Let $(X, d_X)$ and $(Y, d_Y)$ be metric spaces, $(f_n)^∞_{n=m}$ a sequence of functions $f_n : X → Y$, and $f : X → Y$ another function.
Show that $f_n → f$ uniformly on $X$ as $n → ∞$ if and only if for every sequence $(x_n)^∞_{n=m}$ in $X$ we have $d_Y (f_n(x_n),f(x_n)) → 0$ as $n → ∞$.
$Proof:$
$(\Rightarrow)$
Assume: $f_n → f$ uniformly on $X$ as $n → ∞$
WTS: for every sequence $(x_n)^∞_{n=m}$ in $X$ we have $d_Y (f_n(x_n),f(x_n)) → 0$ as $n → ∞$.
$(\Leftarrow)$
Assume: for every sequence $(x_n)^∞_{n=m}$ in $X$ we have $d_Y (f_n(x_n),f(x_n)) → 0$ as $n → ∞$.
WTS: $f_n → f$ uniformly on $X$ as $n → ∞$
Let $\epsilon>0$ by defintion of uniform convergence, $\exists$ $N>m$ s.t. $d_Y (f_n(x),f(x)) < \epsilon$ $\forall x ∈ X, n>N$.
Can I please get help in writing the proof, as I am unsure of how to go about it, thank you!
Assume that the convergence is uniform and let $(x_n)_{n\in\mathbb N}$ be a sequence of elements of $X$. Take $\varepsilon>0$. Then there is some $N\in\mathbb N$ such that, if $x\in X$ and $n\geqslant N$, then $d_Y\bigl(f_n(x),f(x)\bigr)<\varepsilon$. So $n\geqslant N\implies d_Y\bigl(f_n(x_n),f(x_n)\bigr)<\varepsilon$.
Now, suppose that $(f_n)_{n\in\mathbb N}$ does not converge uniformly to $f$. Then, for some $\varepsilon>0$, given $n\in\mathbb N$ there will be some $x_n\in X$ such that $d_Y\bigl(f_n(x_n),f(x_n)\bigr)\geqslant\varepsilon$. But then we don't have $\lim_n d_Y\bigl(f_n(x_n),f(x_n)\bigr)=0$.