show the supremum of the distance function of a compact metric space is finite

Question

Let $X$ be a compact topological space and $(Y,d)$ be a metric space. Show that for every pair of continuous functions $f\colon X\to Y$ and $g\colon X\to Y$, the extended real number $$ B=\sup\{d(f(x_1),g(x_2)):x_1,x_2\in X\}\in [0,\infty] $$ is, in fact, a real number.

So, I know that a continuous image of a compact set is compact and thus, $f(X)$ is compact and metrizable because it is a subset of a metric space. I'm not sure where to go from there, or if I'm even on the right track.

Answer 1

Hint: Take an arbitrary $x_0\in X$ and try to show that there exists an open ball centered at $x_0$ that contains the whole $X$.

Answer 2

$f(X)$ and $g(X)$ are compact, hence (totally) bounded. Pick an element $y_0\in Y$. Since $f(X)$ and $g(X)$ are compact, there exist $w\in f(X)$ and $z\in g(X)$ such that $d(y_0,w)=dist(y_0,f(X))$ and $d(y_0,z)=dist(y_0,g(X))$. Now let $x_1,x_2\in X$. We have:

$$d(f(x_1),g(x_2))\le d(f(x_1),y_0)+d(y_0,g(x_1))\le$$ $$d(f(x_1),w)+d(w,y)+d(y_0,z)+d(z,g(x_2))\le$$ $$diam(f(X))+dist(y_0,f(X))+dist(y_0,g(X))+diam(g(X))<\infty$$
So then the supremum is also bounded.

Answer 3

Consider the map $$ h\colon X\times X\to \mathbb{R} $$ defined by $$ h(x,y)=d(f(x),g(y)) $$ and prove it's continuous. Then apply the fact that $X\times X$ is compact.