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 $a,b,c \in \mathcal{J}$

  1. $d(a,b) \geq 0$
  2. $d(a,b) = 0$ iff $a =b$
  3. $d(a,b) = d(b,a)$
  4. $d(a,b) \leq d(a,c) + d(c,b)$

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.

  • 6
    $\begingroup$ A topological space is metrizable if there is a metric on the space giving the same topology. Remember that in a metric space, we also have open sets. The requirement is then that this metric has exactly the same open sets as the given topology. $\endgroup$ – Tobias Kildetoft May 18 '12 at 21:56
  • $\begingroup$ Here is a reference: encyclopediaofmath.org/index.php/Metrizable_space $\endgroup$ – Jonas Meyer May 18 '12 at 21:58

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.

  • $\begingroup$ I think you need to say "it is sufficient to show that there are distinct positive integers x, y", don't you? (Or not, if every open set around x necessarily includes x -- I suspect that's the case, but I don't know enough about topology to be sure!) $\endgroup$ – j_random_hacker Jan 24 '14 at 12:33

Associated with any metric is a topology where the open sets are arbitrary unions of open balls, i.e. sets of the form $\{y : d(x,y) < \epsilon\}$ for each $x$ in the set and $\epsilon > 0$.

We'd say the topological space $(Z,\mathcal J)$ was metrisable if there was a metric on $Z$ that gave rise to the topology $\mathcal J$. We'd be looking for $d : Z \times Z \to \mathbb R$, not $\mathcal J \times \mathcal J \to \mathbb R$ as you supposed.

For example, the usual topology $\mathcal T$ on $\mathbb R$ arises from the usual metric, so $(\mathbb R, \mathcal T)$ is metrisable. $\mathbb R$ with a different topology may not be metrizable, if the topology cannot arise from a metric.

Proving metrisability can be done by giving a metric, or citing a metrization theorem. Proving non-metrisability usually means finding some property held by all metric spaces and showing that the given topology does not have that property. For example, any two points in a metric space are topologically distinguishable in the sense that the open ball around one of radius $d/2$, where $d$ is the distance between them, does not contain the other, so there is an open set containing one point but not the other. Hence the indiscrete topology on a set with more than one point cannot be metrisable, because its points are not distinguishable in this way.


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