Uniform convergence for bounded multiplied functions problem:
If $\{f_n\}$ and $\{g_n\}$ converge uniformly on a set $E$, and there exist constants $M$ and $N$ such that $|f_n(x)| \le M$ and $|g_n(x)| \le N$ for all $n \in \mathbb{N}$ and all $x \in E$, prove that $\{f_ng_n\}$ converges uniformly on $E$.

what I have done:
I have shown that if function $\{h_n\}$ converges uniformly to $h$ on $E$ and $|h_n(x)| \le B$ for all $n \in \mathbb{N}, x \in E$, then $|h(x)| \le B$ for all $x \in E$.
I am now trying to use triangle inequality to show that there exists an $n_0 \in \mathbb{N}$ such that $|f_n(x) - f(x)||g_n(x) + g(x)| < \epsilon$ for all $n \ge n_0$.
It keeps getting stuck, and so I am wondering if I am going about this the wrong way. If the trick does use triangle inequality, what advice do you have for not getting stuck on these sorts of convergence problems?
Thanks so much!
 A: It's the triangle inequality, but what you're aiming for is incorrect. You need to show that there is an $n_0 \in \mathbb{N}$ so that for any $x \in E$ and $n \geq n_0$
$$
| f_n(x)g_n(x) - f(x)g(x) | < \epsilon.
$$
Note that $| f_n(x)g_n(x) - f(x)g(x) | $ does not equal what you have written if you expand it out. The trick here is to add and subtract $f_n(x) g(x)$ (or $f(x) g_n(x)$, matter of preference) to get
$$
| f_n(x)g_n(x) - f(x)g(x) | = | f_n(x)g_n(x) - f_n(x)g(x) + f_n(x) g(x) - f(x)g(x) |
$$
and then group together the first two terms and second two terms, and then use the triangle inequality:
$$
| (f_n(x)g_n(x) - f_n(x)g(x)) + (f_n(x) g(x) - f(x)g(x)) | 
$$
$$
\leq | f_n(x)g_n(x) - f_n(x)g(x)|+ |f_n(x) g(x) - f(x)g(x)|.
$$
Now you can pull out the $|f_n(x)|$ out of the first one, and the $|g(x)|$ out of the second. By assumption $|f_n(x)| \leq M$ and by what you have shown $|g(x)| \leq N$. So we get
$$
| f_n(x)g_n(x) - f(x)g(x) | \leq | f_n(x)|\cdot |g_n(x) - g(x)|+ |g(x)||f_n(x) - f(x)|
$$
$$
\leq M|g_n(x) - g(x)| + N|f_n(x) - f(x)|.
$$
From here, you can apply the fact that $\{g_n\}$ and $\{f_n\}$ are both uniformly convergent.
I guess the trick to learn out of this is that when you're dealing with a limit of products, you hope the limit is the product of the limits, and it's usually a good idea to add and subtract the product sequence element of the first sequence ($f_n(x)$ in this case) and the limit of the other sequence ($g(x)$ in this case).
