Solutions to Problem Set 19

Math 310-01/02

11-3-2017

1. Prove that

$$\bigcup_{n=1}^\infty \left[3, 5 - \dfrac{2}{n}\right] = [3, 5).$$

I will show that each of the two sets is contained in the other.

Let $\displaystyle x \in
   \bigcup_{n=1}^\infty \left[3, 5 - \dfrac{2}{n}\right]$ . By definition of the union, $x \in \left[3, 5 -
   \dfrac{2}{n}\right]$ for some $n \ge 1$ . Then

$$3 \le x \le 5 - \dfrac{2}{n} < 5.$$

Therefore, $x \in [3, 5)$ . This proves that $\displaystyle \bigcup_{n=1}^\infty
   \left[3, 5 - \dfrac{2}{n}\right] \subset [3, 5)$ .

Conversely, suppose $x \in [3,
   5)$ , so

$$3 \le x < 5.$$

Note that $\displaystyle \lim_{n
   \to \infty} \left(5 - \dfrac{2}{n}\right) = 5$ . Hence, for some $n \ge 1$ ,

$$3 \le x \le 5 - \dfrac{2}{n}.$$

For if, on the contrary, $5 -
   \dfrac{2}{n} < x$ for all $n \ge 1$ , then $\displaystyle \lim_{n \to \infty} \left(5 - \dfrac{2}{n}\right)
   \le x < 5$ .

Therefore, $x \in \left[3, 5 -
   \dfrac{2}{n}\right]$ for some $n \ge 1$ . Hence, $\displaystyle x \in x \in \bigcup_{n=1}^\infty \left[3, 5 -
   \dfrac{2}{n}\right]$ , by definition of the union. This proves that $\displaystyle [3, 5) \subset
   \bigcup_{n=1}^\infty \left[3, 5 - \dfrac{2}{n}\right]$ .

Hence, $\displaystyle
   \bigcup_{n=1}^\infty \left[3, 5 - \dfrac{2}{n}\right] = [3, 5)$ .


All action is involved in imperfection, like fire in smoke. - The Bhagavad Gita


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