Solutions to Problem Set 20

Math 310-01/02

11-3-2017

1. Prove that

$$\bigcap_{n = 1}^\infty \left[0, 1 + \dfrac{1}{n}\right] = [0, 1].$$

Suppose $x \in [0, 1]$ . Then for $n \ge 1$ ,

$$0 \le x \le 1 < 1 + \dfrac{1}{n}.$$

Hence, $x \in \left[0, 1 +
   \dfrac{1}{n}\right]$ . Since this is true for all $n \ge 1$ , it follows that $\displaystyle x \in \bigcap_{n =
   1}^\infty \left[0, 1 + \dfrac{1}{n}\right]$ .

Suppose $\displaystyle x
   \in\bigcap_{n = 1}^\infty \left[0, 1 + \dfrac{1}{n}\right]$ . Then $x \in \left[0, 1 +
   \dfrac{1}{n}\right]$ for all $n \ge 1$ . Hence,

$$0 \le x \le 1 + \dfrac{1}{n} \quad\hbox{for all}\quad n \ge 1.$$

I claim that $x \le 1$ . Suppose on the contrary that $x > 1$ . Now $\displaystyle \lim_{n \to \infty} \left(1 + \dfrac{1}{n}\right)
   = 1$ , so for some $n \ge 1$

$$1 < 1 + \dfrac{1}{n} < x.$$

This contradicts the fact that $x \in \left[0, 1 + \dfrac{1}{n}\right]$ for all $n \ge
   1$ .

Hence, $x \le 1$ , and so $x \in [0, 1]$ .

Therefore, $\displaystyle
   \bigcap_{n = 1}^\infty \left[0, 1 + \dfrac{1}{n}\right] = [0, 1]$ .


2. An equivalence relation is defined on the set $X = \{1, 4, 8, 13, 14, 17, 25, 30,
   31\}$ by $a \sim b$ means that a and b differ by an (integer) multiple of 5. List the elements of the equivalence classes for this relation.

$$\{1, 31\}, \{4, 14\}, \{8, 13\}, \{17\}, \{25, 30\}.\quad\halmos$$


3. For each relation, check each of the axioms for an equivalence relation. For each axiom, if the axiom holds, prove it. If the axiom doesn't hold, give a specific counterexample.

(a) $x \sim y$ on $\real$ means that $|x - y| = 0$ or $|x - y| =
   3$ .

(b) $a \sim b$ on $\real$ means that $a + b \ge 1$ .

(c) $(a, b) \sim (c, d)$ on $\real \times \real$ means $a + b = c + d$ .

(a) For all $x \in \real$ , $|x - x| = 0$ , so $x \sim x$ . Hence, the relation is reflexive.

Suppose $x \sim y$ . There are two cases. First, if $|x - y| = 0$ , Then $|y - x|
   = 0$ , so $y \sim x$ . Second, if $|x
   - y| = 3$ , then $|y - x| = 3$ , so $y \sim x$ . Thus, the relation is symmetric.

I have $2 \sim 5$ , since $|2 - 5| = 3$ . I have $5 \sim 8$ , since $|5 - 8|
   = 3$ . However, $|2 - 8| = 6$ , so $2 \not\sim
   8$ . Therefore, the relation is not transitive.

(b) $-1 + (-1) = -2 \not\ge 1$ , so $-1 \not\sim -1$ . Therefore, the relation is not reflexive.

Suppose $a \sim b$ . Then $a + b \ge 1$ , so $b + a \ge 1$ . Therefore, $b
   \sim a$ . Thus, $\sim$ is symmetric.

Since $0.5 + 0.7 = 1.2 \ge 1$ , I have $0.5 \sim 0.7$ . Since $0.7 +
   0.4 = 1.1 \ge 1$ , I have $0.7 \sim 0.4$ . However, $0.5
   + 0.4 = 0.9 \not\ge 1$ , so $0.5 \not\sim 0.4$ . Therefore, the relation is not transitive.

(c) Since $a + b = a + b$ , I have $(a, b) \sim (a, b)$ . Thus, $\sim$ is reflexive.

If $(a, b) \sim (c, d)$ , then $a + b = c + d$ . Hence, $c + d
   = a + b$ , so $(c, d) \sim (a, b)$ . Thus, $\sim$ is symmetric.

If $(a, b) \sim (c, d)$ and $(c, d) \sim (e, f)$ , then

$$a + b = c + d \quad\hbox{and}\quad c + d = e + f, \quad\hbox{so}\quad a + b = e + f.$$

Therefore, $(a, b) \sim (e, f)$ , so $\sim$ is transitive.

It follows that $\sim$ is an equivalence relation.


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