Showing Uniformly convergent sequence of functions implies function is bounded

Question

Problem: If ${F_n}$ is a sequence of bounded functions from a set $D \subset \mathbb R^p$ into $ \mathbb R^q$ and if ${F_n}$ converges uniformly to $F$ on $D$, then $F$ is also bounded.

Proof(Attempt): Let $\epsilon >0$. Since ${F_n}$ converges uniformly to $F$ on $D$, then there is an $N \in\mathbb R$ such that $||F(x)-F_n(x)||< \epsilon $ whenever $x \in D$ and $n\ge N$.

I'm using Joseph Taylor's Foundation of Analysis textbook. Since ${F_n}$ is bounded we know that $||F_n|| \le M $ for every $x\in D$.
At this point, I'm thinking that I need to use a trick that gets $||F|| \le M$ from the inequality $||F(x)-F_n(x)||< \epsilon $.

I would appreciate advise and hints that will help guide me.

Answer 1

HINT

We have $$ \| F(x) \| = \|F(x)-F_n(x) + F_n(x)\| \le \|F(x)-F_n(x)\| + \|F_n(x)\|.$$

Answer 2

The assumption that each $F_{n}$ is bounded does not necessarily mean that there is an $M>0$ such that $|F_{n}(x)|\leq M$ for all $x\in D$ and $n=1,2,...$

Rather, we need to establish the uniform bound for all $F_{n}$. So the sequence is uniformly Cauchy, so $|F_{n}(x)-F_{m}(x)|<1$ for all $x\in D$ and $n,m\geq N$.

So $|F_{n}(x)|\leq|F_{1}(x)|+\cdots+|F_{N}(x)|+1$ for all $x\in D$ and $n=1,2,...$, now we are to take $M=M_{1}+\cdots+M_{N}+1$, where $|F_{n}(x)|\leq M_{n}$ for all $x\in D$ and $n=1,2,...,N$.