Let $(Y,\tau)$ be a topological space, with $Y\neq 0$ and let $(X,d)$ be a metric space. It is $C_b(Y,X):=\{f:Y\to X|\quad\text{f ist bounded and continuous}\,\}$
$D(f,g)=sup_{y\in Y} d(f(y),g(y))$ is the metric on $C_b(Y,X)$
Task:
Show, that $C_b(Y,X)$ is complete, iff $X$ is complete.
Hello,
I have a question to this task.
"$\Rightarrow$"
Let $C_b(Y,X)$ be complete. Let $n,m\in\mathbb{N}$, then
$d(x_n,x_m)\leq sup_{y\in Y} d(x_n,x_m)=D(x_n,x_m)<\epsilon$.
Since $C_b(Y,X)$ is complete, it exists a $N\in\mathbb{N}$ such that $D(x_n,x_m)<\epsilon$ for every $m,n>N$.
"$\Leftarrow$"
Let $X$ be a complete.
$D(x_n, x_m)=sup_{y\in Y} d(x_n,x_m)<sup_{y\in Y} \epsilon=\epsilon$.
Since $X$ is complete for every $\epsilon>0$ exists an $N\in\mathbb{N}$, such that $d(x_n,x_m)<\epsilon$ for every $m,n>N$.
Is this proof correct? Thanks in advance for every help.
Your method is true but the elements to choose and analyze the cauchy property and convergence are a little inappropriate. I give my proof only for the $\Rightarrow$. The other side is the similar. Also the following needs some details that you can refine. Furthermore, there is a similar proof in the Real Analysis of "Aliprantis".
Let $ C_{b}(Y,X) $ is a complete metric space. And let $ \{f_{m}(y)\}_{m\in \mathbb{N}}$ is cauchy sequence in $X.$ So there exist $m,n \in \mathbb{N}$ such that $d(f_{m}(y),f_{n}(y))<\epsilon$.
Since $ C_{b}(Y,X) $ is a complete metric space and, $D(f_{m},f_{n})=\sup_{y\in Y}d(f_{m}(y),f_{n}(y))<\epsilon $ then, there exists $f\in C_{b}(Y,X)$ such that $f_{m}\to f $ this gives us $D(f_{m},f)<\epsilon\, \Rightarrow\, d(f_{m}(y),f(y))<\epsilon$ so $f(y)\in X $