Metrizable Topological Space

A metrizable space is a topological space $ X $ where there exists a metric $ d $ that induces the topology on $ X $.

A metric $ d $ on a space $ X $ is a function $ d: X \times X \to [0, \infty) $ that satisfies the following properties: non-negativity, symmetry, the triangle inequality, and the condition that $ d(x, y) = 0 $ if and only if $ x = y $.

The topology induced by $ d $ is defined such that the open subsets are unions of open balls, which take the form $ B_r(x) = \{y \in X : d(x, y) < r\} $, where $ r > 0 $ is the radius.

In other words, a topological space $ X $ is metrizable if there exists a metric $ d $ such that the topology generated by the open balls of $ d $ exactly matches the topology of $ X $.

Note: This means the open sets in the topology of $ X $ can be expressed as arbitrary unions of open balls defined by the metric $ d $.

For example, it is well known that a topology that is not Hausdorff cannot be induced by a metric. As a result, not all topological spaces are metrizable.

Practical Example
Notes

Practical Example

Consider the real line $ \mathbb{R} $ with its standard topology.

In this topology, open sets are arbitrary unions of open intervals $ (a, b) $, where $ a, b \in \mathbb{R} $ and $ a < b $.

We define a metric, specifically the standard distance on the real line, as follows:

$$ d(x, y) = |x - y| $$

This is the absolute distance between two points $ x $ and $ y $ on $ \mathbb{R} $.

With this metric, an open ball centered at $ x $ with radius $ r $ is an open interval:

$$ B_r(x) = \{ y \in \mathbb{R} : d(x, y) < r \} = (x - r, x + r) $$

This interval is an open set in the standard topology.

Since every open set in the standard topology of $ \mathbb{R} $ can be represented as a union of open intervals, and open intervals correspond to the open balls generated by the metric $ d(x, y) $, the real line $ \mathbb{R} $ with its standard topology is metrizable.

Example 2

Consider any set $ X $ (finite or infinite) with the discrete topology.

In this topology, every subset of $ X $ is open.

We define the following metric $ d $ on $ X $:

$$ d(x, y) =
\begin{cases}
0 & \text{if } x = y, \\
1 & \text{if } x \neq y.
\end{cases}
$$

This is called the discrete metric.

Let’s now verify whether this space is metrizable.

With this metric, an open ball of radius $ r $ centered at a point $ x $ is:

If $ r \leq 1 $, $ B_r(x) = \{ x \} $
Explanation: When $ r \leq 1 $, only $ d(x, y) = 0 $ satisfies $ d(x, y) < r $, which means $ y = x $. In this case, $ B_r(x) = \{ x \}, $ so the open ball contains only the central point $ x $.
If $ r > 1 $, $ B_r(x) = X $.
Explanation: When $ r > 1 $, both $ d(x, y) = 0 $ (when $ x = y $) and $ d(x, y) = 1 $ (when $ x \neq y $) satisfy $ d(x, y) < r $. As a result, all points in $ X $ meet the condition $ d(x, y) < r $, so $ B_r(x) = X $.

The sets $ \{ x \} $ and $ X $ are open in the discrete topology.

Since every open set in the discrete topology can be described as a union of these open balls, $ X $ with the discrete topology is metrizable.

In this example too, the metric provides an exact description of the topology of the space.

Notes

Here are a few additional notes on metrizable spaces:

If a topological space $ X $ is metrizable and $ Y $ is homeomorphic to $ X $, then $ Y $ is also metrizable.
This theorem states that metrizability is a topological property that remains unchanged under homeomorphisms. In other words, if a space $ X $ is metrizable, then any space that is homeomorphic to it must also be metrizable. So, if I come across a space $ Y $ that is homeomorphic to $ X $, I don’t need to explicitly define or construct a metric for $ Y $—I can immediately conclude that it is metrizable.
Urysohn's Metrization Theorem
A topological space is metrizable if it is regular and has a countable basis. This theorem provides a key criterion for determining whether a given space can be endowed with a metric.

And so forth.