If the space $B(X,Y)$ of bounded function from $X$ into $Y$ is a complete metric space, show that $Y$ is a complete metric space.

Question

Suppose $X$ is a non empty set and $(Y,e)$ is a metric space. If the space $B(X,Y)$ of bounded function from $X$ into $Y$ with its usual supremum metric $s$ is a complete metric space, show that $Y$ is a complete metric space.

NOTE: The supremum metric $s$ is defined as $s(f,g) = \sup \{e(~f(x),g(x)~)~|~x \in X \}$

Solution: The book which I am reading specifies the following arguments.

I understand everything in the argument above except that the ball $b[f;e(a,b)/3]$ ( with center $f$ and radius $e(a,b)/3$) contains no member of $K$.

Suppose $b[f;e(a,b)/3]$ contains a constant function $f(x) = C ~~\forall x \in X$

Then, $\sup \{ e(~f(x),C~) ~|~x \in X\} \le e(a,b)/3$

$\implies e(~f(x),C~) \le e(a,b)/3~~\forall~x \in X$

Where could be the problem letting a constant function be in that ball?

Thanks for reading!

Answer 1

There is some $x_a\in X$ with $f(x_a)=a$ and similarly for $x_b$. If there were some constant function on the ball, we'd have

$$e(a,C)=e(f(x_a),C)\leq\frac13e(a,b)$$ $$e(b,C)=e(f(x_b),C)\leq\frac13e(a,b)$$

This implies

$$e(a,b)\leq e(a,C)+e(b,C)\leq \frac23 e(a,b),$$

contradicting $e(a,b)>0$ (because $a$ and $b$ are assumed to be distinct).