Show that $\sup \bigcup_{n=1}^\infty A_n= \sup_{n \geq 1}\sup A_n$

Let $A_n \subseteq \mathbb{R}, \forall n \geq 1$. If $A= \bigcup_{n=1}^m A_n$, then we know that $\sup A = \max_{1 \leq n \leq m}\sup A_n$.

How about the infinite case, would be true that:

• $\sup A = \sup_{n \geq 1}\sup A_n$, if $A= \bigcup_{n=1}^\infty A_n$?