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Question Number 35703 by scientist last updated on 22/May/18

Answered by arcana last updated on 17/Oct/18

tomando la propiedad de funcion de medida n  A,B conjuntos finitos  n(A∪B)=n(A)+n(B)−n(A∩B)  asi  • n((A∪B)∪(C∪D))=n(A∪B)+n(C∪D)−n((A∪B)∩(C∪D))  =[n(A)+n(B)−n(A∩B)]+[n(C)+n(D)−n(C∩D)]−n(((A∪B)∩C)∪((A∪B)∩D))  (/)(/)(/)(/)(/)(/)(/)(/)(/)(/)(/)(/)(/)(/)(/)(/)(/)(/)(/)(/)(/)  ahora hallamos n([(A∪B)∩C]∪[(A∪B)∩D])  n([(A∪B)∩C]∪[(A∪B)∩D])=n((A∪B)∩C)+  n((A∪B)∩D)−n((A∪B)∩(C∩D))    =[n(A∩C)+n(B∩C)−n(A∩B∩C)]+  [n(A∩D)+n(B∩D)−n(A∩B∩D)]−  [n(A∩C∩D)+n(B∩C∩D)−n(A∩B∩C∩D)]  (/)(/)(/)(/)(/)(/)(/)(/)(/)(/)(/)(/)(/)(/)(/)(/)(/)(/)(/)(/)(/)  por lo anterior podemos reescribir • como  • n((A∪B)∪(C∪D))=n(A)+n(B)+n(C)+n(D)−n(A∩B)−n(C∩D)−  n(A∩C)−n(B∩C)+n(A∩B∩C)]−  n(A∩D)−n(B∩D)+n(A∩B∩D)]+  n(A∩C∩D)+n(B∩C∩D)−n(A∩B∩C∩D)

$$\mathrm{tomando}\:\mathrm{la}\:\mathrm{propiedad}\:\mathrm{de}\:\mathrm{funcion}\:\mathrm{de}\:\mathrm{medida}\:{n} \\ $$$$\mathrm{A},\mathrm{B}\:\mathrm{conjuntos}\:\mathrm{finitos} \\ $$$${n}\left(\mathrm{A}\cup\mathrm{B}\right)={n}\left(\mathrm{A}\right)+{n}\left(\mathrm{B}\right)−{n}\left(\mathrm{A}\cap\mathrm{B}\right) \\ $$$$\mathrm{asi} \\ $$$$\bullet\:{n}\left(\left(\mathrm{A}\cup\mathrm{B}\right)\cup\left(\mathrm{C}\cup\mathrm{D}\right)\right)={n}\left(\mathrm{A}\cup\mathrm{B}\right)+{n}\left(\mathrm{C}\cup\mathrm{D}\right)−{n}\left(\left(\mathrm{A}\cup\mathrm{B}\right)\cap\left(\mathrm{C}\cup\mathrm{D}\right)\right) \\ $$$$=\left[{n}\left(\mathrm{A}\right)+{n}\left(\mathrm{B}\right)−{n}\left(\mathrm{A}\cap\mathrm{B}\right)\right]+\left[{n}\left(\mathrm{C}\right)+{n}\left(\mathrm{D}\right)−{n}\left(\mathrm{C}\cap\mathrm{D}\right)\right]−{n}\left(\left(\left(\mathrm{A}\cup\mathrm{B}\right)\cap\mathrm{C}\right)\cup\left(\left(\mathrm{A}\cup\mathrm{B}\right)\cap\mathrm{D}\right)\right) \\ $$$$\frac{}{}\frac{}{}\frac{}{}\frac{}{}\frac{}{}\frac{}{}\frac{}{}\frac{}{}\frac{}{}\frac{}{}\frac{}{}\frac{}{}\frac{}{}\frac{}{}\frac{}{}\frac{}{}\frac{}{}\frac{}{}\frac{}{}\frac{}{}\frac{}{} \\ $$$$\mathrm{ahora}\:\mathrm{hallamos}\:{n}\left(\left[\left(\mathrm{A}\cup\mathrm{B}\right)\cap\mathrm{C}\right]\cup\left[\left(\mathrm{A}\cup\mathrm{B}\right)\cap\mathrm{D}\right]\right) \\ $$$${n}\left(\left[\left(\mathrm{A}\cup\mathrm{B}\right)\cap\mathrm{C}\right]\cup\left[\left(\mathrm{A}\cup\mathrm{B}\right)\cap\mathrm{D}\right]\right)={n}\left(\left(\mathrm{A}\cup\mathrm{B}\right)\cap\mathrm{C}\right)+ \\ $$$${n}\left(\left(\mathrm{A}\cup\mathrm{B}\right)\cap\mathrm{D}\right)−{n}\left(\left(\mathrm{A}\cup\mathrm{B}\right)\cap\left(\mathrm{C}\cap\mathrm{D}\right)\right) \\ $$$$ \\ $$$$=\left[{n}\left({A}\cap\mathrm{C}\right)+{n}\left(\mathrm{B}\cap\mathrm{C}\right)−{n}\left(\mathrm{A}\cap\mathrm{B}\cap\mathrm{C}\right)\right]+ \\ $$$$\left[{n}\left({A}\cap\mathrm{D}\right)+{n}\left(\mathrm{B}\cap\mathrm{D}\right)−{n}\left(\mathrm{A}\cap\mathrm{B}\cap\mathrm{D}\right)\right]− \\ $$$$\left[{n}\left({A}\cap\mathrm{C}\cap\mathrm{D}\right)+{n}\left(\mathrm{B}\cap\mathrm{C}\cap\mathrm{D}\right)−{n}\left(\mathrm{A}\cap\mathrm{B}\cap\mathrm{C}\cap\mathrm{D}\right)\right] \\ $$$$\frac{}{}\frac{}{}\frac{}{}\frac{}{}\frac{}{}\frac{}{}\frac{}{}\frac{}{}\frac{}{}\frac{}{}\frac{}{}\frac{}{}\frac{}{}\frac{}{}\frac{}{}\frac{}{}\frac{}{}\frac{}{}\frac{}{}\frac{}{}\frac{}{} \\ $$$$\mathrm{por}\:\mathrm{lo}\:\mathrm{anterior}\:\mathrm{podemos}\:\mathrm{reescribir}\:\bullet\:\mathrm{como} \\ $$$$\bullet\:{n}\left(\left(\mathrm{A}\cup\mathrm{B}\right)\cup\left(\mathrm{C}\cup\mathrm{D}\right)\right)={n}\left(\mathrm{A}\right)+{n}\left(\mathrm{B}\right)+{n}\left(\mathrm{C}\right)+{n}\left(\mathrm{D}\right)−{n}\left({A}\cap\mathrm{B}\right)−{n}\left(\mathrm{C}\cap\mathrm{D}\right)− \\ $$$$\left.{n}\left({A}\cap\mathrm{C}\right)−{n}\left(\mathrm{B}\cap\mathrm{C}\right)+{n}\left(\mathrm{A}\cap\mathrm{B}\cap\mathrm{C}\right)\right]− \\ $$$$\left.{n}\left({A}\cap\mathrm{D}\right)−{n}\left(\mathrm{B}\cap\mathrm{D}\right)+{n}\left(\mathrm{A}\cap\mathrm{B}\cap\mathrm{D}\right)\right]+ \\ $$$${n}\left({A}\cap\mathrm{C}\cap\mathrm{D}\right)+{n}\left(\mathrm{B}\cap\mathrm{C}\cap\mathrm{D}\right)−{n}\left(\mathrm{A}\cap\mathrm{B}\cap\mathrm{C}\cap\mathrm{D}\right) \\ $$$$ \\ $$$$ \\ $$

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