Empty Boundary and Clopen Sets

The boundary $\partial A$ of a set $A$ is empty if and only if $A$ is both open and closed (clopen). $$ \partial A = \emptyset \Leftrightarrow A \text{ is clopen} $$

This means that $A$ has no boundary points, which are points that belong to both the closure of $A$ and the closure of its complement.

A Practical Example
The Proof

A Practical Example

Example 1

Consider the set $ A = \emptyset $ in the topological space $\mathbb{R}$ with the standard topology.

We check if the boundary $\partial A = \emptyset$ is empty.

The closure of $A = \emptyset$ is:

$$ \text{Cl}(A) = \emptyset $$

The complement of $A$ is $A^c = \mathbb{R}$.

The closure of the complement is $ \text{Cl}(A^c) = \mathbb{R}$, since $\mathbb{R}$ is already closed.

$$ \text{Cl}(A^c) = \mathbb{R} $$

Thus, the boundary of $A$ is:

$$ \partial A = \text{Cl}(A) \cap \text{Cl}(A^c) = \emptyset \cap \mathbb{R} = \emptyset $$

In this case, the boundary is empty ($\partial A = \emptyset$), so $A$ is clopen.

Indeed, $A = \emptyset$ is open by definition in the standard topology and is also closed because it contains all its accumulation points, being empty.

Example 2

Consider the set $ A = \mathbb{R} $ in the topological space $\mathbb{R}$ with the standard topology.

We check if the boundary $\partial A = \emptyset$ is empty.

The closure of $A = \mathbb{R}$ is:

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

The complement of $A$ is $A^c = \emptyset$.

The closure of the complement is:

$$ \text{Cl}(A^c) = \emptyset $$

Thus, the boundary of $A$ is:

$$ \partial A = \text{Cl}(A) \cap \text{Cl}(A^c) = \mathbb{R} \cap \emptyset = \emptyset $$

In this case, the boundary is empty ($\partial A = \emptyset$), so $A$ is clopen.

Indeed, $A = \mathbb{R}$ is open by definition in the standard topology and is closed because it contains all its accumulation points.

Example 3

Consider the set $A = [0,1)$ in the topological space $\mathbb{R}$ with the standard topology.

We check if the boundary $\partial A = \emptyset$ is empty.

The closure of $A = [0,1)$ is $\text{Cl}(A) = [0,1]$.

The complement of $A$ is $A^c = (-\infty, 0) \cup [1, \infty)$.

The closure of the complement is $\text{Cl}(A^c) = (-\infty, 0] \cup [1, \infty)$.

Thus, the boundary of $A$ is:

$$ \partial A = \text{Cl}(A) \cap \text{Cl}(A^c) = [0,1] \cap \left( (-\infty, 0] \cup [1, \infty) \right) = \{0, 1\} $$

In this case, the boundary is not empty ($\partial A = \{0, 1\} \neq \emptyset$), so $A$ is not clopen.

Indeed, $A = [0,1)$ is open but not closed.

These examples clearly illustrate the theorem: a set $A$ in a topological space has an empty boundary if and only if it is both open and closed (clopen).

The Proof

First, recall that by definition, the boundary $ \partial A $ of a set $A$ is the intersection of the closure of $A$ and the closure of its complement $A^c$

$$ \partial A = \text{Cl}(A) \cap \text{Cl}(A^c) $$

To prove that the boundary $\partial A$ is empty if and only if $A$ is both open and closed (clopen), we proceed with the proof in both directions.

1] If the Boundary is Empty, then the Set A is Both Open and Closed (Clopen)

If $\partial A = \emptyset$, then the intersection of the closure of $ A $ and the closure of its complement $ A^c $ is empty:

$$ \text{Cl}(A) \cap \text{Cl}(A^c) = \emptyset $$

This implies that there are no points that belong both to the closure of $A$ and the closure of the complement of $A$.

Is the set A closed?

A set $A$ is closed if it contains all its accumulation points, that is, if $ A = \text{Cl}(A) $.

Now, since $ \text{Cl}(A) \cap \text{Cl}(A^c) = \emptyset $, every point of $\text{Cl}(A) $ cannot be in $ \text{Cl}(A^c) $.

This implies that every point of $\text{Cl}(A) $ is in $(A^c)^c$:

$$ \text{Cl}(A) \subseteq (A^c)^c $$

Knowing that $(A^c)^c = A$:

$$ \text{Cl}(A) \subseteq A $$

Since $\text{Cl}(A) \subseteq A$ and by definition $ \text{Cl}(A) $ contains $A$, we deduce that $\text{Cl}(A) = A$.

This shows that $A$ is closed.

Is the set A open?

Knowing that the intersection of the closure of $ A $ and the closure of the complement of $ A $ is empty:

$$ \text{Cl}(A) \cap \text{Cl}(A^c) = \emptyset $$

From this, we deduce that the closure of $ A $ is in the complement of $ A $:

$$ \text{Cl}(A^c) \subseteq (A)^c = A^c $$

This implies that $A^c$ is closed because it contains its closure.

If $A^c$ is a closed set, then its complement $(A^c)^c = A $ is open.

Therefore, the set $ A $ is open.

The Set A is Both Open and Closed

In conclusion, the set $ A $ is both closed and open, which means it is a clopen set.

2] If the Set A is Both Open and Closed (Clopen), then the Boundary of A is Empty

In this case, we consider the set $A$ to be both open and closed as the initial hypothesis.

Since $A$ is closed, it contains all its accumulation points:

$$ A = \text{Cl}(A) $$

Since $A$ is open, every point of $A$ is an interior point of $A$.

$$ A = \text{Int}(A) $$

Knowing that $A$ is open, we deduce that its complement $A^c$ is closed, and therefore it contains all its accumulation points:

$$ A^c = \text{Cl}(A^c) $$

By definition, the boundary of a set $A$ is the intersection of the closure of $ A $ and the closure of its complement $ A^c $:

$$ \partial A = \text{Cl}(A) \cap \text{Cl}(A^c) $$

Substituting $ A = \text{Cl}(A) $ and $ A^c = \text{Cl}(A^c) $:

$$ \partial A = A \cap A^c $$

The intersection $A \cap A^c$ between a set and its complement is always empty:

$$ \partial A = \emptyset $$

Therefore, if $A$ is clopen, its boundary $\partial A$ is empty.

3] Conclusion

Thus, we have proved that $$ \partial A = \emptyset \Leftrightarrow A \text{ is clopen} $$

And so on.