prove $d(f,g):=\sup_{x \in X} d(f(x),g(x))$ is finite.

Question

Let $X$,$Y$ be metric space. define $A_b:=\{f:X \to Y | f(X) \text{ is bounded}\}$. for all $f,g \in A_b$, prove $d(f,g):=\sup_{x \in X} d(f(x),g(x))$ is finite.

Here is my attempt:

Since $f,g \in A_b$, for all $x,y \in X ,\enspace \sup \{d(f(x),f(y))\}$ and $\sup \{d(g(x),g(y))\}$ is finite.

Now I have, for all $x,y \in X$, $$\sup \{d(f(x),f(y))\}+\sup \{d(g(x),g(y))\} = \sup \{d(f(x),f(y))+d(g(x),g(y))\} \in \mathbb{R}$$

Then I would like to apply triangle inequality to the equation above to have $\sup \{d(f(x),g(x))\}$ is finite. i.e., I would like to use $$ d(x,z)\leq d(x,y)+d(y,z)$$ but we can't promise there exists a point that $M \in f(X) \cap g(X)$, thus I'm stuck here...

Any hints or solution will be greatly appreciated, thanks in advance.

Answer 1

Hint: $f(X) \cup g(X)$ is also bounded.