Supremum of Unions

Question

Let $\{A_n\}$be a collection of sets, where $A_n$ are subsets of $\mathbb R$.

show that: $\sup(\bigcup_n A_n) = \sup(\bigcup_n(\sup A_n))$, $\inf(\bigcup_n A_n) = \inf(\bigcup_n(\inf A_n))$.

Thanks.

Answer 1

Let us assume that the suprema on both sides of the equation exist. Note that I don't assume the sets to be countable.

Let $M$ be any upper bound of the set $\bigcup_{n} A_n$. Then for every $n\in \mathbb{N}$ one has that $x \leq M$ for all $x \in A_n$. Since a supremum is by definition the smallest upper bound of a set, we get that $\sup(A_n) \leq M$ for every $n \in \mathbb{N}$. Consequently $M$ is an upper bound for the set $\{\sup(A_n) \mid n\in\mathbb{N}\}$, so the definition of a supremum gives $\sup\{\sup(A_n) \mid n\in\mathbb{N}\} \leq M$. If we now choose $M = \sup(\bigcup_n A_n)$ (the supremum is a particular upper bound) we obtain $$\sup\{\sup(A_n) \mid n\in\mathbb{N}\} \leq \sup\left(\bigcup_n A_n\right).$$

Conversely, let $N$ be any upper bound of $\{\sup(A_n) \mid n\in\mathbb{N}\} $. Then for every $n \in \mathbb{N}$, we have $\sup(A_n) \leq N$. Consequently, $N$ is an upper bound for all sets $A_n$, and by the mere definition of the union, it is also an upper bound for $\bigcup_n A_n$. By definition of a supremum, $\sup\{\bigcup_n A_n\} \leq N$. Finally, if we now let $N$ be the supremum of $\{\sup(A_n) \mid n\in\mathbb{N}\}$, we obtain $$\sup\left(\bigcup_n A_n\right) \leq \sup\{\sup(A_n) \mid n\in\mathbb{N}\}.$$

The proof for the infima is analogous.