By definition of continuous function, we need the function to satisfied certain properties for all $\epsilon > 0$. Can I prove the function is continuous by fixing $\epsilon$ to be any $real > 0$? This problem arise when I try to prove a sequence continuous function $f_n$ uniformly converge to another function $f$.
Given $f_n$ uniformly converges to $f$ ,and $f_n$ continuous for any n, show $f$ is continuous at $x_0$: $d_{Y}(f(x),f(x_0)) <= d_{Y}(f(x),f(x_0)) + d_{Y}(f(x_0),f_n(x_0)) + d_{Y}(f(x),f_n(x))$. It works well when the fn uniform convergent to f because there exist a function $f_n$ that f is able to "use" for any $\epsilon >0$. (for any $\epsilon >0$, there exists $N(\epsilon >0)$ s.t. both $d_{Y}(f(x_0),f_n(x_0)) < \epsilon$ and $d_{Y}(f(x),f_n(x)) <\epsilon$).
But if $f_n$ only pointwise converges to $f$,and because we fix $\epsilon$, we are able to find $N^* := max(N(\epsilon, x_0),N(\epsilon, x))$ such that $d_{Y}(f(x),f(x_0))$ is also bounded within $3\epsilon$ whenever $d(x,x_0) < \delta(f_n)$