Let $A$ be a set, let $f,g : A → R$, and let $(f_n)_{n∈N}$ and $(g_n)_{n∈N}$ be sequences of functions from $A$ to $R$. Show that if $f_n → f$ uniformly and $g_n→g$ uniformly, then $f_n + g_n →f + g$ uniformly.
My thought was to use the Triangle Inequality for the uniform norm and the squeeze theorem to solve this. So far I have that $||f_n-f|| + ||g_n-g||\ge||(f_n +g_n) -(f+g)||$ using the uniform norm theorem. In order to show that this is true for $||(f_n +g_n) -(f+g)||$ I need to find some $h_n$ such that $||(f_n +g_n) -(f+g)|| \ge ||h_n - h|| \to 0$ but also such that $h_n \to f+ g $.
For reference the uniform norm theorem states that $f_n \to f$ and $||f_n-f||_A \to 0$ are logically equivalent.
This is where I am having trouble. (Finding the $h_n$ to make the hypothesis true) Any help would be appreciated.
$f_n \to f$ uniformly in $A$ if for any $\epsilon>0$ there exists $N_1(\epsilon)$ such that $$n\ge N_1(\epsilon) \implies |f_n(x)-f(x)|< \epsilon, \forall x\in A. $$
$g_n \to g$ uniformly in $A$ if for any $\epsilon>0$ there exists $N_2(\epsilon)$ such that $$n\ge N_2(\epsilon) \implies |g_n(x)-g(x)|< \epsilon, \forall x\in A. $$
So, for any $\epsilon>0$ there exists $N(\epsilon)=\max\{N_1(\epsilon),N_2(\epsilon)\}$ such that $$n\ge N(\epsilon) \implies |(f_n+g_n)(x)-(f+g)(x)|<|f_n(x)-f(x)+ |g_n(x)-g(x)2 \epsilon, \forall x\in A. $$
What can you get from this?