Proving a metric space is a complete metric space.

Question

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.

Answer 1

I guess that if $f\in B(X)$, then $f:X\to \mathbb R$. Moreover, the compactness of $X$ looks not important. Let $(f_n)$ be a Cauchy sequence. Since $f_n(x)$ is a Cauchy sequence for all $x$, it will converges to some $f(x)$. Let $\varepsilon >0$. Since $(f_n)$ is a Cauchy sequence, there is $N\in\mathbb N$ s.t. for all $x\in X$,

$$|f_N(x)-f_{N+r}(x)|<\varepsilon .$$ Taking $r\to \infty $ gives $$|f_N(x)-f(x)|<\varepsilon,$$ for all $x\in X$. Finally, if $x\in X$, $$|f(x)|\leq |f_N(x)-f(x)|+|f_N(x)|\leq \varepsilon +\sup_{x\in X}|f_N|<\infty .$$

Therefore, $f\in B(X)$.

Answer 2

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.