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.
Suppose that there is a sequence $(r_n)^∞ _{n=m}$ in $R$ with $r_n → 0$ as $n → ∞$ and such that $d_Y (f_n(x), f(x)) ≤ r_n$ for all $x ∈ X$ and $n ≥ m$.
Show that $f_n → f$ uniformly on $X$ as $n → ∞$.
Proof:
We will make $r_n$ arbitrarily small, that is assume that, let $r_n>0$
There is an $N$ s.t. $d_Y (f_n(x),f(x)) < r_n$ $\forall x ∈ X, n>N$.
It appears to me that $r_n$ is taking the place of $\epsilon$ from the original definition of uniform convergence, but I am having a difficulty utilizing it in the proof.
Now we need to find this $N$ but I am not quite sure as to how I should proceed.
Take $\varepsilon>0$. Since $\lim_nr_n=0$, there is some $N\in\mathbb N$ such that $n\geqslant N\implies r_n<\varepsilon$. So$$(\forall n\in\mathbb{N})(\forall x\in X):n\geqslant N\implies d_Y\bigl(f_n(x),f(x)\bigr)\leqslant r_n<\varepsilon.$$