Closure of the Complement and Complement of the Interior of a Set
The closure of the complement of a set A is equal to the complement of the interior of set A. $$ \text{Cl}(X - A) = X - \text{Int}(A) $$
This relation highlights an interesting duality between the concepts of closure and interior in the context of set complements in a topological space.
A Practical Example
Consider the topological space \( X = \mathbb{R} \) with the standard topology, where open sets are open intervals and their unions.
Let's take an example set \( A \subseteq X = \mathbb{R} \) consisting of the closed interval \( A = [1, 2] \).
To verify this property, we'll divide the exercise into two parts: calculating the closure of the complement of A and the complement of the interior of A.
1] The Closure of the Complement of A
The complement of A in \( \mathbb{R} \) is X - A
$$ X - A = \mathbb{R} - [1, 2] = (-\infty, 1) \cup (2, \infty) $$
To find the closure of the complement of set A, we need to add the accumulation points of \( (-\infty, 1) \cup (2, \infty) \).
In this case, the complement of \( A \) is a union of open sets, and the points 1 and 2 are accumulation points because every neighborhood of 1 contains points from \( (-\infty, 1) \) and every neighborhood of 2 contains points from \( (2, \infty) \).
Therefore, the closure of the complement of A is:
$$ \text{Cl}(X - A) = \text{Cl}((-\infty, 1) \cup (2, \infty)) = (-\infty, 1] \cup [2, \infty) $$
2] The Complement of the Interior of A
The interior of A = [1, 2] is the open interval contained within A. So:
$$ \text{Int}(A) = (1, 2) $$
The complement of the interior of A is X - Int(A)
$$ X - \text{Int}(A) = \mathbb{R} - (1, 2) = (-\infty, 1] \cup [2, \infty) $$
3] Conclusion
The closure of the complement of A and the complement of the interior of A refer to the same interval:
$$ \text{Cl}(X - A) = (-\infty, 1] \cup [2, \infty) $$
$$ X - \text{Int}(A) = (-\infty, 1] \cup [2, \infty) $$
Thus, the equality \(\text{Cl}(X - A) = X - \text{Int}(A)\) holds true.
The Proof
Consider a set \( A \subseteq X \) in the topological space X.
The closure of the complement of \( A \) is the set of all points in the complement of A plus all its accumulation points.
$$ \text{Cl}(X - A) $$
The complement of the interior of A is the set of all points that are not in the interior of A.
$$ X - \text{Int}(A) $$
To prove that \(\text{Cl}(X - A) = X - \text{Int}(A)\), we consider two inclusions:
- \( \text{Cl}(X - A) \subseteq X - \text{Int}(A) \)
If a point x belongs to Cl(X-A), then every neighborhood of x contains at least one point from X-A. This implies that x cannot be an interior point of A, because otherwise there would be a neighborhood of x completely contained in A. Therefore, the point x does not belong to the interior Int(A), which means x is in the complement of the interior, or X-Int(A). - \( X - \text{Int}(A) \subseteq \text{Cl}(X - A) \)
If a point x belongs to X-Int(A), then x is not an interior point of A. This means that every neighborhood of x contains at least one point that is not in A, or a point that belongs to X-A. Therefore, the point x belongs to the closure of X-A.
Once both inclusions are proven, we can conclude that:
$$ \text{Cl}(X - A) = X - \text{Int}(A) $$
This relation demonstrates a duality between the concepts of closure and interior in the context of set complements.
And so on.
