Urysohn’s Metrization Theorem
If a topological space is regular and has a countable base, then it is metrizable.
In simpler terms, any regular topological space with a countable base can be described by a metric that produces exactly the same topology.
- A space is regular if, for every point and every closed set not containing it, there exist two disjoint open sets that separate them. Intuitively, this means points and closed sets can be "kept apart" by suitable open neighborhoods.
- A space has a countable base if there exists a countable collection of open sets from which all other open sets can be generated. In other words, a finite or countable number of "building blocks" is enough to construct the entire topology.
So if a space is sufficiently well-behaved (regular) and not too vast or complex (countable base), it can always be represented using a distance function.
The converse is not true. If a space is metrizable, this does not necessarily imply that it has a countable base or that it is regular by definition. In other words, some spaces can be described by a distance function even though they do not satisfy the precise conditions required by the theorem. Urysohn's theorem tells us when a metric can exist - it doesn't claim that every metric must preserve the same topological properties.
Understanding the Theorem
Urysohn's metrization theorem explains under what conditions a topological space can be expressed in terms of a distance function. It shows when we can move from abstract notions of "open" and "closed" sets to a concrete metric framework.
The essential idea is that topology doesn't always begin with distance. Sometimes, we only specify which sets are open, and from that information the entire structure of the space can be determined.
The key question is: when can a topology be represented by a metric? In other words, when does there exist a function $ d(x,y) $ that measures distance in a way consistent with the given topology?
Urysohn provided a precise answer: a topological space is metrizable if it satisfies two fundamental conditions:
- It is regular. That is, for every point and every closed set disjoint from it, there exist open sets that separate them. This ensures a well-defined notion of separation among points in the space.
- It has a countable base, meaning there exists a countable family of open sets that generates the entire topology. This condition prevents the space from being "too large" in a topological sense.
If both of these requirements are met, then the space is metrizable - that is, it can be fully described by some metric.
Why does this matter? The theorem bridges the gap between topology and geometry. It shows that certain "well-structured" topological spaces can be treated as metric spaces, meaning we can meaningfully talk about distance, convergence, and continuity just as we do in ordinary geometry. Many familiar spaces from analysis and physics - such as the real line, the plane, and Euclidean spaces - fit within this framework.
A Concrete Example
The real line ℝ, equipped with the standard topology (formed by open intervals), satisfies both of Urysohn's conditions:
- It is regular.
- It has a countable base (the intervals with rational endpoints).
Therefore, by Urysohn's theorem, the real line ℝ is metrizable. In fact, the usual distance function $ d(x, y) = |x - y| $ works perfectly - it generates exactly the same topology.
Note. The standard topology on ℝ is defined so that open sets are unions of open intervals of the form (a, b).
A point $ x $ belongs to an open set $ A $ if there exists a small interval $ (x - \varepsilon, x + \varepsilon) $ entirely contained in $ A $. This topology arises directly from the Euclidean distance $ |x - y| $ - it's the "natural" one used in analysis, where limit, continuity, and convergence have their usual geometric meanings.
Example 2
Now consider the real line ℝ with the discrete topology.
This is also a metrizable space, because we can define the discrete metric as follows:
$$ d(x, y) =
\begin{cases}
0, & \text{if } x = y \\ \\
1, & \text{if } x \ne y.
\end{cases}
$$
However, its base is not countable, since each point forms its own distinct open set, and the set of real numbers is infinite and uncountable.
This confirms that the converse of the theorem does not hold: a metrizable space does not necessarily have a countable base.
Note. The discrete topology is the "finest" possible topology on a set: every subset is both open and closed. In particular, each point forms its own open set, that is,
$$ {x} \text{ is open for every } x \in \mathbb{R}. $$
In this kind of topology, there is no notion of continuity between points - each one stands entirely on its own. It's a very simple topology, yet "too detailed" to be generated by a countable base when the underlying set (like ℝ) is uncountable.
And so on.
