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Exercise 2.1.15. The power set $P(N)$ is the collection of all subsets of the set $N$ of natural numbers. For example, the following are considered as three points in $P(N)$: $E = \{2, 4, 6, 8, ...\} = $ all even numbers, $P = \{2, 3, 5, 7, ...\} = $ all prime numbers, $S = \{1, 4, 9, 16, ...\} = $ all square numbers. For $A, B \in P(N)$, define $d(A, B) = \begin{cases} 0, & \text{if } A = B, \\\frac{1}{min((A - B) \cup (B - A))}, & \text{if } A \neq B. \end{cases}$ For example, $(E - P) \cup (P - E) = \{3, 4, 5, ...\} $ implies $d(E, P) = \frac{1}{3}$. Prove 1. $d(A, C) \le max\{d(A, B), d(B, C)\}$. 2. $d$ is a metric. 3. $d(N - A, N - B) = d(A, B)$. The last property means that the complement map $A \to N - A$ is an isometry.

          Exercise 2.1.15. The power set $P(N)$ is the collection of all subsets of the set $N$ of natural numbers. For example, the following are considered as three points in $P(N)$:
$E = \{2, 4, 6, 8, ...\} = $ all even numbers,
$P = \{2, 3, 5, 7, ...\} = $ all prime numbers,
$S = \{1, 4, 9, 16, ...\} = $ all square numbers.
For $A, B \in P(N)$, define
$d(A, B) = \begin{cases} 0, & \text{if } A = B, \\\frac{1}{min((A - B) \cup (B - A))}, & \text{if } A \neq B. \end{cases}$
For example, $(E - P) \cup (P - E) = \{3, 4, 5, ...\} $ implies $d(E, P) = \frac{1}{3}$. Prove
1. $d(A, C) \le max\{d(A, B), d(B, C)\}$.
2. $d$ is a metric.
3. $d(N - A, N - B) = d(A, B)$.
The last property means that the complement map $A \to N - A$ is an isometry.

Exercise 2.1.15. The power set P(N) is the collection of all subsets of the set N of natural numbers. For example, the following are considered as three points in P(N):
E = {2, 4, 6, 8, ...} = all even numbers,
P = {2, 3, 5, 7, ...} = all prime numbers,
S = {1, 4, 9, 16, ...} = all square numbers.
For A, B ∈ P(N), define
d(A, B) = 0, if A = B,
(1)/(min((A - B) ∪ (B - A))), if A ≠ B.
For example, (E - P) ∪ (P - E) = {3, 4, 5, ...} implies d(E, P) = (1)/(3). Prove
1. d(A, C) ≤ max{d(A, B), d(B, C)}.
2. d is a metric.
3. d(N - A, N - B) = d(A, B).
The last property means that the complement map A → N - A is an isometry.

Added by Sandra C.

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