Let $a<c<b$. Let {$f_n$} be a sequence of functions converging uniformly on $[a,c]$ and $[c,b]$. Prove that {$f_n$} converges uniformly on $[a,b]$
My attempt:
Intuitively, I see that {$f_n$} converges to a limit function {$f_1$} on $[a,c]$ and {$f_2$} on $[c,b]$.
Now, since the point $c$ is included in both cases, can I conclude that {$f_1$}={$f_2$}?
How to formally write the proof?
I'll take a shot at this
Here I'm letting $f:[a,b]\rightarrow\mathbb{R}$ be defined as $$ f(x) = \left\{ \begin{array}{lr} f_a(x) & : x \in [a,c)\\ f_b(x) & : x \in [c,b] \end{array} \right. $$ where I replace $f_1$ with $f_a$ and $f_2$ with $f_b$ from your answer (I did this because $f_1$ and $f_2$ are actually functions within the sequence)
Since we have $f_{n}\rightarrow f_a$ and $f_n\rightarrow f_b$ uniformly on $[a,c]$ and $[c,b]$ respectively, this means that we have that for $\epsilon>0$ there exist an $N_1,N_2\in\mathbb{N}$ such that $n\geq N_1$, then $$|f_{n}(x)-f(x)|<\epsilon$$ for $x\in[a,c)$. Similarly we have for $n\geq N_2$, then $$|f_{n}(x)-f(x)|<\epsilon$$ for $x\in[c,b]$.
Thus letting $N=\max\{N_1,N_2\}$, we have that for $n\geq N$, then $$|f_{n}(x)-f(x)|<\epsilon$$ for $x\in[a,b]$ thus $f_n\rightarrow f$ uniformly on $[a,b]$