A Basis for the Metric Topology
In a metric space \((X,d)\), the collection of all open balls $$ \mathcal{B}=\{B_d(x,\varepsilon)\mid x\in X,\ \varepsilon>0\} $$ forms a basis for a topology on \(X\). The topology generated in this way is called the metric topology.
This result is one of the fundamental links between metric spaces and topology. It shows that every open set in a metric space can be built from open balls, making them the basic geometric objects from which the entire topology is constructed.
A collection of sets \(\mathcal{B}\) is a basis for a topology if every open set can be expressed as a union of sets belonging to \(\mathcal{B}\).
To qualify as a basis, the collection must satisfy two properties:
- Covering property. Every point \(x\in X\) must belong to at least one basis element.
- Intersection property. Whenever a point \(x\) belongs to the intersection of two basis elements, there must be a third basis element containing \(x\) and lying entirely inside that intersection.
The theorem states that the family of all open balls satisfies both properties. As a consequence, open balls generate the metric topology.
Recall that the open ball centered at \(x\) with radius \(\varepsilon>0\) is defined by $$ B_d(x,\varepsilon)=\{y\in X\mid d(x,y)<\varepsilon\}. $$ In other words, it consists of all points whose distance from \(x\) is strictly less than \(\varepsilon\).
From a geometric point of view, the theorem tells us that every open set can be assembled from open balls in much the same way that a building is assembled from bricks.
An Example
Consider the following subset of the plane \(\mathbb{R}^2\):
$$ A=\{(x_1,x_2)\in\mathbb{R}^2\mid 1
This set consists of all points whose distance from the origin is greater than 1 and less than 2.
Geometrically, it is an annulus, that is, the region between two concentric circles of radii 1 and 2.
Since the boundary circles are excluded, the set \(A\) is open.
Can this annulus be written as a union of open balls?
According to the basis theorem, the answer is yes.
Choose any point inside the annulus and draw an open ball around it with a radius small enough to keep the entire ball inside the set.
For example, consider the point \(p_1=(1.5,0)\) and the radius \(\varepsilon_1=0.3\):
$$ B_d((1.5,0),0.3) = \{(x_1,x_2)\in\mathbb{R}^2 \mid d((1.5,0),(x_1,x_2))<0.3\}. $$
This open ball lies entirely within the annulus.
Now consider another point, \(p_2=(-1.5,0)\), together with the radius \(\varepsilon_2=0.4\):
$$ B_d((-1.5,0),0.4) = \{(x_1,x_2)\in\mathbb{R}^2 \mid d((-1.5,0),(x_1,x_2))<0.4\}. $$
This ball is also completely contained in the annulus.
By continuing this process, we can place open balls throughout the annulus, always choosing radii small enough to avoid crossing its boundaries.
Collectively, these open balls cover every point of \(A\).
In symbolic form:
$$ A=\bigcup_i B_d(x_i,\varepsilon_i). $$
Each center \(x_i\) lies in the annulus, and each radius \(\varepsilon_i\) is chosen so that $$ B_d(x_i,\varepsilon_i)\subseteq A. $$
This example illustrates a general principle: every open set in a metric space can be represented as a union of open balls.
That is why open balls are regarded as the fundamental building blocks of the metric topology.
Proof
We now prove that the collection of all open balls in a metric space \((X,d)\) forms a basis for a topology on \(X\).
It is enough to verify the two basis properties.
Every Point Belongs to a Basis Element
Let \(x\in X\).
For any radius \(\varepsilon>0\), the open ball \(B_d(x,\varepsilon)\) belongs to the collection \(\mathcal{B}\).
Moreover, $$ d(x,x)=0<\varepsilon, $$ so \(x\in B_d(x,\varepsilon)\).
Therefore, every point of \(X\) belongs to at least one open ball, and the covering property is satisfied.
The Intersection Property
Let \(B_1\) and \(B_2\) be two open balls, and suppose that $$ x\in B_1\cap B_2. $$
We must show that there exists an open ball centered at \(x\) that is contained entirely within the intersection.
Because \(x\in B_1\), there is a radius \(\delta_1>0\) such that
$$ B_d(x,\delta_1)\subseteq B_1. $$
Similarly, because \(x\in B_2\), there is a radius \(\delta_2>0\) such that
$$ B_d(x,\delta_2)\subseteq B_2. $$
Now choose the smaller of the two radii:
$$ \delta=\min\{\delta_1,\delta_2\}. $$
Then every point of \(B_d(x,\delta)\) belongs to both \(B_1\) and \(B_2\), which implies
$$ B_d(x,\delta)\subseteq B_1\cap B_2. $$
Thus, the intersection property is satisfied.
Since both basis properties hold, the family of all open balls is a basis for a topology on \(X\).
This topology is precisely the metric topology induced by the metric \(d\).
