Uniform convergence for sections of functions

Question

I have a statement I cannot figure out whether it's true or not. Let $(X,d_X)$, $(Y,d_Y)$ and $(Z,d_Z)$ be compact metric spaces, and $f\colon X\times Y\to Z$ be continuous. Let $(x_n)_n\in X^\mathbb N$ be a sequence in $X$ such that $x_n\to x$. I am trying to figure out whether the sequence of functions $h_n\colon Y\to Z$ defined by \begin{equation} h_n(y)=f(x_n,y) \end{equation} converges uniformly to $h(y)=f(x,y)$. Pointwise convergence is clear, and I know that $f$ is uniformly continuous. The problem I can't decide on, specifically, is whether the "$N$" in the statement for convergence ($\forall \epsilon>0\; \exists N\in\mathbb N$ s.t. ...). of the $h_n$ functions depends on both $\epsilon$ and $y$, or can be made independent of the latter.

Answer 1

Yes, it is true. Take $\varepsilon>0$. Now, take $\delta>0$ such that$$d_X(x_1,x_2)<\delta\wedge d_Y(y_1,y_2)<\delta\implies d_Z\bigl(f(x_1,y_1),f(x_2,y_2)\bigr)<\varepsilon.$$Now, if $p\in\mathbb N$ is such that $n\geqslant p\implies d_X(x_n,x)<\delta$, then\begin{align}n\geqslant p&\implies d_X(x_n,x)<\delta\wedge d_Y(y,y)=0<\delta\\&\implies d_Z\bigl(f(x_n,y),f(x,y)\bigr)<\varepsilon\end{align}