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.
Best Answer
A key property that all metric spaces have is the Hausdorff axiom. Namely given any $x,y \in X$ such that $x \neq y$ there are open sets $U$ about $x$, $V$ about $y$ such that $U \cap V = \emptyset$.
To see how this holds in metric space take any two distinct points $v$ and $w$. They are distinct so by definition of a metric $d(v,w) = \epsilon$ for some $\epsilon > 0$. By the Archimedean property of the reals there exists $n \in \Bbb{N}$ such that
$$ 0 < \frac{1}{n} < \epsilon$$
from which it follows that $B_{\frac{1}{n}}(v)$ does not intersect $B_{\epsilon - \frac{1}{n}} (w)$. Since $v,w$ were any two distinct points in your metric space this proves the claim.
So to show that your topological space above cannot be turned into a metric space it is sufficient to show that there are positive integers $x,y \in Z$ such that any two open neighbourhoods about $x$ and $y$ respectively must have non-trivial intersection. Consider $1$ and $2$ that are in $\Bbb{Z}$. Then it is easy to see that the only open neighbourhood about $1$ is $O_1 = \{1,2,3, \ldots \}$ while the only open neighbourhood about $2$ is $O_2 = \{2,3,\dots \}$ or even $O_1$ as well. However it is clear that
$$O_1 \cap O_1 \neq \emptyset, \hspace{5mm} O_1 \cap O_2 \neq \emptyset$$
from which it follows that the topology $\mathcal{J}$ that you put on $Z$ is not Hausdorff and hence is not metrisable.