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Question-205472




Question Number 205472 by SANOGO last updated on 21/Mar/24
Answered by TheHoneyCat last updated on 01/Apr/24
This will be true if (and only if) the A_n  are all  disjoint.  If they are not, take x∈A_i ∩A_k  (with k≠i)  Σ_(n∈N) I_A_n  (x)≥I_A_i  (x)+I_A_k  (x)≥2  So the innequality has to be false since  ∀X  I_X :X→{0,1}    But if they are for any x  either x∈∪_(n∈N) A_n  so ∃!n_x : x∈A_n_x                so Σ_(n∈N) I_A_n  (x)=I_A_n_x   (x)=1  either x∉∪_(n∈N) A_n  so                    Σ_(n∈N) I_A_n  (x)=0  so yeah... in that case only, it works.
$$\mathrm{This}\:\mathrm{will}\:\mathrm{be}\:\mathrm{true}\:\mathrm{if}\:\left(\mathrm{and}\:\mathrm{only}\:\mathrm{if}\right)\:\mathrm{the}\:{A}_{{n}} \:\mathrm{are}\:\mathrm{all} \\ $$$$\mathrm{disjoint}. \\ $$$$\mathrm{If}\:\mathrm{they}\:\mathrm{are}\:\mathrm{not},\:\mathrm{take}\:{x}\in{A}_{{i}} \cap{A}_{{k}} \:\left(\mathrm{with}\:{k}\neq{i}\right) \\ $$$$\underset{{n}\in\mathbb{N}} {\sum}\mathbb{I}_{{A}_{{n}} } \left({x}\right)\geqslant\mathbb{I}_{{A}_{{i}} } \left({x}\right)+\mathbb{I}_{{A}_{{k}} } \left({x}\right)\geqslant\mathrm{2} \\ $$$$\mathrm{So}\:\mathrm{the}\:\mathrm{innequality}\:\mathrm{has}\:\mathrm{to}\:\mathrm{be}\:\mathrm{false}\:\mathrm{since} \\ $$$$\forall{X}\:\:\mathbb{I}_{{X}} :{X}\rightarrow\left\{\mathrm{0},\mathrm{1}\right\} \\ $$$$ \\ $$$$\mathrm{But}\:\mathrm{if}\:\mathrm{they}\:\mathrm{are}\:\mathrm{for}\:\mathrm{any}\:{x} \\ $$$$\mathrm{either}\:{x}\in\underset{{n}\in\mathbb{N}} {\cup}{A}_{{n}} \:\mathrm{so}\:\exists!{n}_{{x}} :\:{x}\in{A}_{{n}_{{x}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{so}\:\underset{\mathrm{n}\in\mathbb{N}} {\sum}\mathbb{I}_{{A}_{{n}} } \left({x}\right)=\mathbb{I}_{{A}_{{n}_{{x}} } } \left({x}\right)=\mathrm{1} \\ $$$$\mathrm{either}\:{x}\notin\underset{{n}\in\mathbb{N}} {\cup}{A}_{{n}} \:\mathrm{so} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{n}\in\mathbb{N}} {\sum}\mathbb{I}_{{A}_{{n}} } \left({x}\right)=\mathrm{0} \\ $$$$\mathrm{so}\:\mathrm{yeah}…\:\mathrm{in}\:\mathrm{that}\:\mathrm{case}\:\mathrm{only},\:\mathrm{it}\:\mathrm{works}. \\ $$

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