Question

Arată că $\bigcap_{n=1}^{\infty}{\left(-\frac{1}{n},\frac{1}{n}\right)=\left\{0\right\}$

Answer 1

$\displaystyle \bigcap\limits_{n=1}^{\infty}\Big(-\dfrac{1}{n},\dfrac{1}{n}\Big) = \\ \\ = \Big(-\dfrac{1}{1},\dfrac{1}{1}\Big)\cap \Big(-\dfrac{1}{2},\dfrac{1}{2}\Big)\cap\Big(-\dfrac{1}{3},\dfrac{1}{3}\Big)\cap ... \cap\Big(-0.000...1,0.000...1\Big) = \\ \\ = \Big((-1,1)\cap(-0.5,0.5)\Big)\cap(-0.333...,0.333...)\cap...\cap(-0.000...1,0.000...1) =\\ \\ \boxed{(-1,1)\cap(-0.5,0.5) = (-0.5,0.5)}$

$= \Big((-0.5,0.5)\cap(-0.333...,0.333...)\Big)\cap...\cap(-0.000...1,0.000...1) = \\ \\ \boxed{(-0.5,0.5)\cap(-0.333...,0.333...) = (-0.333...,0.333...)} \\ \\ =(-0.333...,0.333...)\cap...\cap(-0.000...1,0.000...1) = \\ \\ \text{Observam ca toate intervalele se absorb de la stanga la dreapta.}\\ \\ \text{Cum limita acestei intersectii este }(0_-,0_+).\\ \\ \text{Asta inseamna ca intervalele se absorb repetat pana la }(0_-,0_+).$

$\displaystyle\Rightarrow \bigcap\limits_{n=1}^{\infty}\Big(-\dfrac{1}{n},\dfrac{1}{n}\Big) = (0_-,0_+) = \{0\}$

Answer 2

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