# Can a metric space over integers induce a topology?

Questions to get a better grasp of basic topology:

A metric space is an ordered pair $(M,d)$ where $M$ is a set and $d$ is a metric on $M$, i.e., a function

$$d \colon M \times M \to \mathbb{R}$$

such that for any $x, y, z \in M$, the following holds:

1. $d(x,y) = 0 \Leftrightarrow x = y$ (identity of indiscernibles)
2. $d(x,y) = d(y,x)$ (symmetry)
3. $d(x,z) \le d(x,y) + d(y, z)$ (subadditivity or triangle inequality)

Now let $M = \{1,2,3\}$ and $d(x, y) = |x - y|$. I claim that all three properties are satisfied (feel free to check).

A metric space is not a topological space. However, every metric space gives rise to a topological space in a rather natural way. This is the well known construction that takes a metric space $X$ and constructs the topology on $X$ where a set $U$ is open precisely when for every $x \in U$ there exists some $e>0$ such that the open ball $B_e(x)$ is contained in $U$.

Let $U = \{1,2\}$ with $1,2 \in X$. Is $U$ open? For $x=1$ and $e = 1$ it holds that $B_1(2) \subseteq U$, given the following definition:

$$B_r(x) := \{y \in M : d(x,y) < r\}$$ via Wikipedia: Metric space

because $\{\} \subseteq \{1,2\}$. For $x=2$ and $e=1$, the same holds. $U=\{1,2\}$ is one example. What are other open sets? I claim $U=\{1\}$, $U=\{1,3\}$ and actually any element of $\mathcal P(X)$.

Formally, let $X$ be a set and let $τ$ be a family of subsets of $X$. Then $τ$ is called a topology on X if:

1. Both the empty set and $X$ are elements of $τ$
2. Any union of elements of $τ$ is an element of $τ$
3. Any intersection of finitely many elements of $τ$ is an element of $τ$

If $τ$ is a topology on $X$, then the pair ($X$, $τ$) is called a topological space.

Let $X = \{1,2,3\}$ and $τ = \{\{\}, \{1,2,3\}, \{1\}, \{1,2\}, \{1,3\}\}$. I claim that all three properties are satisfied (feel free to check). So $(X, τ)$ is a topological space.

Questions:

1. Is $(M, d)$ a metric space (here)?
2. Is $U$ an open set for every $U \in \mathcal P(X)$ (here)?
3. Is $(X, τ)$ a topological space (here)?
4. Assuming my construction was right, the answer is yes. Otherwise, please give me an explicit example of a metric space over integers with some induced topology.
• (1) Yes, that's what the notation means. (2) What do you mean by unique "in terms of the given metric space"? (3) No, not every subset of a metric space is an open set. But every subset of the integers (using the distance metric) is an open set. (4) Yes, that's what the notation means. – arctic tern Jul 22 '16 at 22:34
• every metric space induces a topology, when we take the set of open balls as a basis. – Jorge Fernández Hidalgo Jul 22 '16 at 22:36
• For an example of a non-discrete topology copy the structure of the metric space over Q to Z. – Jorge Fernández Hidalgo Jul 22 '16 at 22:37
• In the topology $\tau$ induced by the usual metric on a finite set of integers, every point is open. That is, every singleton $\{x\} = B_1(x)$ is open; thus every subset of the finite set is open. So $\tau$ is the discrete topology. – BrianO Jul 22 '16 at 22:37
• And you "claim" the answers to 1 and 4 are yes in the statements preceding your questions. – Aweygan Jul 22 '16 at 22:39

1. Yes, the set $M=\{1,2,3\}$ equipped with the function $d(x,y) = |x-y|$ is a metric space, because $d$ satisfies all the metric space requirements.
2. (If you mean $U\subseteq M$ rather than $U\subseteq X$.) Yes, every subset of $M$ is an open set. In particular, the singletons are open (because you can take a ball of radius $\frac{1}{2}$ around any point), and $M$ is finite, so every subset of $M$ can be written as a countable (in fact, finite!) union of (open!) singletons and is therefore open.
3. Yes, $(X, \tau)$ is a topological space. (It is not the one induced by the metric $d$, because for example $\{2\}$ is not open in $\tau$.)
• LGTM except for question 4. $U = \{2\}$, so $x=2$ and $e=1$. Then $B_1(2) = \{2\}$ and $B_1(2) \subseteq U$. So $U$ is open?! – meisterluk Jul 22 '16 at 23:09
• @meisterluk $\{2\}$ is open in the topology induced by the metric $d$ (for the reason you just gave), but not in the topology $\tau$ (since $\{2\} \notin \tau$). – Adriano Jul 23 '16 at 2:57