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 $S_i$, $i \in \left\{{1, \ldots, n}\right\}$, $n \in \N_{>0}$, denote sets within the domain of $f$.

Define $S = \displaystyle \bigcup_{i \mathop = 1}^{n} S_i$.

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

We need to prove that $f$ is bounded on $S$.

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

Let $b = \max \left({b_1, \ldots, b_n}\right)$ where $\max$ is the maximum operation.

We note that $b \ge b_i$ for every $i$ in $\left\{{1, \ldots, n}\right\}$.

Now, by definition of bound, a number greater than or equal to a bound is also a bound.

Therefore, since $b_i$ is a bound for $f$ on $S_i$ and $b \ge b_i$, $b$ is a bound for $f$ on $S_i$ for every $i$.

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 $K \ge \vert f(s) \vert$ for every element $s$ in $S$.

Pick a set $S_i$.

Since $K \ge \vert f(s) \vert$ is true of every element $s$ of $S$, this is also true of every element of $S_i$ as $S_i$ is a subset of $S$.

Consequently, $K$ 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 $\left\{{1, \ldots, n}\right\}$.