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.



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 $f_n$...)

  • $\begingroup$ Thanks for your reply! Why would you need to show that $f:=\lim\limits_{n\to\infty}f_n(x)$ exists? Since $f_n$ is Cauchy it converges and has a limit right? $\endgroup$ – titusAdam Mar 14 '18 at 13:33
  • $\begingroup$ Firstly, you have no x on the left and an x on the right of this equation, so this is a bit imprecise. Secondly, no, that's really part of the argument. You only know that Cauchy sequences in the real numbers always have a limit. $\endgroup$ – Luke Mar 14 '18 at 16:03
  • $\begingroup$ I don't think I understand what you mean with $x$ on the left and $x$ on the right. In this example we're talking about $B(\Omega)$, the collection of *real valued• bounded sequences, so shouldn't Cauchy sequences have a limit? Furthermore, suppose that we're not talking about Cauchy sequences in the real numbers, like you said. Then why are you allowed to say that by completeness of $\mathbb{R}$ there is a limit $\lim_{n\to\infty}f_n(x) = f(x)$ pointwise? Isn't this the same as saying Cauchy sequences have a limit in the real numbers? $\endgroup$ – titusAdam Mar 15 '18 at 7:41
  • $\begingroup$ "so shouldn't Cauchy sequences have a limit?" Well, they do, but in a proof you will also have to justify that. With the x on left and right: Always check if the objects in your equations are functions (Like $f, f_n$), objects in the space $B(\Omega)$, or numbers (Like $f(x), f_n(y),...$) in $\mathbb{R}$. $\endgroup$ – Luke Mar 15 '18 at 14:19
  • $\begingroup$ Hmm okay thanks! Could you show explicitly how you would solve step $5$? I can't figure it out myself. $\endgroup$ – titusAdam Mar 15 '18 at 15:13

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