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Let $X$ be a set. Let $l^{\infty}(X,N)$ be all bounded functions on the form $f: X\longrightarrow N$. Let $d(f,g)=\sup\{n(f(x),g(x): x\in X)\}$ be a metric on $l^{\infty}(X,N)$, where $n$ is metric on $N$.

How to show that if $N$ is complete then $l^{\infty}(X,N)$ is complete?

Thank you.

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Did you want to write $n(f(x),g(x)): x\in X$ instead of $n(f(x),g(x): x\in X)$ (the position of brackets)? – Martin Sleziak Nov 8 '12 at 11:45
up vote 1 down vote accepted

I'll help you get started.

Start with a Cauchy sequence $(f_n)$ in $l^{\infty}(X,N)$. Then for any $\epsilon>0$ we have $N>0$ such that for any $n,m\geq N$ we have $\sup_{x\in X}n(f_n(x),f_m(x))<\epsilon$. Consequentially, the sequence $(f_n(x))$ is Cauchy in $N$ for any $x\in X$ (Why?) and converges, say to $y$. We have to find an $f\in l^{\infty}(X,N)$ such that $f(x)=y$. Well, we have a natural candidate, namely the function whose value at $x$ is the limit of the sequence $(f_n(x))$ in $N$. Now just check that $f$ is bounded and that the sequence $(f_n)$ converges to $f$ and you are done.

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