I am learning a bit about Topology through independent study. I am using Bert Mendelson's "Introduction to Topology - 3rd Edition".
I have a question on one of the book's example and related exercise.
Example 7, pg. 72
Let $Z$ be the set of positive integers. For each positive integer $n$, let $O_n = \{n, n+1, n+2, \cdots\}$. Let $\mathcal{J} = \{\emptyset,O_1,O_2,\cdots\}$, then $(Z,\mathcal{J})$ is a topological space.
Excercise 1 on pg. 74 asks us to prove that the topological space defined in example 7 is "non-metrizable".
The book so far has no specific definition of metrizable vs. non-metrizable topological spaces. However, it does mention that "some topological spaces cannot have risen from a metric space", citing example 7 as one of these cases.
Here is my question:
Does metrizable refer to the ability of defining some metric space of $\mathcal{J}$? In other words, is there some function $d:\mathcal{J} \times \mathcal{J} \rightarrow \mathcal{R}$, satisfying the conditions:
Let $a,b,c \in \mathcal{J}$
- $d(a,b) \geq 0$
- $d(a,b) = 0$ iff $a =b$
- $d(a,b) = d(b,a)$
- $d(a,b) \leq d(a,c) + d(c,b)$
In this case if we define a function:
For $a,b \in \mathcal{J}$
$d'(a,b) = \left\{ \begin{array}{c c l} 0 & \text{if } & a = b \\ 1 & \text{if } & a \neq b \end{array}\right.$
Will $(\mathcal{J},d')$ not be a metric space?
I would appreciate any insight.