Metric Spaces Are Hausdorff Spaces

Every metric space is a Hausdorff space. If a topological space is not Hausdorff, it cannot arise from a metric.

The property of being a Hausdorff space ensures that for any two distinct points, there exist two disjoint neighborhoods separating them.

In simpler terms, this means that in a metric space (a space where distances between points are well-defined), we can always find open sets that separate any two distinct points.

Note. The Hausdorff property must hold for every pair of distinct points in the space, not just for some.

A Practical Example
Proof

A Practical Example

Consider the Euclidean plane $\mathbb{R}^2$ with the standard Euclidean distance, denoted $d(x, y)$, where $x = (x_1, x_2)$ and $y = (y_1, y_2)$ are points in the plane. The distance is defined as:

$$ d(x, y) = \sqrt{(x_1 - y_1)^2 + (x_2 - y_2)^2}. $$

In this context, $\mathbb{R}^2$ with the Euclidean distance forms a metric space.

Metric spaces like $\mathbb{R}^2$ satisfy the Hausdorff property, meaning we can "separate" any two distinct points using disjoint open sets.

Take two points $ A = (x_1, y_1) $ and $ B = (x_2, y_2) $ in $\mathbb{R}^2$ with $A \neq B$.

The Euclidean distance between $A$ and $B$ is $d(A, B) > 0$ because the points are distinct.

Now, choose a radius smaller than half the distance between $A$ and $B$:

$$ r = d(A, B) / 2 $$

Construct two circles with radius $r$:

$ U = \{ P \in \mathbb{R}^2 : d(P, A) < r \} $, the set of points within $r$ of $A$,
$ V = \{ P \in \mathbb{R}^2 : d(P, B) < r \} $, the set of points within $r$ of $B$.

Notice that $U$ and $V$ do not overlap:

$$ U \cap V = \varnothing $$

This happens because every point in $U$ is closer to $A$ than $B$, and every point in $V$ is closer to $B$ than $A$.

Since this reasoning applies to any two distinct points $ A $ and $ B $ in $\mathbb{R}^2$, we can conclude that $\mathbb{R}^2$ with the Euclidean distance satisfies the Hausdorff property.

Thus, $\mathbb{R}^2$ is a Hausdorff space.

Example 2

Consider the set $\mathbb{R}$ of real numbers with the finite complement topology.

In the finite complement topology, a set $U \subseteq \mathbb{R}$ is open if $U$ is empty ($\varnothing$) or if its complement $\mathbb{R} \setminus U$ is finite (i.e., it contains only finitely many elements).

In other words, a set is open if it includes "almost all" points of $\mathbb{R}$, except for a finite number.

Now, take two distinct points $x, y \in \mathbb{R}$.

Try to find disjoint open sets containing $x$ and $y$.

Let $U$ be an open set containing $x$. To be open in the finite complement topology, the complement of $U$ ($\mathbb{R} \setminus U$) must be finite, meaning $U$ must contain "almost all" points of $\mathbb{R}$, except for a finite number.

Similarly, let $V$ be an open set containing $y$, with its complement $\mathbb{R} \setminus V$ also finite.

Since both $U$ and $V$ contain "almost all" points of $\mathbb{R}$, their intersection $U \cap V$ cannot be empty because both sets must include infinitely many shared points.

Note. For example, take $x = 1$ and $y = 2$. Try to "separate" $x$ and $y$ using disjoint open sets in the finite complement topology.

Define an open set $U$ containing $x = 1$. For $U$ to be open in this topology, it must include all points of $\mathbb{R}$ except a finite number. For example, exclude $y = 2$ and a few other nearby points. Thus, $U$ will contain almost all real numbers: $$ U = \mathbb{R} \setminus (2-\epsilon, 2+\epsilon) $$
Define an open set $V$ containing $y = 2$. Similarly, $V$ must include all points of $\mathbb{R}$ except a finite number. For instance, exclude $x = 1$ and a few nearby points: $$ V = \mathbb{R} \setminus (1-\epsilon, 1+\epsilon) $$

Because of how open sets are defined, both $U$ and $V$ include "almost all" points of $\mathbb{R}$, so their intersection $U \cap V$ cannot be empty. It will contain infinitely many points: $$ U \cap V = \mathbb{R} \setminus \left[(2 - \epsilon, 2 + \epsilon) \cup (1 - \epsilon, 1 + \epsilon)\right] \ne \emptyset $$ This shows that in $\mathbb{R}$ with the finite complement topology, it is impossible to find disjoint open sets that separate $x = 1$ and $y = 2$.

Since we cannot find disjoint open sets $U$ and $V$ around two distinct points $x$ and $y$, the space $(\mathbb{R}, \text{finite complement topology})$ is not a Hausdorff space.

As a result, this topological space cannot be represented by a metric space.

Proof

Consider two distinct points $ x $ and $ y $ in a metric space $(X, d)$.

Since they are distinct, the two points are separated by some positive distance $\varepsilon > 0$.

Define two open spherical neighborhoods, or open balls, around $x$ and $y$ that do not overlap.

Let these neighborhoods be $ U $ and $ V $:

$ U $ is the set of points less than $\varepsilon/2$ away from $x$,
$ V $ is the set of points less than $\varepsilon/2$ away from $y$.

We now show that $U$ and $V$ do not overlap.

Assume, for contradiction, that there is a point $z$ belonging to both $U$ and $V$. This would mean:

$ z $ is within $\varepsilon/2$ of $x$,
$ z $ is within $\varepsilon/2$ of $y$.

By adding these distances and using the triangle inequality, the distance between $ x $ and $ y $, $ d(x, y) $, would be less than $\varepsilon$:

$$ d(x,y) < d(x,z) + d(z,y) < \epsilon/2 + \epsilon/2 = \epsilon $$

This contradicts our initial assumption that $ d(x, y) = \varepsilon$.

Thus, no point can belong to both $U$ and $V$.

This proves that in a metric space, we can always find disjoint open sets $ U $ and $ V $ around $ x $ and $ y $.

Therefore, every metric space is Hausdorff.

And so on.