Question

Let \(S\) be an arbitrary set. Prove: (a) \(\partial S\) is closed. (b) \(S^{0}\) is open. (c) The exterior of \(S\) is open. (d) The limit points of \(S\) form a closed set. (e) \(\overline{(\bar{S})}=\bar{S}\).

Short answer

Question: Prove that the boundary of a set S is closed, the interior of S is open, the exterior of S is open, the limit points of S form a closed set, and the closure of the closure of S is equal to the closure of S. Answer: To prove these properties, we first need to define essential terms such as boundary, interior, exterior, limit points, and closures. By showing that the boundary of a set S is the intersection of the closures of S and its complement, we can prove it is closed. The interior of S is open as it contains all points with open neighborhoods entirely contained in S. Similarly, the exterior of S is open because it consists of all points with open neighborhoods contained in the complement of S. The set of limit points of S is closed since its complement is open. Lastly, by showing the closure of the closure of S is a subset of the closure of S and vice versa, we can prove they are equal.

Step-by-step solution

The boundary of a set \(S\), denoted as \(\partial S\), is the set of points where every neighborhood of the point has non-empty intersection with both \(S\) and its complement \(S^c\). In other words, \(\partial S = \overline{S} \cap \overline{S^c}\), where \(\overline{S}\) is the closure of \(S\). #Step 2: Prove boundary is closed#

We need to show that the complement of \(\partial S\) is open. Let \(x \notin \partial S\). Then \(x \notin (\overline{S} \cap \overline{S^c})\). By De Morgan's Law, \(x \in (\overline{S^c})^c \cup (\overline{S})^c\). Since the union of open sets is open, it follows that \((\overline{S^c})^c \cup (\overline{S})^c\) (which is the complement of \(\partial S\)) is open. Therefore, \(\partial S\) is closed. #b) Proving the interior of S is open# #Step 1: Define the interior of S#

The interior of a set \(S\), denoted as \(S^0\), is the set of points that have an open neighborhood entirely contained in \(S\). #Step 2: Prove the interior is open#

Let \(x \in S^0\). By definition, there exists an open neighborhood \(N\) of \(x\) such that \(N \subseteq S\). Since \(N\) is an open neighborhood, \(x\) is an interior point of \(N\). Since \(N \subseteq S\), it follows that \(S^0\) is open. #c) Proving the exterior of S is open# #Step 1: Define the exterior of S#

The exterior of a set \(S\), denoted as \(S^e\), is defined as the set of points that have an open neighborhood entirely contained in \(S^c\) (the complement of \(S\)). #Step 2: Prove the exterior is open#

Using a similar argument as in part (b), let \(x \in S^e\). There exists an open neighborhood \(N\) of \(x\) such that \(N \subseteq S^c\). Since \(N\) is an open neighborhood, \(x\) is an interior point of \(N\). Therefore, \(S^e\) is open as it is a union of open neighborhoods. #d) Proving the limit points of S form a closed set# #Step 1: Define limit points#

A point \(x\) is a limit point of a set \(S\) if every neighborhood of \(x\) contains points of \(S\) distinct from \(x\) itself. #Step 2: Prove limit points form a closed set#

Let \(L\) be the set of limit points of \(S\). We need to show that the complement of \(L\), denoted as \(L^c\), is open. If \(x \in L^c\), then \(x\) is not a limit point of \(S\). This means that there exists an open neighborhood \(N\) of \(x\) such that \(N \cap S = \{x\}\) or \(N \cap S = \emptyset\). In both cases, \(N \subseteq L^c\). Hence, \(L^c\) is open, and it follows that \(L\) is closed. #e) Proving the closure of the closure of S is equal to the closure of S# #Step 1: Define closure of S#

The closure of a set \(S\), denoted as \(\bar{S}\), is the union of \(S\) and its limit points. It can also be expressed as the intersection of all closed sets containing \(S\). #Step 2: Prove the closure of the closure of S is equal to the closure of S#

We will show that \(\overline{(\bar{S})} \subseteq \bar{S}\) and \(\bar{S} \subseteq \overline{(\bar{S})}\). (i) \(\overline{(\bar{S})} \subseteq \bar{S}\): Let \(x \in \overline{(\bar{S})}\). Then, \(x\) is either in \(\bar{S}\) or is a limit point of \(\bar{S}\). If \(x \in \bar{S}\), then obviously \(x \in \bar{S}\). If \(x\) is a limit point of \(\bar{S}\), then every neighborhood of \(x\) contains a point in \(\bar{S}\). Since \(\bar{S}\) contains all limit points of \(S\), \(x\) is either in \(S\) or is a limit point of \(S\). In either case, \(x \in \bar{S}\). Therefore, \(\overline{(\bar{S})} \subseteq \bar{S}\). (ii) \(\bar{S} \subseteq \overline{(\bar{S})}\): Let \(x \in \bar{S}\). Then, \(x\) is either in \(S\) or is a limit point of \(S\). If \(x \in S\), then \(x \in \bar{S}\) and hence \(x \in \overline{(\bar{S})}\). If \(x\) is a limit point of \(S\), then every neighborhood of \(x\) contains a point in \(S\). Since \(S \subseteq \bar{S}\), every neighborhood of \(x\) contains a point in \(\bar{S}\). Therefore, \(x\) is also a limit point of \(\bar{S}\), which means \(x \in \overline{(\bar{S})}\). Thus, \(\bar{S} \subseteq \overline{(\bar{S})}\). Combining (i) and (ii), we conclude that \(\overline{(\bar{S})}=\bar{S}\).

Key concepts

Boundary of a Set

Understanding the boundary of a set is crucial in topology. The boundary of a set, denoted by \(\partial S\), encompasses the points residing at the 'edge' of a set. More formally, a point is on the boundary if every neighborhood of that point intersects both the set \(S\) and its complement \(S^c\). So, \(\partial S = \overline{S} \cap \overline{S^c}\), with \(\overline{S}\) representing the closure of \(S\). This expression highlights that the boundary elements are more than just nearby the set; they are perfectly poised between the set and everything outside it. - A set’s boundary is crucial: it defines the points closest to not being either fully in or out of the set.- The boundary \(\partial S\) is always closed since its complement, the union of two open sets, is open. Boundary identification aids in understanding a set’s overall closure and how it behaves under topological transformations.

Interior of a Set

The interior of a set \(S\), represented as \(S^0\), consists of all points that are 'comfortably' inside \(S\), meaning those that can be surrounded by a neighborhood completely contained within \(S\). This concept can help understand the 'heart' of a set, removing any ambiguity caused by boundary points.- Every point in the interior has a fully contained neighborhood, affirming it is part of \(S\) without 'touching' its edges.- This makes the interior inherently open — any point selected from within it will always have some 'wiggle room' inside \(S\). For example, consider a circle excluding its boundary as its interior. Only paths entirely within the circle show that these points belong to its interior. Appreciating the interior is essential for insights into which points firmly lie within a set.

Limit Points

A limit point of a set \(S\) is a point where every neighborhood contains at least one point from \(S\) other than the limit point itself. Thus, even though the point might not belong to the set explicitly, it is 'theoretically' enveloped by the set. This is significant when analyzing the set's closure.- Limit points are either inside the set or infinitesimally close, making them critical for exploring sequence or set behavior.- Collectively, these points create a closed set, given that the complement lacks 'approaching' points, ensuring any neighborhood is open. Recognizing limit points requires understanding neighborhood behavior, as these points signify how widely a set’s influence spreads.

Open and Closed Sets

In topology, 'open' and 'closed' sets are not opposites but complementary concepts helping understand how sets and spaces behave under various operations.

Open Sets

Open sets are defined by the property that every point within them has a neighborhood lying completely inside. This characteristic is crucial for continuity and defines spaces where boundary limits are not part of the set.

• Open balls in metric spaces serve as perfect examples.
• They allow seamless transition entirely within the space, ensuring no edge boundaries.

Closed Sets

Closed sets, clarified by containing all their limit points, encapsulate sets where boundary inclusion is ensured.

• Such sets maintain their completeness, denoting that any sequence within the set has a limit within the set.

Both open and closed sets are pivotal for defining topological properties, as they lay the foundation for continuity, boundary definitions, and comprehending space behaviors in mathematical analysis.