Let $X$ be a compact metric space, and let $B(X)$ be the set of real-valued bounded
functions on $X$. For any $f, g ∈ B(X)$, define $$d_B(f, g) :=\sup _{x\in X}\left | f(x)-g(x) \right |$$
Suppose, we already know $(B(X),d_B)$ is a metric space.
Prove that $B(X)$ is a complete metric space.
My idea (Using the compactness of $X$.)
Since $X$ is a compact metric space, any sequence $x_n$ in $X$ must
converge to some point $a$ in $X$ because $X$ is sequentially compact.Then, for any function $f$ in $B(X)$, $f(x_n)$ will converge to
$f(a)$.Thus, any sequence $f(x_n)$ converges to some constant function in
$B(X)$.Hence, $B(X)$ is a complete metric space.
But, I am not sure if this idea works to prove the statement because I didn't even use a condition of Cauchy sequence.
Best Answer
Compactness of $X$ does not play a rôle. If $(f_n)_n $ is a Cauchy sequence in $B(X)$, note that for each fixed $x \in X$, the sequence $(f_n(x))_n$ is Cauchy in $\Bbb R$. Now use completeness of $\Bbb R$ to have a candidate $f \in B(X)$ to converge to, and finally show that it does converge to that $f$.
Note that you're given a sequence of fucntions, not a sequence of points of $X$. The domain is actually not that relevant.