Minimum Over Unions: A Distribution Property

This post explains how the infimum distributes over unions of sets, and how this leads to clean formulas for minima when additional assumptions are present. We also highlight a standard counterexample showing where the minimum identity fails for infinite unions.

1. Main Result

Given a family of nonempty subsets $\{A_i\}_{i\in I}$ of $\mathbb{R}$, the following distribution identity holds:

$$\inf\Bigl(\bigcup_{i\in I} A_i\Bigr) =\inf_{i\in I}\bigl(\inf A_i\bigr).$$

Proof. Let
$$x = \inf(\cup_{i\in I} A_i),\quad y = \inf_{i\in I} \inf A_i.$$

Since $A_i \subset \cup_{i\in I} A_i$ for all $i \in I$, thus, $\inf A_i\geq x$ for all $i \in I$. Therefore $y \geq x$.

For any $a \in \cup_{i \in I} A_i$, there exists $A_i$ containing $a$, thus $\inf A_i \leq a$. Thus, $y\leq a$ for any $a$, i.e. $y$ is a lower bound for $\cup_{i \in I} A_i$, hence $y \leq x$.

Thus $x = y$.

2. Consequences

(i) Finite index set

When $I$ is finite, an infimum of a finite set of real numbers is actually a minimum. Therefore:

$$\inf\Bigl(\bigcup_{i\in I} A_i\Bigr) =\min_{i\in I} \inf(A_i).$$

(ii) Closed and lower-bounded sets

If $I$ is finite and each $A_i$ is closed and bounded below, then each $\inf A_i$ is attained, so:

$$\inf A_i = \min A_i$$
and thus $(\bigcup_{i\in I} A_i$ is also closed and bounded below, thus
$$\inf\Bigl(\bigcup_{i\in I} A_i\Bigr) = \min\Bigl(\bigcup_{i\in I} A_i\Bigr).$$
Hence:

$$\min\Bigl(\bigcup_{i\in I} A_i\Bigr) =\min_{i\in I} \min A_i.$$

3. Counterexample for Infinite Index Sets

For infinite $I$, the following identity does not hold

$$\min\Bigl(\bigcup_{i\in I} A_i\Bigr) =\inf_{i\in I} \min A_i.$$

The problem is that the minima may fail to exist even if each $A_i$ has one. Indeed, let us consider:

$$A_i = \{1/i\}.$$

Each $A_i$ has a minimum equal $1/n$, thus the right-hand-side is $0$.

However, the union is:

$$\bigcup_{i\ge 1} A_i = \{1, 1/2, 1/3, \dots\}.$$

Its infimum is $0$, but the min does not exist.

Conclusion

The identity

$$\inf\Bigl(\bigcup_{i\in I} A_i\Bigr) = \inf_{i\in I}\bigl(\inf A_i\bigr)$$

is always valid. However, replacing infimum by minimum requires additional assumptions—most notably finiteness of the index set and closedness of the sets involved. Infinite unions can easily destroy the existence of a minimum, even when every individual set has one.