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If exists a set $X$ such that $\mathscr{B}(X;M)$ is complete, then $M$ is a complete metric space.

$\mathscr{B}(X;M)$ is the set of all bounded functions from $X\rightarrow M.$ In $\mathscr{B}(X;M)$ we do consider the sup metric.

I've tried a path which got me stuck. I'll post it here:

Suppose that $M$ is not complete and by contradition suppose $\mathscr{B}(X;M)$ complete. Then there exists a Cauchy sequence $(x_n)$ in $M$ which is not convergent there. For each $n\in \mathbb N$, consider $f_n$ such that $f_n(X)= \{x_n\}$. Then, $(f_n)$ is a cauchy sequence in $\mathscr{B}(X;M)$, because $(x_n)$ is a Cauchy sequence in $M$. Since $\mathscr{B}(X;M)$ is complete, we do know that $f_n \rightarrow f\in \mathscr{B}(X;M)$.

What now? How do i proceed (if possible)? Any help with an alternative way?

user2345678
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    What about considering a Cauchy sequence in $M$ and transferring it to a sequence of constant functions ? – Maxime Ramzi Nov 02 '17 at 17:59
  • Your proof is pretty much correct, but the proof by contradiction is superfluous. Get rid of it by just taking a sequence in $M$ and promoting it to a sequence int $B$, as Max suggests above. – ziggurism Nov 02 '17 at 18:00
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    You need to require $X \neq \varnothing$. $\mathscr{B}(\varnothing, M) = {\varnothing}$ is complete for every $M$. – Daniel Fischer Nov 02 '17 at 18:09
  • @DanielFischer That's true. The book does not state this hyphotesis, but i think that it is assumed, since it does make the problem more interesting. – user2345678 Nov 02 '17 at 18:15

2 Answers2

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Your idea works.

Let $(x_n)$ be a cauchy sequence in $M$. Take constant functions $f_n:X \to M$ given by $f_n(y)=x_n$ for all $y \in X$.

First note that ince $B(X,M)$ is complete, we ha $\|f_n-f_m\|=\|x_n-x_m\|$ so this is a cauchy sequence, so we have $f_n \to f$ by assumption.

Clearly $\sup_{y \in X}\|f_n(y)-f(y)\|:=\sup_{y \in X} \|x_n-f(y)\|\to 0$ as $n \to \infty$, so $\sup_{y \in X} \|f(y)\|$ is the limit.

Andres Mejia
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The limit of $f_n$ is $f$ i.e for every $c>0$ there exists $N$ such that $n>N$ implies that $d(f_n(x),f(x))=d(x_n,f(x))<c$ for every $x\in X$ i.e $x_n$ converges towards $f(x)$.

Tsemo Aristide
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