uniform cty of fn on union of closed sets

Question

Given $A_1, A_2 \subset \mathbb R$ closed and $f\colon A_1 \cup A_2 \to \mathbb R$ such that $f\vert_{A_1}$ uniformly cts and $f\vert_{A_2}$ uniformly cts, is it true that $f$ is uniformly cts on $A_1 \cup A_2$? I don't believe it to be true and believe that there may exist a counterexample with $A_1, A_2$ unbounded. Just looking for someone to confirm or deny my suspicions. Not looking for a counterexample or proof.

Answer 1

Its true if $A_1$ and $A_2$ are bounded or if there exists some number $\delta_0 > 0$ such that if $x, y \in A_1 \cup A_2$ and $|x - y| < \delta_0$ then $x, y \in A_1$ or $x, y \in A_2$.

For an example where this is not the case, let $$A_1 = \left[0,\frac12\right] \cup \left[2, 2 + \frac34\right] \cup \dots \cup \left[ 2n, 2n + 1 -\frac{1}{2^{n + 1}} \right] \cup \cdots $$ and $$A_2 = \left[1,1 + \frac12\right] \cup \left[3, 3 + \frac34\right] \cup \dots \cup \left[ 2n + 1, 2n + 2 -\frac{1}{2^{n + 1}} \right] \cup \cdots $$

Now set $f$ to be identically $0$ on $A_1$ and identically $1$ on $A_2$. Clearly $f|_{A_1}$ and $f|_{A_2}$ are uniformly continuous, but we have the sequences $$ x_n = 2n + 1 -\frac{1}{2^{n + 1}}, y_n = 2n + 1 $$ for which $|f(x_n) - f(y_n)| = |0 - 1| = 1$ and $|x_n - y_n| = 2^{-(n + 1)} \to 0$.