- #1

Terrell

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- 26

## Homework Statement

The book I'm using provided a proof, however I'd like to try my hand on it and I came up with a different argument. I feel that something might be wrong.

Proposition: Let ##<X,d>## be a metric space, ##<Y,D>## a complete metric space. Then ##<C(X,Y), \sup D>## is a complete metric space. For emphasis, the metric of ##C(X,Y)## is ##\sup D[f(x),g(x)]## such that ##x \in X##. Also, ##C(X,Y)## is the space of bounded and continuous functions from ##X## to ##Y##.

## Homework Equations

N\A

## The Attempt at a Solution

Let ##(f_k(x))## be a Cauchy sequence in the complete metric space ##Y##. Hence, ##(f_k(x))\rightarrow g(x)## for some ##g(x) \in Y##. Keep in mind that this means, given ##\epsilon \gt 0##, for ##M\in\Bbb{N}##, ##\forall k\geq M##, we have ##D[g(x),f_k(x)] \lt \epsilon##, ##\forall x\in X##. Since ##f,g\in C(X,Y)## is bounded, then ##D[g(z),f_k(z)]=\sup\{D[g(x), f_k(x)] \vert x \in X\}## for some ##z \in X## and ##\forall k \in \Bbb{N}##. But note that ##\forall k\geq M##, ##\sup\{D[g(x), f_k(x)] \vert x \in X\} \lt \epsilon##. So ##C(X,Y)## must be complete.

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