Mapping is Bounded on Union iff Bounded on Each Component/Real-Valued Function

Theorem
Let $f$ be a real-valued function.

Then:
 * $f$ is bounded on the union of a finite number of sets within the domain of $f$

iff:
 * $f$ is bounded on each of the sets.

Proof
Let $S$ denote the union of sets.

Let $\left \langle{S_i}\right\rangle_{i \in \left\{{1, \ldots, n}\right\}}$, denote the sets that are the component sets of that union.

Suppose first that $f$ is bounded on each of the sets $S_i$.

Let $b_i$ be an upper bound for $f$ on $\left \langle{S_i}\right\rangle_{i \in \left\{{1, \ldots, n}\right\}}$.

By definition of the maximum operation, the set $\left\{{b_i: i \in \left\{{1, \ldots, n}\right\}}\right\}$ is bounded above by $b = \max \left({b_1, \ldots, b_n}\right)$.

$b$ serves as an upper bound for $f$ on each of the sets $S_i$ because $b \ge b_i$.

Since $b$ is an upper bound for $f$ on every set $S_i$, $b$ is an upper bound for $f$ on the union of the sets $S_i$.

In other words, $f$ is bounded on $S$.

This finishes the "if" part of the proof.

Now, suppose that $f$ is bounded on $S$.

We need to prove that $f$ is bounded on each of the sets $S_i$, $i \in \left\{{1, \ldots, n}\right\}$.

Since $f$ is bounded on $S$, there is an upper bound $K$ that satisfies $\vert f(s) \vert \le K$ for every element $s$ in $S$.

Pick a set $S_i$.

Since every element $s$ of $S$ satisfies $\vert f(s) \vert \le K$, this is also true for every element of $S_i$.

This is true because every element of $S_i$ is an element of $S$ as $S_i$ is a subset of $S$.

Accordingly, $K$ is an upper bound for $f$ on $S_i$.

Since $i$ is arbitrary, we conclude that $f$ is bounded on $S_i$ for every $i \in \left\{{1, \ldots, n}\right\}$.