Question

Prove: (a) A boundary point of a set \(S\) is either a limit point or an isolated point of \(S\). (b) \(\quad\) A set \(S\) is closed if and only if \(S=\bar{S}\).

Short answer

### Short Answer A boundary point of a set is either a limit point (every neighborhood contains at least one point in the set distinct from it) or an isolated point (exists a neighborhood that intersects the set only in it). A set is considered closed if it equals its closure, meaning it contains all its limit points.

Step-by-step solution

Part (a): Prove a boundary point is either a limit point or an isolated point

Let \(x\) be a boundary point of set \(S\). Since \(x\) is a boundary point, every neighborhood of \(x\) intersects both \(S\) and \(S^c\). We will consider two cases: Case 1: \(x\) is a limit point of \(S\). Since \(x\) is a limit point, every neighborhood of \(x\) contains at least one point in \(S\) distinct from \(x\). This case is consistent with the definition of a boundary point. Case 2: \(x\) is an isolated point of \(S\). Since \(x\) is an isolated point, there exists a neighborhood of \(x\) that intersects \(S\) only in \(x\). This case is also consistent with the definition of a boundary point, since the neighborhood still intersects both \(S\) and \(S^c\). Thus, a boundary point of a set \(S\) is either a limit point or an isolated point of \(S\).

Part (b): Prove a set is closed iff it equals its closure

Let \(\bar{S}\) be the closure of a set \(S\). By definition, \(\bar{S}\) is the smallest closed set containing \(S\). We must prove two statements: 1. If \(S\) is closed, then \(S = \bar{S}\). 2. If \(S = \bar{S}\), then \(S\) is closed. Proof of statement 1: Since \(S\) is closed and \(\bar{S}\) is the smallest closed set containing \(S\), it follows that \(S \subseteq \bar{S}\). Since \(\bar{S} \subseteq S\) by definition, we have \(S = \bar{S}\). Proof of statement 2: If \(S = \bar{S}\), then \(S\) contains all its limit points. By definition, \(S\) is closed. Thus, a set \(S\) is closed if and only if \(S = \bar{S}\).

Key concepts

Boundary Point

In real analysis, the concept of a boundary point helps us understand the edges of a set. A point \( x \) is a boundary point of a set \( S \) if every neighborhood around \( x \) intersects both \( S \) and the complement of \( S \), denoted as \( S^c \). This means, no matter how close we zoom in to the point \( x \), it touches both \( S \) and everything outside \( S \).

There are two possibilities for such points:

• Limit Point: If \( x \) is a limit point, then in every neighborhood around \( x \), there are infinitely many points from \( S \), distinct from \( x \) itself. This keeps \( x \) grounded within its boundary role.
• Isolated Point: On the other hand, if \( x \) is an isolated point, there is a neighborhood around \( x \) that contains \( x \) as the only point from \( S \).
  
  This might sound contradictory to being a boundary point. However, such a neighborhood still snips both into \( S \) at \( x \) and into \( S^c \) just around it.

Limit Point

Limit points are crucial elements in the analysis of space and sets. A point \( x \) is called a limit point of the set \( S \) if every neighborhood around \( x \) contains at least one point from \( S \) other than \( x \) itself. This property essentially anchors \( x \) close to the set \( S \) by proximity.

Limit points show us how a set behaves as we go infinitely closer, even possibly to the border of the set. For instance:

• Consider \( \mathbb{R} \), the set of real numbers. If \( S \) is the interval \( (0,1) \), then both 0 and 1 are limit points of \( S \) though they are not in the actual interval \( S \).
• This reflects that limit points facilitate understanding how sets "approach" and "interact" with their surroundings.

Closed Set

A set \( S \) is defined to be closed if it contains all its limit points. This means nothing gets left "floating" outside \( S \) if \( S \) could potentially grab it as per its boundary or limit connection.

Closed sets are important in analysis because they include their boundary, allowing for the formulation of well-contained and clearly defined mathematical spaces. This can be further comprehended by considering:

• The entire space is closed because it naturally contains all its limit aspects within itself.
• If you take the set of integers \( \mathbb{Z} \) within \( \mathbb{R} \), \( \mathbb{Z} \) is closed because there are no 'new' limit points in the real number space adjoining \( \mathbb{Z} \).

Closure of a Set

The closure of a set \( S \), denoted \( \bar{S} \), is an extension of \( S \) that includes all the limit points of \( S \). It provides the smallest closed set enclosing \( S \). The closure helps bridge any gap between elements originally in the set and those infinitely approached or encircled by it.

The closure precisely captures this intuition:

• Think of \( S \) as a dried leaf and \( \bar{S} \) as a thin rim of water around it. The water forms finely, slipping into every space \( S \) can reach, aggregating all boundary-line interactions into a unified form.
• If \( S = \bar{S} \), it means that \( S \) is closed. All its surrounding points fit back snugly within \( S \).