If $E_n = (-1 + \sin\frac{1}{2n}, 1 - \sin\frac{1}{2n} )$, then what would $\bigcup_{n=1}^\infty E_n$ be equal to? Would it equal the interval (-1, 1)?.
Thanks
If $E_n = (-1 + \sin\frac{1}{2n}, 1 - \sin\frac{1}{2n} )$, then what would $\bigcup_{n=1}^\infty E_n$ be equal to?
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elementary-set-theory
1 Answers
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Yes, $\sin \frac{1}{2n}$ starts at $\sin \frac{1}{2}$ and decreases monotonically to zero, so $E_1 \subseteq E_2 \subseteq \cdots$.
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1I think you mean that it starts at $\sin(½)$ – 2017-02-28