Show that for every indexed family of subsets $A_n\subset\overline{\rm \mathbb{R}}$ for $n\in \mathbb{N}$
sup$(\bigcup_{n\in\mathbb{N}}A_n)=$sup$\{$sup$A_n|n\in\mathbb{N}\}$
I wanted to Show the equality by induction I assume that the equality holds for a finite Union of subsets with n elements on the left side and the suprema of the suprema of those n sets on the right side. But my teacher told me it is not possible could anyone of you tell me why it is not possible and a hint how I could prove the equality?
Here's a hint for how to prove equality.
For all $n\in\mathbb{N}$ we have that $\sup A_{n}\leq\sup\left(\bigcup_{n\in\mathbb{N}}A_{n}\right)$ for the simple reason that $A_{n}\subseteq\bigcup_{n\in\mathbb{N}}A_{n}$. This implies that $\sup\left(\bigcup_{n\in\mathbb{N}}A_{n}\right)$ is an upper bound for $\{\sup A_{n}\mid n\in\mathbb{N}\}$ and hence
$$\sup\{\sup A_{n}\mid n\in\mathbb{N}\}\leq\sup\left(\bigcup_{n\in\mathbb{N}}A_{n}\right)$$
To prove the reverse inequality I would recommend using contradiction. Assume that $\sup\{\sup A_{n}\mid n\in\mathbb{N}\}<\sup\left(\bigcup_{n\in\mathbb{N}}A_{n}\right)$. What goes wrong?