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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?

RM777
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  • Induction will tell you that the statement is true for every finite union of the $A_{n}$, but that is not enough. You need to show that it holds for a countably infinite union. – Robert Thingum Nov 13 '18 at 20:16

1 Answers1

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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?

Robert Thingum
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