Boundary of a Set is Always Closed

The boundary of a set is always closed because it is the intersection of the closure of $A$ and the closure of the complement of $A$: $$ \partial A = Cl(A) \cap Cl(X - A) $$

The boundary of a set $A$ in a topological space $X$, denoted by $\partial A$, is defined as the intersection of the closure of $A$ and the closure of the complement of $A$: $\partial A = Cl(A) \cap Cl(X - A)$.

Since the intersection of closed sets is always closed, it follows that $\partial A$ is always closed.

A Practical Example
The Proof

A Practical Example

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

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

The closure of $A$, denoted $Cl(A)$, is the set $[0, 1]$, which includes all the points of $A$ plus its limit points (in this case, 0 and 1).

The complement of the open set $A$ in $\mathbb{R}$ is the closed set:

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

The closure of the complement of $A$ is

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

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

The boundary of $A$, denoted $\partial A$, is the intersection of $Cl(A)$ and $Cl(\mathbb{R} - A)$:

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

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

Therefore, in this example, the boundary of $A$ is $\{0, 1\}$, which is a closed set in $\mathbb{R}$.

The Proof

The proof relies on some fundamental properties of closed sets and topology.

In any topological space $X$, the closure of a set $A$, denoted by $\overline{A}$ or $Cl(A)$, is a closed set.

By definition, $Cl(A)$ is the smallest closed set that contains $A$.

The complement of a set $A$ in $X$ is the set $X - A$. If $A$ is closed, then $X - A$ is open, and vice versa.

The boundary of a set $A$, denoted by $\partial A$, is the intersection of the closure of $A$ and the closure of its complement.

$$ \partial A = Cl(A) \cap Cl(X - A) $$

In a topological space, the intersection of closed sets is still a closed set. This is a fundamental property of closed sets.

Using these properties, we can prove that $\partial A$ is always closed:

$Cl(A)$ is closed by definition.
$Cl(X - A)$ is closed by definition, given that $X - A$ is an open set and the closure of an open set is a closed set.
The intersection of two closed sets, $Cl(A) \cap Cl(X - A)$, is closed.

Therefore, $\partial A = Cl(A) \cap Cl(X - A)$ is always a closed set in any topological space.

And so forth.