Question

Prove: If \(S\) is open and \(S=A \cup B\) where \(\bar{A} \cap B=A \cap \bar{B}=\emptyset,\) then \(A\) and \(B\) are open.

Short answer

Question: Prove that sets A and B are open when given that S is open, S=A ∪ B, and 𝐴¯ ∩ B=A ∩ 𝐵¯=∅. Answer: We need to show that for every point x in A or B, there exists an open neighborhood around x contained entirely in either A or B. Since S is open and S=A ∪ B, this implies that for each point x in S, there exists an open neighborhood around x in S. By showing that the open neighborhood can be restricted to either A or B using the conditions 𝐴¯ ∩ B=A ∩ 𝐵¯=∅, we can prove that A and B are open.

Step-by-step solution

Show that for every point \(x \in S,\) there exists an open neighborhood around \(x\) contained in \(S.\)

Let \(x \in S,\) as \(S = A \cup B,\) we have \(x \in A\) or \(x \in B.\) Since \(S\) is open, there exists an open neighborhood \(U \subset S\) around \(x.\)

Show that if \(x \in A,\) then the neighborhood around \(x\) is contained entirely in \(A.\)

Suppose \(x \in A.\) We know that \(U \subset S\) and \(S = A \cup B\) while \(\bar{A} \cap B = \emptyset.\) Thus, no point in \(U\) can belong to \(B.\) Consequently, \(U \cap B = \emptyset\) and \(U \subseteq A.\) That means for every \(x \in A,\) there exists an open neighborhood around \(x\) contained entirely in \(A.\)

Show a similar property for \(x \in B.\)

Suppose \(x \in B.\) We know that \(U \subset S\) and \(S = A \cup B\) while \(A \cap \bar{B} = \emptyset.\) Thus, no point in \(U\) can belong to \(A.\) Consequently, \(U \cap A = \emptyset\) and \(U \subseteq B.\) That means for every \(x \in B,\) there exists an open neighborhood around \(x\) contained entirely in \(B.\)

Conclude that \(A\) and \(B\) are open.

We have shown that for every point \(x \in A\) or \(x \in B,\) there exists an open neighborhood around \(x\) contained entirely in either \(A\) or \(B.\) As this condition is true for all points in \(A\) and \(B,\) we can conclude that both \(A\) and \(B\) are open.

Key concepts

Topology

Topology is a fascinating area of mathematics. It's all about understanding spaces and how they work. Think of it like geometry, but with a lot more flexibility. In geometry, we look at shapes and sizes closely. However, topology doesn't worry about sizes that much. Instead, it focuses on the way the parts of a space connect with each other. This connection is what makes topology so unique.

A key idea in topology is that it considers spaces on a broader scale. Shapes can stretch or bend but without tearing or gluing points together. This allows topologists to analyze properties that remain unchanged even under such transformations. This is why topology is often called 'rubber-sheet geometry'. Examples of topics in topology include the study of open sets, closed sets, compactness, and continuity.

One area where topology becomes very important is in understanding how different sets within a space, like neighborhoods, behave. This is linked closely to how we define open and closed sets in topology.

Union of Sets

The union of sets is a simple but essential concept in set theory. It applies not only in set theory, but also comes handy in topology. When we talk about the union of two sets, say set \(A\) and set \(B\), we are combining all elements that are in either \(A\) or \(B\).

In formula terms, the union is expressed as \(A \cup B\). This union set includes all elements regardless if they belong to one or both of the original sets. You can think of it as putting all members of \(A\) and \(B\) into one large set. This concept is crucial in topology since many topological properties like openness are preserved under unions.

In our problem, set \(S\) was defined as \(A \cup B\), indicating that the elements of \(S\) are either in \(A\) or in \(B\). Understanding the union helps us to see why \(S\) is considered open, even as we prove \(A\) and \(B\) themselves are open.

Neighborhood in Topology

A neighborhood in topology refers to an environment or a surrounding area around a point within a space. More formally, a neighborhood of a point \(x\) is any open set that contains \(x\). It’s like having your point \(x\) in the middle of a little cookie-cutter. Even as you shift or expand the shape around \(x\), it stays securely inside it.

This concept becomes pivotal when determining the openness of a set. For a set to be open in topology, each point in the set must have a neighborhood contained entirely within the set. In simple words, you should be able to wiggle a bit around any point without stepping outside the boundaries of the set.

In our step-by-step solution, neighborhoods were used to show openness for sets \(A\) and \(B\). By proving that for every point in \(A\) and \(B\), there are neighborhoods entirely within these sets, we could establish \(A\) and \(B\)'s openness. Understanding neighborhoods and how they fit within sets becomes crucial in topology's exploration of various spaces.