Menu Close

Question-159300




Question Number 159300 by saly last updated on 15/Nov/21
Answered by Derrick last updated on 15/Nov/21
preve  a)montrons par double inclusion  soit x  x∈f^(−1)  (A∪B)⇔f(x)∈(A∪B)                               ⇔f(x)∈A ou f(x)∈B                     ⇔x∈f^(−1) (A) ou x∈f^(−1) (B)                    ⇔x∈(f^(−1) (A)∪f^(−1) (B)
$${preve} \\ $$$$\left.{a}\right){montrons}\:{par}\:{double}\:{inclusion} \\ $$$${soit}\:{x} \\ $$$${x}\in{f}^{−\mathrm{1}} \:\left({A}\cup{B}\right)\Leftrightarrow{f}\left({x}\right)\in\left({A}\cup{B}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Leftrightarrow{f}\left({x}\right)\in{A}\:{ou}\:{f}\left({x}\right)\in{B} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Leftrightarrow{x}\in{f}^{−\mathrm{1}} \left({A}\right)\:{ou}\:{x}\in{f}^{−\mathrm{1}} \left({B}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Leftrightarrow{x}\in\left({f}^{−\mathrm{1}} \left({A}\right)\cup{f}^{−\mathrm{1}} \left({B}\right)\right. \\ $$$$ \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *