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Question Number 205472 by SANOGO last updated on 21/Mar/24

Answered by TheHoneyCat last updated on 01/Apr/24

$$\mathrm{This}\mathrm{will}\mathrm{be}\mathrm{true}\mathrm{if}\left(\mathrm{and}\mathrm{only}\mathrm{if}\right)\mathrm{the}{A}_n\mathrm{are}\mathrm{all}$$$$\mathrm{disjoint}.$$$$\mathrm{If}\mathrm{they}\mathrm{are}\mathrm{not},\mathrm{take}x\in A_i\cap A_k(\mathrm{with}k\ne i)$$$$\sum _{n\in \mathbb{N}}{\mathbb{I}}_{A_n}\left(x\right)⩾{\mathbb{I}}_{A_i}\left(x\right)+{\mathbb{I}}_{A_k}\left(x\right)⩾2$$$$\mathrm{So}\mathrm{the}\mathrm{innequality}\mathrm{has}\mathrm{to}\mathrm{be}\mathrm{false}\mathrm{since}$$$$\mathrm{\forall }X{\mathbb{I}}_X:X\to \{0,1\}$$$$$$$$\mathrm{But}\mathrm{if}\mathrm{they}\mathrm{are}\mathrm{for}\mathrm{any}x$$$$\mathrm{either}x\in \underset{n\in \mathbb{N}}{\cup }{A}_n\mathrm{so}\mathrm{\exists }!n_x:x\in A_{n_x}$$$$\mathrm{so}\sum _{\mathrm{n}\in \mathbb{N}}{\mathbb{I}}_{A_n}\left(x\right)={\mathbb{I}}_{A_{n_x}}\left(x\right)=1$$$$\mathrm{either}x\notin \underset{n\in \mathbb{N}}{\cup }{A}_n\mathrm{so}$$$$\sum _{n\in \mathbb{N}}{\mathbb{I}}_{A_n}\left(x\right)=0$$$$\mathrm{so}\mathrm{yeah}...\mathrm{in}\mathrm{that}\mathrm{case}\mathrm{only},\mathrm{it}\mathrm{works}.$$

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