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Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces with $(Y,d_Y)$ bounded. Let $C(X,Y)$ denote the set of all continuous functions from $X$ to $Y$. Let $d$ be the uniform metric on $C(X,Y)$, i.e. $d(f,g) = \sup_{x \in X} d_Y(f(x),g(x))$.

i) Show that if $(Y,d_Y)$ is complete then $(C(X,Y),d)$ is complete.

ii) Consider the map $R: C([0,1],\mathbb R) \to C((0,1), \mathbb R)$ which takes a map on $[0,1]$ to its restriction to $(0,1)$. Is the image of $R$ complete?

I've attempted the first part, but I just can't seem to get through the definitions so any help would be great. As for the second part, I really don't know how to approach this.

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Sketch:

  1. Suppose that $(f_n)$ is a Cauchy sequence in $C(X,Y)$.
    Informally, being a Cauchy sequence means that 'it wants to converge to a point but that point might be missing from the space'.
    Using the definition of $d$, show that $f_n(x)$ is a Cauchy sequence for all $x$, then use that $Y$ is complete to get $f(x)$.
    Finally, verify that $d(f_n,f)\to 0$ if $n\to\infty$.
  2. By the previous statement, as $\Bbb R$ is complete, $C([0,1],\Bbb R)$ is complete.
    Let $(g_n)$ be a Cauchy sequence in the image of the restriction map $R$, so that $g_n=R(f_n)$ for some $f_n:[0,1]\to\Bbb R$.
    By continuity of $f_n$, we can derive that $R$ is an isometry, i.e. it preserves distance: $d(f,g)=d(R(f),\,R(g))$.
    It follows that $(f_n)$ is also Cauchy, living in a complete space, so it converges to some $f$, and then of course $g_n\to R(f)$ will hold.
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  • $\begingroup$ could you help me put together a proof of the first part? I'm struggling with getting the specifics right $\endgroup$ – Tom Oct 28 '13 at 0:26
  • $\begingroup$ which part? $\ $ $\endgroup$ – Berci Oct 29 '13 at 21:44

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