# 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 \( A \).

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

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.

## 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 \( 0 \in \partial A \).

**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 \( 1 \in \partial A \).

**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 \( 0.5 \notin \partial A \).

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.