Show that $(B(X), D)$ is complete.

Question

Exercise: Let $\Omega$ be a nonempty set. Let $B(\Omega)$ be collection of bounded real-valued functions on $\Omega$. Define $D:B(\Omega)\times B(\Omega)\to[0,\infty)$ by $$D(f,g) = \sup_{w\in\Omega}\left|f(w) - g(w)\right|$$ Show that $(B(\Omega), D)$ is complete.

What I've tried: I know that $(B(\Omega), D)$ is complete if every Cauchy sequence in $B(\Omega)$ converges to a point in $B(\Omega)$ in the metric $D$. Pick an arbitrary Cauchy sequence $(f_n)$ in $B(\Omega)$. I want to show that $(f_n)$ converges to a point in $B(\Omega)$ in the metric $D$.

Let $\lim\limits_{n\to\infty}f_n = f(x)$. Since $(f_n)$ is a sequence of real valued bounded functions, $(f_n)$ is real value and bounded for every $n\geq 1$. This means that $\lim\limits_{n\to\infty} f_n$ is real valued and bounded. Or, equivalently, for every $n\geq 1$ we have that $f_n\leq B$ for some value $B$. Since $f(x) = \lim_{n\to\infty}f_n$ we have $f(x) \leq B$.

Now finally, I need to show that $D(f_n,f)\to 0$ as $n\to\infty$. $D(f_n,f) = \sup_{w\in\Omega}\left|f_n(w) - f(w)\right|$. Since $f_n$ is Cauchy, we know that there exists an $N\in\mathbb{N}$ such that $\sup_{w\in\Omega}\left|f_n(w) - f_m(w)\right|<\epsilon$, whenever $m,n\geq N$. Now since $\lim_{n\to\infty}f_n$ satisfies $m > N$, we have that there exists an $N$ such that $\sup_{w\in\Omega}\left|f_n(w) - f\right|<\epsilon$ for $n\geq N$.

Question: What would be a more correct and rigorous proof to show that $(B(X), D)$ is complete? I think my proof is far too intuitive and really not sufficient.

Thanks!

Answer 1

Your proof has some correct parts, but it does not really follow the right logic. In particular, where does your function $f$ come from? You must show that it exists. I'll give you a hint:

1. Start with the assumption: Let $(f_n)_{n \in \mathbb{N}}$ be a Cauchy sequence in $B(\Omega)$.

2. Now you have to get to this $f$. To define it pointwise, you should show the following thing: It follows from 1 that at each fixed $x \in \Omega$, $f_n(x)$ is a Cauchy sequence in $\mathbb{R}$.

3. By completeness of $\mathbb{R}$, there is thus a limit $\lim_{n \to \infty} f_n(x) = f(x)$ pointwise.

4. Now you have the $f$ and can show as you do that it is bounded and thus in $B(\Omega)$.

5. For the last step, your proof that $f_n$ converges to $f$ is unclear to me. Try to really look at the $D$-norm of $|f_n - f|$ and find an argument why it goes to zero (Hint: This comes from point 2, and of course the Cauchy property of the $f_n$...)