Problems on Metric Spaces

Question

Im doing some self studying on Topology, and in the chapter on metric spaces sometimes the examples just completely skips all the steps and just states the result for example one of the examples goes like this:

Consider the power set $P(\mathbb{N})$, for some set $C,D$ $\in$ $ P(\mathbb{N})$ (Note, the set of all positive even numbers, positive odd numbers, and prime numbers are examples of the members in the power set $P(\mathbb{N})$). Let us define $d(C,D)$ as:

$d(C,D)=0$, if $C=D$, and $\frac{1}{min((C-D)\cup (D-C))}$ if $C \ne D$.

Indeed $d$ is a metric and it can be shown with more effort that $d(C,E) \le \max\{d(C,D), (D,E)\}$. Then it says, can you show that $d(\mathbb{N}-C, \mathbb{N}-D)=d(C,D)?$

I am not so sure why this is the case, and have trouble showing what they ask, can anyone help me please? It will be a great help in further understanding this chapter.

Answer 1

I suppose you don't have problems in justifying that

• $d(A,B) \geq 0$ and $d(A,B) = 0$ iff $A=B$.
• $d(A,B) = d(B,A)$.

From (as Pedro M. claims in his comment) $$(A\setminus C) \cup (C \setminus A) \subseteq (A\setminus B) \cup (B \setminus A) \cup (B \setminus C) \cup (C \setminus B)$$ it follows that \begin{align} \min((A\setminus C) \cup (C \setminus A)) &\geq \min((A\setminus B) \cup (B \setminus A) \cup (B \setminus C) \cup (C \setminus B))\\ &= \min \{ \min((A\setminus B) \cup (B \setminus A)), \min((B \setminus C) \cup (C \setminus B)) \}, \end{align} whence $$\frac{1}{\min((A\setminus C) \cup (C \setminus A))} \leq \max\Bigl\{ \frac{1}{\min((A\setminus B) \cup (B \setminus A))}, \frac{1}{\min((B \setminus C) \cup (C \setminus B))} \Bigr\}.$$ Therefore $d(A,C) \leq d(A,B) + d(B,C)$.

Now, from (as claimed by Peter Melech) $$(\mathbb N \setminus C) \setminus (\mathbb N \setminus D) = D \setminus C$$ it follows that $$C \setminus D \cup D \setminus C = ((\mathbb N \setminus D)\setminus(\mathbb N \setminus C)) \cup ((\mathbb N \setminus C)\setminus(\mathbb N \setminus D)),$$ and so, $$\min(C \setminus D \cup D \setminus C) =\min((\mathbb N \setminus C)\setminus(\mathbb N \setminus D) \cup (\mathbb N \setminus D)\setminus(\mathbb N \setminus C)),$$ yielding $d(C,D) = d(\mathbb N \setminus C, \mathbb N \setminus D)$.