Intersection between the Boundary and the Interior of a Set

The intersection between the boundary $ \partial A $ and the interior $ \text{Int}(A) $ is empty: $$ \partial A \cap \text{Int}(A) = \emptyset $$

This statement illustrates the relationship between the boundary and the interior of a set in a topological context.

A Numerical Example
Proof

A Numerical Example

Consider the topological space $\mathbb{R}$ with the standard topology, where the open sets are open intervals.

Take the set $A = (0, 1)$, the open interval between 0 and 1.

The interior of $A$ is the set of points that have a neighborhood entirely contained within $A$. In this case, every point in $A$ is an interior point.

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

The closure of $A$ includes all points of $A$ plus its boundary points, 0 and 1.

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

The complement of $A$ in $\mathbb{R}$ is:

$$ \mathbb{R} - A = (-\infty, 0] \cup [1, \infty) $$

The closure of the complement of $A$ is:

$$ \text{Cl}(\mathbb{R} - A) = (-\infty, 0] \cup [1, \infty) $$

Since the complement of $A$ is already closed, its closure remains the same.

The boundary of $A$ consists of the points 0 and 1.

$$ \partial A = \text{Cl}(A) \cap \text{Cl}(\mathbb{R} - A) $$

$$ \partial A = [0, 1] \cap ((-\infty, 0] \cup [1, \infty)) $$

$$ \partial A = \{0, 1\} $$

The intersection between the boundary and the interior of $A$ is empty.

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

As shown, there are no common points between the boundary of $A$ ($\partial A = \{0, 1\}$) and the interior of $A$ ($\text{Int}(A) = (0, 1)$).

This numerical example confirms that the intersection between the boundary of a set $A$ and its interior is always empty:

$$ \partial A \cap \text{Int}(A) = \emptyset $$

Proof

This property can be proven using the definitions of topological sets.

By definition, the boundary of a set $A$ is:

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

The boundary of $A$ ($\partial A$) contains those points that are on the edge of $A$ but not completely inside $A$.

By definition, the interior $ \text{Int}(A) $ of the set $A$ is the set of all points in $A$ that have a neighborhood completely contained in $A$. The interior of $A$ contains points that are fully within $A$, meaning there is a neighborhood around each point in $ \text{Int}(A) $ that is entirely within $A$.

Consider a point $x \in \partial A$.

By definition, $x$ must satisfy:

$x \in \text{Cl}(A)$, so $x$ is either a boundary point of $A$ or a point in $A$.
$x \in \text{Cl}(X - A)$, so $x$ is either a boundary point of $X - A$ or a point in $X - A$.

Since $x \in \text{Cl}(A)$ and $x \in \text{Cl}(X - A)$, every neighborhood of $x$ will intersect both $A$ and $X - A$.

Therefore, $x$ cannot be a point that is completely within $A$, meaning it cannot be a point in $ \text{Int}(A) $.

Now consider a point $y \in \text{Int}(A)$.

By definition, there exists a neighborhood of $y$ entirely contained within $A$.

This means that $y$ cannot be a boundary point of $X - A$ and therefore cannot be in $\text{Cl}(X - A)$.

Therefore, $y$ cannot be a point in $\partial A$.

From these observations, we conclude that there are no common points between $\partial A$ and $\text{Int}(A)$, meaning their intersection is an empty set.

$$ \partial A \cap \text{Int}(A) = \emptyset $$

And so on