Intersection of Boundary and Set in Topology

The intersection between the boundary $ \partial A $ of a set and the set itself $ A $ is empty if and only if the set is open: $$ \partial A \cap A = \emptyset \Leftrightarrow A \text{ is open} $$

In other words, a set $ A $ is open if and only if none of its points lie on its boundary.

This means that the set $ A $ and its boundary are disjoint when $ A $ is open.

A Practical Example
The Proof

A Practical Example

Consider the open interval $(0, 1)$ in a simple topological space such as the real line $\mathbb{R}$, with the usual topology.

$$ A = (0, 1) $$

The set $ A $ is open in the usual topology on $\mathbb{R}$.

The boundary of $ A $ is the intersection between the closure of $ A $ and the closure of the complement $ \mathbb{R}-A $:

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

The closure of $ A $ is the closed interval $[0, 1]$.

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

The closure of the complement of $ A $ is

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

Thus, the boundary of $ A $ is:

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

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

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

Now that we have the boundary of $ A $, we can find the intersection $ \partial A \cap A $.

In this case, the intersection of the boundary $ \partial A $ with the set $ A $ is empty because $ A $ is open:

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

This example demonstrates that the open interval $(0, 1)$ is an open set and its boundary has no points in common with the set.

Example 2

Consider the closed interval $ B = [0, 1]$ in the usual topology on $\mathbb{R}$.

$$ B = [0, 1] $$

The set $ B $ is a closed set.

The boundary of $ B $ consists of points that belong to both the closure of $ B $ and the closure of the complement of $ B $:

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

The closure of $ B $ is $[0, 1]$.

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

The closure of the complement of $ B $, $\overline{\mathbb{R} \setminus B}$, is

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

Thus, the boundary of $ B $ is:

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

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

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

In this case, the intersection of the boundary $ \partial B $ with the set $ B $ is not empty because $ B $ is a closed set.

$$ \partial B \cap B = \{0, 1\} \cap [0, 1] = \{0, 1\} $$

This example confirms that the closed interval $[0, 1]$ is not open and its boundary shares points with the set.

The Proof

The proof of this property requires demonstrating the double implication.

(⇒) If $ \partial A \cap A = \emptyset $, then $ A $ is open

Assume the hypothesis that the intersection between the boundary and the set itself is empty:

$$ \partial A \cap A = \emptyset $$

We need to show that $ A $ is open.

If $ \partial A \cap A = \emptyset $, then no point of $ A $ lies on the boundary of $ A $.

This implies that every point in $ A $ has a neighborhood entirely contained within $ A $.

By definition, a set is open if every point in it has a neighborhood entirely contained within the set.

Therefore, the set $ A $ is open.

(⇐) If $ A $ is open, then $ \partial A \cap A = \emptyset $

Assume the hypothesis that $ A $ is open.

Now we need to show that the intersection of the boundary and the set itself is empty:

$$ \partial A \cap A = \emptyset $$

If $ A $ is open, then every point in $ A $ has a neighborhood contained within $ A $.

Thus, no point of $ A $ can be on the boundary of $ A $, since each point in $ A $ has a neighborhood that does not intersect $ X - A $.

Therefore, $ \partial A \cap A = \emptyset $.

Conclusion

This proves that $ \partial A \cap A = \emptyset $ if and only if $ A $ is open.

And so on.