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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!

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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 $f_n$...)

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  • $\begingroup$ 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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