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91Ó°ÊÓ

Show that the set \(E^{\prime}\) of points of accumulation of any set \(E\) must be closed.

Short Answer

Expert verified
The set of accumulation points \(E^{\prime}\) of any set \(E\) is closed because it contains all its limit points.

Step by step solution

01

Understanding Points of Accumulation

First, recognize that a point of accumulation of a set \(E\) is a point \(x\) such that every neighborhood of \(x\) contains at least one point of \(E\) distinct from \(x\). The set of these points for set \(E\) is denoted as \(E^{\prime}\).
02

Definition of a Closed Set

Recall that a set is closed if it contains all its limit points. In other words, if \,\(x_n\) is a sequence in a closed set and \,\(x_n\) converges to \,\(x\), then \,\(x\) must be in the set.
03

Prove \(E^{\prime}\) Contains Its Limit Points

Suppose \,\(x\) is a limit point of \,\(E^{\prime}\). This means that every neighborhood \,\(U\) of \,\(x\) must contain a point of \,\(E^{\prime}\) other than \,\(x\). This is true if and only if for every neighborhood \,\(U\), there exists a point \,\(y eq x\) in both \,\(U\) and \,\(E^{\prime}\).
04

Neighborhood and Sequence Argument

For \,\(x\) to be a limit point of \,\(E^{\prime}\), construct a sequence \,\(x_n\) from points in \,\(E^{\prime}\) that converges to \,\(x\). Since \,\(x_n\) is in \,\(E^{\prime}\) and converges to \,\(x\), the nature of accumulation points means \,\(x\) must be in \,\(E^{\prime}\).
05

Conclusion: \(E^{\prime}\) is Closed

Since every limit point \,\(x\) of \,\(E^{\prime}\) is contained in \,\(E^{\prime}\), it follows that \,\(E^{\prime}\) contains all its limit points and is therefore closed by definition.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Points of Accumulation
When we discuss the points of accumulation, we refer to a fascinating aspect of a set in mathematical analysis. Consider a set \(E\) within a space. A point \(x\) is called a point of accumulation of \(E\) if every neighborhood around \(x\), no matter how small, contains another point from \(E\) that differs from \(x\) itself. This feature gives us a unique perspective about how sets behave within a given space.

To visualize this, picture standing at a point \(x\) on the number line. Imagine drawing a tiny circle around you—this is your neighborhood. If no matter how small your circle gets, there’s still always another point from \(E\) lying within it, then you are indeed in the midst of points of accumulation.

Remember, the collection of all such points for a set \(E\) is denoted as \(E^{\prime}\). This subset forms the basis of understanding the "closeness" principle and how sets interrelate through these accumulation points.
Limit Points
The term "limit points" revolves around the idea of convergence within sets. If you have a sequence in mathematics, think of it as a list of numbers getting progressively closer to a particular point—this point is known as a limit point. In the context of a set, a limit point of \(E\) is a point such that every neighborhood around it contains other points of \(E\).

Consider, for instance, the sequence of numbers \(1, 0.5, 0.25, 0.125, \ldots\). As these numbers get smaller, they seem to be inching closer to \(0\) yet never quite touch it. Here, \(0\) would be a limit point. In a practical sense, if you’re observing a crowd slowly gathering closer to a place, that place might be viewed as the "limit point" for the group.

Thus, in terms of closed sets, a set is considered closed when it encompasses all its limit points. If within the set \(E^{\prime}\), all limit points occur, then \(E^{\prime}\) itself must be closed, fitting into the broader framework of set theories and topological concepts.
Neighborhood
The notion of a neighborhood helps us understand the behavior of points around a given location in a set. In mathematical terms, a neighborhood of a point \(x\) means a set that contains an open set which in turn contains \(x\). Think of it as drawing a bubble around a point within which certain properties are guaranteed.

In the context of limit points or points of accumulation, a neighborhood serves as the testing ground. For instance, if you want to determine if \(x\) is a limit point, you'll check if every neighborhood around \(x\) snuggly fits other points belonging to the set \(E\).

This idea is crucial because neighborhoods don't just help in identifying accumulation points; they also facilitate understanding local properties of functions and sequences. By employing neighborhoods, we delve deeper into how close sets are structured and operate—a central idea to mastering the concepts of closed sets and continuous functions.

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Most popular questions from this chapter

In many applications of open sets and closed sets we wish to work just inside some other set \(A\). It is convenient to have a language for this. A set \(E \subset A\) is said to be open relative to \(A\) if \(E=A \cap G\) for some set \(G \subset \mathbb{R}\) that is open. A set \(E \subset A\) is said to be closed relative to \(A\) if \(E=A \cap F\) for some set \(F \subset \mathbb{R}\) that is closed. Answer the following questions. (a) Let \(A=[0,1]\) describe, if possible, sets that are open relative to \(A\) but not open as subsets of \(\mathbb{R}\). (b) Let \(A=[0,1]\) describe, if possible, sets that are closed relative to \(A\) but not closed as subsets of \(\mathbb{R}\). (c) Let \(A=(0,1)\) describe, if possible, sets that are open relative to \(A\) but not open as subsets of \(\mathbb{R}\). (d) Let \(A=(0,1)\) describe, if possible, sets that are closed relative to \(A\) but not closed as subsets of \(\mathbb{R}\).

Give an example of a set that has the set $$ E=\\{0\\} \cup\\{1,1 / 2,1 / 3,1 / 4,1 / 5, \ldots\\} $$ as its set of accumulation points.

Give examples of closed sets that are countable and closed sets that are uncountable.

Let \(A\) be a set and \(B=\mathbb{R} \backslash A\). Show that every boundary point of \(A\) is also a boundary point of \(B\).

Let \(\mathcal{F}\) be a family of (nondegenerate) intervals; that is, each member of \(\mathcal{F}\) is an interval (open, closed or neither) but is not a single point. Suppose that any two intervals \(I\) and \(J\) in the family have no point in common. Show that the family \(\mathcal{F}\) can be arranged in a sequence \(I_{1}, I_{2}, \ldots\)

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