Uniform continuity of functions and restrictions.

Question

Hi I have this exercise on my real analysis course, I haven´t seen topology course:

Let $A_{i} \subset \mathbb{R}$ an interval for all $i \in \mathbb{N}$. Is it true or false?

a) If $f_{1}: A_{1} \longrightarrow \mathbb{R}$ and $f_{2}: A_{2} \longrightarrow \mathbb{R}$ are uniformly continous and $f: A_{1} \cup A_{2} \longrightarrow \mathbb{R}$ is continous, then $f$ is uniformly continous.

b) Let $f_{i}: A_{i} \longrightarrow \mathbb{R}$ uniformly continous for all $i \in \mathbb{N}$ and $f: \displaystyle{\bigcup_{i \in \mathbb{N}}{A_{i}}} \longrightarrow \mathbb{R}$ continous, then $f$ is uniformly continous.

Since they seem very similar I suppose I need induction on $i \in \mathbb{N}^{\geq 2}$ for both of them. However I am not abble to prove any case. For the fisrst $i = 2$ I do not have any theorem to use or a definition wich implies union of intervals on my textbook. Do you have any hint?

Answer 1

The first statement is false. Take $A_1=[0,1)$, $A_2=[1,2]$, $f_1(x)=0$, and $f_2(x)=1$. Then $f_1$ and $f_2$ are uniformly continuous, but $f$ isn't even continuous.

Answer 2

Assuming that you define $f$ to be $f(x) = f_i(x) \iff x\in A_i$ and that $A_i \cap A_j = \varnothing $, then b) is false. Consider $f_i(x) = \frac 1x \; \forall x$ and $A_1 = (1,2], A_i = (1/i,\inf {A_{i-1}}]$ (Edit: notice the domain of $f$ is an interval. I previously defined it erroneously and made it disconnected). Then $f_i$ is uniformly continuous but $f$ is not.

You have that each $f_i$ is uniformly continuous, but $f$ is not bounded and therefore not uniformly continuous.