Question

Prove that \(\bar{A}\) is contained in any closed superset of \(A\) and is the intersection of all such supersets.

Short answer

\(\bar{A}\) is contained in any closed superset of \(A\) and is the intersection of all such supersets.

Step-by-step solution

Understanding the Problem

We need to prove two statements about the closure of a set \(\bar{A}\). First, we need to show that \(\bar{A}\) is contained in any closed superset of \(A\). Second, we need to show that \(\bar{A}\) is the intersection of all such closed supersets.

Definitions

Recall that the closure of a set \(A\), denoted \(\bar{A}\), consists of all points in \(A\) together with all limit points of \(A\). A closed set is a set that contains all its limit points.

Prove \(\bar{A} \subseteq S\) for any closed superset \(S \supseteq A\)

Suppose \(S\) is a closed set such that \(A \subseteq S\). Since \(S\) contains all limit points of \(A\) and is a superset of \(A\) itself, it must contain \(\bar{A}\). Thus, \(\bar{A} \subseteq S\).

Prove \(\bar{A}\) is the intersection of all closed supersets of \(A\)

Let \( \mathcal{C} \) be the collection of all closed supersets of \(A\). We want to show that \( \bar{A} = \bigcap_{S \in \mathcal{C}} S\). We have already shown \(\bar{A} \subseteq S\) for every \( S \in \mathcal{C}\), so it follows that \( \bar{A} \subseteq \bigcap_{S \in \mathcal{C}} S\). Conversely, any point in \(A\) or any limit point of \(A\) must be in every closed superset of \(A\), hence in \( \bigcap_{S \in \mathcal{C}} S\). Therefore, \( \bar{A} = \bigcap_{S \in \mathcal{C}} S\).

Key concepts

Limit Points

In topology, a **limit point** or **accumulation point** of a set is a point that can be "approached" by other points of the set in a neighborhood. These points are crucial because they help define the closure of a set. If a point is a limit point of a set \(A\), then every neighborhood of the point contains at least one other point from \(A\). This concept ensures that the points close to the set \(A\) are also taken into considerations when talking about its closure.Limit points are essential when forming the closure of a set. The closure \(\bar{A}\) of a set \(A\) is simply the union of \(A\) and the set of its limit points. In mathematical terms, this is expressed as the set \(A\) plus all its limit points:\[ \bar{A} = A \cup \text{Limit points of } A \] which encompasses all the elements that define the boundary of \(A\). Including limit points ensures that \(\bar{A}\) is the smallest closed set containing \(A\). When discussing closed supersets, these limit points are automatically included to ensure completeness of the set.

Closed Set

A **closed set** in topology is particularly defined by the inclusion of all its limit points. This means, if a set \(S\) is closed, then it contains every point where the points of \(A\) accumulate. Think of it as an unbreakable boundary around the set; no points can exist just outside it which belong inside its closure.Closed sets have several important properties:

• They contain all their limit points, making them very complete and well-defined.
• The complement of a closed set is an open set, providing a duality that is fundamental in topology.
• They remain closed under intersections, meaning the intersection of any collection of closed sets is also a closed set.

Understanding closed sets helps in visualizing and grasping how the closure \(\bar{A}\) operates as it is essentially the smallest closed set which includes set \(A\). This ties directly into why \(\bar{A}\) must be contained in any closed superset \(S\) of \(A\), since \(S\) contains \(A\) and all its limit points by definition.

Intersection of Sets

The **intersection of sets** involves finding a new set containing only those elements common to all sets being considered. In terms of closure, the intersection of all closed supersets of a set \(A\) provides a powerful way to precisely define \(\bar{A}\).For example, if \(S_1, S_2, \ldots, S_n\) are closed sets each containing \(A\), then the intersection \( S_1 \cap S_2 \cap \ldots \cap S_n \) also contains \(A\) and is closed. This intersection is tighter or smaller than any single \(S_i\) since it gathers only the common elements, particularly those in \(\bar{A}\). By taking the intersection of all such closed supersets, we effectively isolate the closure \(\bar{A}\):\[ \bar{A} = \bigcap_{S \supseteq A, \text{ closed}} S \]This formula represents \(\bar{A}\) as the smallest possible set that still satisfies all conditions of being a closed set containing \(A\). This results in a precise closure that is made solely from the shared elements among these sets, thus conforming to the definition of \(\bar{A}\).