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Question from Real Analysis by Haaser and Sullivan

Let X be the set of all continuous functions from [a,b] into $R^n$ and let $d$ be defined by $$d(f,g)=max(|f(t)-g(t)|:t\in[a,b]) $$ Show that (X,d) is a complete metric space.

What I need to show is that for every Cauchy sequence in X then the cauchy sequence converges to a point in X. Now I know $R^n$ is a complete metric space. So if $f_n(t)$ is a Cauchy sequence then $f_n(t)$ converges to some point say $f(t)\in R^n$. Now this means that $$d(f_n(t),f(t))\lt\partial.$$ Since $f$ continuous and a $0\lt\partial$ exists then $\forall \epsilon \gt 0$ $$d(f_n,f)\lt\epsilon.$$ Hence (X,d) is a complete metric space.

If someone could explain the difference between uniformly continuous and continuous? Also in general how one shows uniform continuity vs. continuity?

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    Your equation defining $d(x,y)$ doesn't have $x$ or $y$ appearing in the right hand side... I assume you meant $d(f,g)$?2012-11-29
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    @ZachL. Yes thank you for noting that.2012-11-29
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    The last paragraph is not related to the rest and should be postponed to another question. About the proof you propose: nothing guarantees a priori that the limit $f$ is continuous, in fact to show that it is is the heart of the proof you are asked to write down.2012-12-02

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