Boundary is the Intersection of the Closure of the Set and the Closure of the Complement

If $A$ is a subset of a topological space $X$, the boundary $\partial A$ of the set $A$ is defined as the set of points that belong to both the closure of $A$ and the closure of the complement of $A$.  $$\partial A = \text{Cl}(A) \cap \text{Cl}(X \setminus A)$$

In other words, the boundary of a set $A$ is the intersection of the closure of $A$ and the closure of its complement.

The intersection of these two sets, $\text{Cl}(A) \cap \text{Cl}(X \setminus A)$, identifies the points that are simultaneously in the closure of $A$ and the closure of the complement of $A$.

These points form the boundary of $A$ because they lie "on the edge" between $A$ and its complement. 

A Practical Example

Consider the set $A$ as the open interval $(0, 1)$ on the real line $\mathbb{R}$.

In this case, the closure of $(0, 1)$ includes all points between 0 and 1, including the endpoints.

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

The closure of the complement of $(0, 1)$ is

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

The intersection of these two sets is the boundary of the set $A$.

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

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

Thus, the boundary of the interval $(0, 1)$ is $\{0, 1\}$, which are the points exactly at the interface between the interval and the rest of the real line.

The Proof

By definition, the boundary $\partial A$ of $A \subseteq X$ is the set of all points $x \in X$ such that every neighborhood of $x$ contains at least one point of $A$ and at least one point of $X \setminus A$.

$$\partial A = \{ x \in X \mid \forall U \in \mathcal{N}(x), U \cap A \neq \emptyset \text{ and } U \cap (X \setminus A) \neq \emptyset \}$$

Where $\mathcal{N}(x)$ denotes the family of neighborhoods of $x$.

We will divide the proof into two parts:

1] Proof that $\partial A \subseteq \text{Cl}(A) \cap \text{Cl}(X \setminus A)$

By the definition of boundary, every point $x \in \partial A$ has a neighborhood $U$ that intersects both the set $A$ and its complement $X \setminus A$.

This implies that every point $x \in \partial A$

• has a neighborhood that intersects $A$, so $x \in \text{Cl}(A)$
• has a neighborhood that intersects the complement $X \setminus A$, so $x \in \text{Cl}(X \setminus A)$

Therefore, $x \in \text{Cl}(A) \cap \text{Cl}(X \setminus A)$.

This shows that the boundary of $A$ is a subset of $\text{Cl}(A) \cap \text{Cl}(X \setminus A)$

$$\partial A \subseteq \text{Cl}(A) \cap \text{Cl}(X \setminus A)$$

2] Proof that $\text{Cl}(A) \cap \text{Cl}(X \setminus A) \subseteq \partial A$

Every point $x \in \text{Cl}(A) \cap \text{Cl}(X \setminus A)$

• has a neighborhood that intersects the set $A$ because $x \in \text{Cl}(A)$.
• has a neighborhood that intersects the complement $X \setminus A$ because $x \in \text{Cl}(X \setminus A)$

Thus, every neighborhood of $x$ contains at least one point of $A$ and at least one point of $X \setminus A$.

This means that $x \in \partial A$.

Therefore, the intersection $\text{Cl}(A) \cap \text{Cl}(X \setminus A)$ is a subset of the boundary $\partial A$

$$\text{Cl}(A) \cap \text{Cl}(X \setminus A) \subseteq \partial A$$

3] Conclusion

The two parts of the proof confirm that:

• $\partial A \subseteq \text{Cl}(A) \cap \text{Cl}(X \setminus A)$
• $\partial A \supseteq \text{Cl}(A) \cap \text{Cl}(X \setminus A)$

Thus, the boundary of $A$ is equal to the intersection of the closures of $A$ and its complement.

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

This concludes the proof.

And so on.