Uniform convergence of $(f_n)_{n \in \Bbb{N}}$ on $D_1 \cup D_2$

Question

I want to show using the Cauchy criterion that if $(f_n)_{n \in \Bbb{N}}$ is uniformly convergent on both $D_1$ and $D_2$, then it must be uniformly convergent on $D_1 \cup D_2$

Here's what I have:

$(f_n)_{n \in \Bbb{N}}$ is uniformly convergent on $D_1$, therefore, $\forall \varepsilon \gt 0, \exists n_0$ such that if $ n,m \ge n_0$ then $\vert f_n(x) - f_m(x) \vert \lt \frac{\varepsilon}2$.

Similarly, $(f_n)_{n \in \Bbb{N}}$ is uniformly convergent on $D_2$, therefore, $\forall \varepsilon \gt 0, \exists k_0$ such that if $ p,q \ge k_0$ then $\vert f_p(x) - f_q(x) \vert \lt \frac{\varepsilon}2$.

Hence letting all $n, m, p, q \ge max\{n_0, k_0\}$:

$$\vert \vert f_n(x)- f_m(x)\vert -\vert f_p(x) - f_q(x)\vert \vert \le \vert f_n(x)- f_m(x)\vert + \vert f_p(x) - f_q(x)\vert \lt \frac{\varepsilon}2 + \frac{\varepsilon}2 = \varepsilon$$

Then it would follow that $$\vert \vert f_n(x)- f_m(x)\vert -\vert f_p(x) - f_q(x)\vert \vert \lt \varepsilon $$

Which I want to think implies that $(f_n)_{n \in \Bbb{n}}$ is convergent on $D_1 \cup D_2$

I don't know if what I did was correct, or if I made a mistake somewhere, so I would appreciate any feedback on my take and some hints on what direction my proof should go assuming it went wrong.

Answer 1

Let $N=\max \{n_0,k_0\}$, take any $x \in D_1 \cup D_2$ then we will have $|f_{n_1}(x)-f_{n_2}(x)|< \frac{\epsilon}{2}$ for all $n_1, n_2 \geq N$, because if $x \in D_1$ then the fact that $f$ is uniformly convergent on $D_1$ is used (the inequality you've written in the 3rd paragraph), similar if $x \in D_2$.

Answer 2

Let me suggest one possible way to prove desired. We know that $f_n \to f$ uniformly on $A$, iif $\lim\limits_{n \to \infty}\sup\limits_{x\in A}|f_n(x)-f(x)|=0$.

So, we can consider 2 sequences $x_n=\sup\limits_{x\in A}|f_n(x)-f(x)|$ and $y_n=\sup\limits_{x\in B}|f_n(x)-f(x)|$. Knowing, that $x_n \to 0$ and that $y_n \to 0$, we need $z_n=\max(x_n,y_n)\to 0$. Now this is more easy to achieved, because for $\forall \varepsilon \gt 0$ we simply take $N=\max(N_1,N_2)$ from limit definition for $x_n$ and $y_n$