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

Theorem
A real-valued function $f$ 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, and $S_i$, $i \in \left\{{1, \ldots, n}\right\}$, 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 a bound for $f$ on $S_i$, $i \in \left\{{1, \ldots, n}\right\}$.

Every $b_i$ is bounded by the maximum $b = max\left({b_1, \ldots, b_n}\right)$.

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

Since $b$ is a bound for $f$ on every set $S_i$, $b$ is a 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 a 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 a bound for $f$ on $S_i$.

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