Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.