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$$M = \bigcap_{n\in N}{I_n},$$ $$I_n = \bigcup_{k = 1}^{3^n - 2} \left(\frac{k}{3^n},\frac{k+1}{3^n}\right)$$

1 Answers1

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Hint

First prove that if $M=\bigcap_{n\in\Bbb N}I_n$, then $$ \sup M=\inf_n\sup I_n\\ \inf M=\sup_n\inf I_n $$

Mostafa Ayaz
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