space of all lipschitz maps is a polish metric space Suppose that $(X, d_X)$ and $(Y, d_Y )$ are Polish metric spaces. Let $L(X, Y )$ denote the set of all Lipschitz maps from $X$ to $Y$ with the pointwise convergence topology. 
Show that $L(X, Y )$ is a polish space. 
Hint: Consider $d_L(f, g) = \Sigma_{i ∈ ω} 2^{−(i+1)}d_Y (f(x_i), g(x_j ))$. 
I am having trouble showing that this metric is complete. And also to show that $L(X,Y)$ is separable ( I tried to use the fact of both $X$ and $Y$ are Polish and pointwise convergence topology, but it did not work). 
Any thoughts or help will be much appreciated.
Thank you in advance! 

I assume the correct definition of $d_L$ is $d_L(f,g) = \sum_i 2^{-(i+1)} d_Y(f(x_i), g(x_i))$.
Suppose $\{f_n\}$ is a Cauchy sequence in $L(X,Y)$.  Hints:


*

*Show that for each $i$, $\{f_n(x_i)\}$ is a Cauchy sequence in $Y$.  Therefore...

*If $f$ is going to be the limit of $f_n$, this tells you what $f(x_i)$ has to be, for each $i$.

*If $f$ is going to be Lipschitz (in particular, continuous), then what does $f(x)$ have to be for $x$ which are not among the $x_i$?  (Is this well defined?)

*Now you have a candidate $f$ for the limit of $f_n$.  Verify that this function $f$ is Lipschitz.  Finally verify that $f_n \to f$ in the $d_L$ metric.
