_1>0 : d(x,x_0)<\delta_1 \implies |f(x)-f(x_0)|<\epsilon \implies f(x)>f(x_0)-\epsilon,$$
$$\delta_2>0 : d(x,x_0)<\delta_2 \implies |g(x)-g(x_0)|<\epsilon \implies g(x)<g(x_0)+\epsilon.$$
Thus,
$$d(x,x_0)<\delta=\min\{\delta_1,\delta_2\}\implies g(x)<g(x_0)+\epsilon=f(x_0)-\epsilon<f(x).$$
That is, $B(x_0,r)\subset U.$
To ii)
Show that $F^c$ is open using $i).$
To iii)
Show that $G^c$ is open using $i)$ and the intersection of two adequate sets.