Boundary of a Set

The boundary of a subset $ A $ in a topological space $ X $ is the set of points that belong to the closure of $ A $ but not to its interior. \[ \partial A = \text{Cl}(A) - \text{Int}(A) \]

Where $ \text{Cl}(A) $ is the closure of $ A $, which includes all the points in $ A $ and its limit points.

Meanwhile, $ \text{Int}(A) $ is the interior of $ A $, consisting of all points in $ A $ that have a neighborhood entirely contained within $ A $.

example of boundary of a set

It's essential to reiterate that the concept of a boundary is not an intrinsic property of a set; it depends on the topology employed.

Thus, the boundary of a set can vary when a different topology is utilized.

In other words, the boundary of a set $ A $ consists of the points that are "close" to both the set $ A $ and its complement $ X \setminus A $.

A Practical Example
The Boundary Theorem
Notes

A Practical Example

Consider the set $ A = (0, 1) $ as a subset of the real line $ \mathbb{R} $ in the standard topology.

Let's determine the boundary of $ A $.

1] Calculate the Closure of A

The closure of $ A $, denoted $ \text{Cl}(A) $, includes all the points in $ A $ and its limit points.

For $ A = (0, 1) $, the closure is the closed interval $[0, 1]$ because every point in the interval $(0, 1)$ is a limit point, and the points 0 and 1 are the endpoints of the interval.

$$ \text{Cl}(A) = [0, 1] $$

2] Calculate the Interior of A

The interior of $ A $, denoted $ \text{Int}(A) $, consists of all points in $ A $ that have a neighborhood entirely contained within $ A $.

For $ A = (0, 1) $, the interior is the same set $(0, 1)$ because every point within the open interval has a neighborhood entirely within $(0, 1)$.

$$ \text{Int}(A) = (0, 1) $$

3] Calculate the Boundary of A

The boundary of $ A $, denoted $ \partial A $, is defined as the set difference between the closure of $ A $ and its interior:

$$ \partial A = \text{Cl}(A) - \text{Int}(A) $$

Using the results from the previous steps:

$$ \partial A = [0, 1] - (0, 1) = \{0, 1\} $$

Therefore, in the standard topology on $ \mathbb{R} $, the boundary of the set $ A = (0,1) $ is the set of points $\{0, 1\}$.

These two points are close to both the interior and exterior of $ A $, thus representing the boundary of the set.

example of set boundary

The Boundary Theorem

A point $ x $ in a topological space $ X $ belongs to the boundary $ x \in \partial A $ of a subset $ A $ if and only if every neighborhood of $ x $ intersects both $ A $ and $ X - A $.

According to this theorem, to determine if a point $ x $ belongs to the boundary of a set $ A $, you check if every neighborhood of $ x $ intersects both $ A $ and $ X - A $.

Example

Consider the set $ A = (0, 1) $ on the real line $ \mathbb{R} $ with the standard topology.

The closure and interior of this set are:

$$ \text{Cl}(A) = [0, 1] $$

$$ \text{Int}(A) = (0, 1) $$

Thus, the boundary of the set $ A $ consists only of the points 0 and 1.

$$ \partial A = \text{Cl}(A) - \text{Int}(A) = [0, 1] - (0, 1) = \{0, 1\} $$

Let's verify if the points 0 and 1 belong to the boundary using the previous theorem.

1] Check Point 0

Consider a neighborhood of zero, $ (0-\epsilon, 0+\epsilon) $ with $ \epsilon > 0 $.

This neighborhood intersects $ A $ because every neighborhood of 0 contains points in $ (0, 1) $.

The neighborhood also intersects $ X - A $, which is $ (-\infty, 0] \cup [1, \infty) $, as it contains points less than 0.

Since every neighborhood of 0 intersects both $ A $ and $ X - A $, we conclude that $ 0 \in \partial A $.

Neighborhood of point 0

2] Check Point 1

Consider a neighborhood of 1, for instance, the interval $ (1-\epsilon, 1+\epsilon) $ with $ \epsilon > 0 $.

This neighborhood intersects $ A $ because every neighborhood of 1 contains points in $ (0, 1) $.

The neighborhood also intersects $ X - A $, which is $ (-\infty, 0] \cup [1, \infty) $, as it contains points greater than 1.

Since every neighborhood of 1 intersects both $ A $ and $ X - A $, we conclude that $ 1 \in \partial A $.

Neighborhood of point 1

3] Check a Point Within (0,1)

Consider any point within the interval (0,1).

For example, the interval $ (0.5-\epsilon, 0.5+\epsilon) $ with $ \epsilon > 0 $.

This neighborhood intersects $ A $ because every neighborhood of 0.5 contains points in $ (0, 1) $.

However, this neighborhood does not intersect $ X - A $ because all points within this neighborhood belong to $ A $.

Since there is a neighborhood of 0.5 that does not intersect $ X - A $, we conclude that $ 0.5 \notin \partial A $.

Neighborhood of 0.5

In conclusion, using the theorem, we verified that the points 0 and 1 belong to the boundary of $ A = (0, 1) $, while an interior point like 0.5 does not. This confirms that the boundary of $ A $ is $\{0, 1\}$.

Notes

Some observations and additional notes on set boundaries:

Boundary $ \partial A $ is a subset of $ A $ if and only if $ A $ is closed:
\[ \partial A \subseteq A \Leftrightarrow A \text{ è chiuso} \]
Intersection of the boundary and the set A is empty if and only if A is open
In other words, a set $ A $ is open if and only if none of its points lie on its boundary.
\[ \partial A \cap A = \emptyset \Leftrightarrow A \text { is open} \]
Boundary $ \partial A $ is empty if and only if $ A $ is both open and closed (clopen):
\[ \partial A = \emptyset \Leftrightarrow A \text{ is clopen} \]
Boundary $ \partial A $ coincides with the intersection of the closure of $ A $ and the closure of $ X-A $:
\[ \partial A = \text{Cl}(A) \cap \text{Cl}(X - A) \]
Boundary $ \partial A $ is always a closed set

The intersection of two closed sets is always a closed set, which is a fundamental property in topological spaces. Since $\partial A$ is defined as the intersection of the two closed sets $\text{Cl}(A)$ and $\text{Cl}(X - A)$ \[ \partial A = \text{Cl}(A) \cap \text{Cl}(X - A) \] we can therefore conclude that $\partial A$ is always closed.
Intersection of the boundary $ \partial A $ and the interior $ \text{Int}(A) $ is empty:
\[ \partial A \cap \text{Int}(A) = \emptyset \]
Union of the boundary $ \partial A $ and the interior $ \text{Int}(A) $ is equal to the closure of $ A $:
\[ \partial A \cup \text{Int}(A) = \text{Cl}(A) \]

And so on.