The Complementarity Property between the Interior and Closure of a Set
In topology, the complementarity property between the interior and closure of a set states that the interior of the complement of a set \( A \) is equal to the complement of the closure of \( A \). $$ \text{Int}(X - A) = X - \text{Cl}(A) $$
A Practical Example
Let's consider a simple topological space: the real line \(\mathbb{R}\) with the standard topology where the open sets are open intervals.
The interval \( A = [0, 1] \) is a closed interval.
$$ A = [0, 1] $$
The complement of \( A \) in the real line is:
$$ \mathbb{R} - A = (-\infty, 0) \cup (1, \infty) $$
The interior of \( \mathbb{R} - A \), denoted Int(\(\mathbb{R} - A\)), is the set of all interior points of \( \mathbb{R} - A \).
Since \((- \infty, 0) \cup (1, \infty)\) are already open, the interior of the complement \( \mathbb{R} - A \) is:
$$ \text{Int}(\mathbb{R} - A) = (-\infty, 0) \cup (1, \infty) $$
The closure of \( A \), denoted Cl(\(A\)), is the set \( A \) along with its accumulation points.
Since \( A \) is already a closed interval, the closure of \( A \) is:
$$ \text{Cl}(A) = [0, 1] $$
The complement of the closure of \( A \) in the real line is:
$$ \mathbb{R} - \text{Cl}(A) = \mathbb{R} - [0, 1] = (-\infty, 0) \cup (1, \infty) $$
Now, let's compare the results:
- \(\text{Int}(\mathbb{R} - A) = (-\infty, 0) \cup (1, \infty)\)
- \(\mathbb{R} - \text{Cl}(A) = (-\infty, 0) \cup (1, \infty)\)
The results are the same. From this, we deduce that the interior of the complement of a set \( A \) is equal to the complement of the closure of \( A \).
$$ \text{Int}(\mathbb{R} - A) = \mathbb{R} - \text{Cl}(A) $$
This confirms the complementarity property between the interior and closure of a set.
The example above clearly shows how the complementarity property between the interior and closure applies in a concrete case.
Proof
Let's consider a set \( A \) in a topological space \( X \).
We need to prove that \( \text{Int}(X - A) = X - \text{Cl}(A) \).
To prove that the interior of the complement of a set is the complement of the closure of that set, we use the fundamental properties of the interior and closure of a set in a topological space.
Starting with the definitions of interior and closure:
- \(\text{Int}(B)\): the interior of a set \( B \) is the set of all interior points of \( B \).
- \(\text{Cl}(A)\): the closure of a set \( A \) is the set \( A \) together with its accumulation (limit) points.
The proof proceeds in two parts:
1] Prove that \(\text{Int}(X - A) \subseteq X - \text{Cl}(A)\):
Let's take an element \( x \) from the interior of \( X - A \).
$$ x \in \text{Int}(X - A) $$
For every point \( x \) in \( X - A \), there exists a neighborhood \( U \) of \( x \) such that \( U \subseteq X - A \).
This means that \( U \cap A = \emptyset \).
Since \( U \cap A = \emptyset \), \( x \) cannot be an accumulation point of \( A \).
If \( x \) were an accumulation point of \( A \), every neighborhood of \( x \) would intersect \( A \), which contradicts the fact that \( U \cap A = \emptyset \).
Therefore, \( x \notin \text{Cl}(A) \), implying that \( x \in X - \text{Cl}(A) \).
Thus, \(\text{Int}(X - A) \subseteq X - \text{Cl}(A)\).
2] Prove that \(X - \text{Cl}(A) \subseteq \text{Int}(X - A)\)
Let's take any element \( x \) from \( X - \text{Cl}(A) \).
$$ x \in X - \text{Cl}(A) $$
Consequently, \( x \) is not in \( \text{Cl}(A) \).
$$ x \notin \text{Cl}(A) $$
This means that there exists a neighborhood \( U \) of \( x \) such that \( U \cap A = \emptyset \).
This implies that \( U \subseteq X - A \), meaning that \( x \in \text{Int}(X - A) \).
Therefore, \( X - \text{Cl}(A) \subseteq \text{Int}(X - A) \).
3] Conclusion:
Since we have proven both inclusions:
$$ \text{Int}(X - A) \subseteq X - \text{Cl}(A) $$
$$ X - \text{Cl}(A) \subseteq \text{Int}(X - A) $$
We conclude that:
$$ \text{Int}(X - A) = X - \text{Cl}(A) $$
This concludes the proof.
