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 the following properties of sets: (a) \(\partial S\) is closed (b) \(S^0\) is open (c) The exterior of \(S\) is open (d) Limit points of \(S\) form a closed set (e) \(\overline{(\bar{S})}=\bar{S}\) Answer: (a) To prove that the boundary of a set, \(\partial S\), is closed, we showed that its complement, \(\partial S^c\), is open: \(\partial S^c = (\operatorname{int} S) \cup (\operatorname{int} S^c)\). Since the union of open sets is open, we concluded that \(\partial S\) is closed. (b) To show that the interior of a set, \(S^0\), is open, we considered an arbitrary point in \(S^0\) and showed that there exists an open neighborhood of that point contained entirely in \(S^0\). This holds for all points in \(S^0\), confirming that it is open. (c) To prove that the exterior of a set, \(S\), is open, we demonstrated that there exists an open neighborhood of any point in the exterior that is disjoint from \(S\) and contained entirely in the exterior. This holds for all points in the exterior, establishing that it is open. (d) To show that the limit points of a set, \(S\), form a closed set, we proved that the complement of the set of limit points, \((L)^c\), is open. For any point not being a limit point of \(S\), there exists an open neighborhood disjoint from \(S\), which implies that \(U \subseteq (L)^c\). Consequently, \((L)^c\) is open and \(L\) is closed. (e) Finally, to prove that \(\overline{(\bar{S})}=\bar{S}\), we demonstrated that the closure of the closure of a set, \(S\), is equal to the closure of \(S\). By showing that the closure of \(S\), \(\bar{S}\), is the smallest closed set containing \(S\), and since it contains all its boundary points, we concluded that the property holds, \(\overline{(\bar{S})} = \bar{S}\).

Step-by-step solution

(a) Proving \(\partial S\) is closed

To prove that the boundary of \(S, \partial S\), is closed, we need to show that its complement, \(\partial S^c\), is open. By definition, we have \(\partial S = \bar{S} \cap \bar{S^c}\), where \(\bar{S}\) and \(\bar{S^c}\) denote the closures of \(S\) and \(S^c\), respectively. Then we have: $$\partial S^c = (\bar{S} \cap \bar{S^c})^c = \bar{S}^c \cup \bar{S^c}^c = (\operatorname{int} S) \cup (\operatorname{int} S^c)$$ Since the union of open sets is open, and both \(\operatorname{int} S\) and \(\operatorname{int} S^c\) are open by definition, \(\partial S^c\) is open. Therefore, \(\partial S\) is closed.

(b) Proving \(S^0\) is open

We want to show that the interior of the set \(S, S^0\), is open. Recall that by definition, a point \(x\) is included in \(S^0\) if there exists an open neighborhood \(U\) of x such that \(U \subseteq S\). Let \(x\) be an arbitrary point in \(S^0\). Since \(x \in S^0\), there exists an open neighborhood \(U\) of x contained in \(S\). Now for any point \(y \in U\), there exists another open neighborhood of \(y\), call it \(V\), such that \(V \subseteq U \subseteq S\). This means that \(y\) is an interior point of \(S\), so \(U \subseteq S^0\). Since this holds for any \(x \in S^0\), it follows that \(S^0\) is open.

(c) Proving the exterior of \(S\) is open

To prove that the exterior of \(S\) is open, we must show that for each point \(x\) in the exterior, there exists an open neighborhood of \(x\) contained entirely in the exterior. By definition, a point \(x\) is in the exterior of \(S\) if there exists an open neighborhood \(U\) containing \(x\) that is disjoint from \(S\). Let \(x\) be an arbitrary point in the exterior of \(S\). Then there exists an open neighborhood \(U\) of \(x\) such that \(U \cap S = \emptyset\). For any point \(y \in U\), \(y\) is not in \(S\) and there exists an open neighborhood \(V \subseteq U\) of \(y\) for which \(V \cap S = \emptyset\). Since \(V\) is disjoint from \(S\), \(y\) is also in the exterior of \(S\). This means that \(U\) is contained entirely in the exterior of \(S\). Since this holds for any \(x\) in the exterior, it follows that the exterior of \(S\) is open.

(d) Proving the limit points of \(S\) form a closed set

To prove that the set of limit points of \(S\), denoted as \(L\), form a closed set, we need to show that \((L)^c\) is open. By definition, a point \(x\) is a limit point of \(S\) if every open neighborhood of \(x\) intersects \(S\). So, if \(x\) is not a limit point of \(S\), there exists an open neighborhood \(U\) of \(x\) that is disjoint from \(S\). Suppose \(x\) is not a limit point of \(S\). Since \(U\) is disjoint from \(S\), for every point \(y \in U\), there exists an open neighborhood \(V \subseteq U\) of \(y\) such that \(V \cap S = \emptyset\). This implies that \(y\) is not a limit point of \(S\). Thus, \(U \subseteq (L)^c\). Since this holds for every \(x \in (L)^c\), it follows that \((L)^c\) is open and \(L\) is closed.

(e) Proving \(\overline{(\bar{S})}=\bar{S}\)

To prove this property, we need to show that the closure of the closure of \(S\) is equal to the closure of \(S\). Recall that the closure of a set \(S\), denoted by \(\bar{S}\), is the smallest closed set containing \(S\). Since \(\bar{S}\) is closed, it contains all its limit points, and therefore contains all its boundary points. Recall that \(\partial S = \bar{S} \cap \bar{S^c}\), so for any boundary point \(x\) of \(S\), \(x \in \bar{S}\). This means that \(S \cup \partial S \subseteq \bar{S}\), and since the union of \(S\) and its boundary is the closure of \(S\), we have \(\bar{S} = S \cup \partial S\). Since \(\bar{S}\) is closed, its boundary points are also contained in \(\bar{S}\). Therefore, we have that \(\overline{(\bar{S})} = \bar{S}\), proving the desired property.

Key concepts

Boundary of a Set

In real analysis, the boundary of a set is a pivotal concept that determines the dividing line between the points inside the set and the points outside. Mathematically, the boundary of a set S, denoted as &partial;S, includes all those points that are arbitrarily close both to S and to the space outside of S. To visualize this, think of the boundary as the edge of a bubble surrounding the set. What makes it interesting is that the boundary can contain points that are neither strictly inside nor strictly outside the set.

The exercise you're working on shows that the boundary of a set is closed. This fact can be summarized by saying that if you take any point not on that edge, which forms the complement of the boundary, you can find a little open bubble around that point that doesn't touch the boundary. This is technical speak for why we consider the boundary as a closed set.

Interior of a Set

Moving on to the interior of a set, denoted S⁰, which refers to the most 'cozy' part of a set. Imagine that you're standing inside a room; everywhere you can step without touching the walls is the interior. In mathematical terms, if there is a small open set surrounding a point that is completely contained within S, that point is part of the interior of S. The solution presented in your exercise confirms that interiors are open sets by nature. They're like open invitations; there's room around every point for you to step into without stepping out of the set.

Exterior of a Set

The exterior of a set is essentially the opposite of the interior. It includes all the points that can 'wander freely' without ever bumping into the set. The points in the exterior are surrounded by an open space that does not intersect with the set at all. Your exercise proof showcases that the exterior is an open set, much like the interior. If you step outside that aforementioned room, you can keep on walking without ever hitting the walls—that's the exterior for you.

Limit Points

Diving into limit points, these are points that, even if they aren't within a set, they're never lonely because other points of the set keep them company at any distance. More technically, limit points have other points of the set snug in every open set that surrounds them. In your exercise, the proof that the set of all limit points is closed tells you that limit points can't be escaped from; once you're a limit point, you're always a limit point—there's no open set that you can step into to leave all other set members behind.

Closed Set

When something is closed in real analysis, it's not about doors or lids; it means the set hugs all its limit points close. A closed set contains every approach you make towards it, without ever letting you step outside. In simpler words, if you're walking on the edge, or limit points, of the set, you're still considered to be within the set. This is critical because it tells us how we can lay boundaries to confine our mathematical points.

Open Set

Now, let's unpack what an open set is. Just like a park with no fences, an open set gives you the freedom to 'roam' around any point within the set without unintentionally stepping out. Every point has its own little open neighborhood entirely contained within the set, like having a personal bubble of space. This is an inviting concept because it assures that there are no edge points—you're either fully in, with room to spare, or you're out.

Closure of a Set

Lastly, the term closure of a set, denoted ̅S, encompasses the set itself plus all its boundary points; it's like putting a fence around everything that belongs to the set, as well as everything that's right next door. In essence, the closure is the smallest closed set that completely contains the original set. The exercise solution states that the closure of the closure is the closure itself, which makes sense: once you've enclosed everything, including the limit points, enclosing it again won't change anything—you've already captured everything the first time around.