Understanding Set Operations: B - A

This problem asks about the nature of the resulting set when we perform the Set Difference (B - A) where A is an Open Set and B is a Closed Set. This involves understanding fundamental concepts in Set Theory and Topology.

Key Concepts in Topology

In mathematics, sets can be classified as open or closed depending on their topological properties. An Open Set contains a neighborhood around each of its points. A Closed Set is one that contains all its limit points, or equivalently, whose complement is open.

Performing Set Operations: The Difference B - A

The Set Difference B - A includes all elements belonging to set B but not to set A. This operation can be related to the complement of a set. If X is the universal set containing A and B, the complement of A relative to X is \(A^c = X - A\). Then, \(B - A\) is equivalent to the intersection of B and \(A^c\):

\(B - A = B \cap A^c\)

Applying Set Operations Properties

Consider the properties of sets under Set Operations in a topological space:

• Property 1: The complement of an Open Set is always a Closed Set. Since A is an Open Set, its complement \(A^c\) is a Closed Set.
• Property 2: The intersection of any collection of Closed Sets (finite or infinite) is always a Closed Set.

Conclusion: Nature of B - A

We have established that \(B - A = B \cap A^c\). We are given that B is a Closed Set, and we have shown that \(A^c\) is a Closed Set (because A is open by definition in Topology). Therefore, \(B - A = B \cap A^c\) is the intersection of two Closed Sets (B and \(A^c\)). By the property mentioned above regarding Set Operations, the intersection of two closed sets is a closed set.

Thus, B - A is a Closed Set.

Illustrative Example of Set Operations

Let's consider an example on the real number line (\(\mathbb{R}\)) with its usual topology:

Let A = \((0, 1)\) (an open interval, which is an open set).

Let B = \([0, 2]\) (a closed interval, which is a closed set).

The set difference is \(B - A = [0, 2] - (0, 1) = [0, 0] \cup [1, 2] = \{0\} \cup [1, 2]\). The resulting set \(\{0\} \cup [1, 2]\) is a Closed Set in \(\mathbb{R}\). This example supports our general conclusion about these Set Operations.

In summary, when you perform the Set Operations of subtracting an open set A from a closed set B, the result B - A is always a Closed Set.