Question

Define a point \(p\) in a metric space \(X\) to be a condensation point of a set \(E \subset X\) if every neighborhood of \(p\) contains uncountably many points of \(E\). Suppose \(E \subset R^{k}, E\) is uncountable, and let \(P\) be the set of all condensation points of \(E\). Prove that \(P\) is perfect and that at most countably many points of \(E\) are not in \(P\). In other words, show that \(P^{\epsilon} \cap E\) is at most countable. Hint: Let \(\left\\{V_{n}\right\\}\) be a countable base of \(R^{k}\), let \(W\) be the union of those \(V_{n}\) for which \(E \cap V_{n}\) is at most countable, and show that \(P=W^{c}\).

Short answer

In summary, we have shown that \(P\), the set of all condensation points of a given uncountable set \(E \subset R^k\), is perfect and that at most countably many points of \(E\) are not in \(P\). To achieve this, we used the given hint of the countable base of \(R^k\) by defining the set \(W\) as the union of those sets from the countable base, for which \(E\) intersected with them is at most countable. We proved that \(P\) is closed, and then we showed that \(P=W^{c}\), leading us to the conclusion that the intersection of \(P^{\epsilon}\) and \(E\) is at most countable.

Step-by-step solution

Defining and Proving Closed

Let \(\{V_n\}\) be a countable base of \(R^{k}\). Define the set \(W\) as the union of those \(V_n\) for which \(E \cap V_{n}\) is at most countable. To prove that \(P\) is closed, we need to show that every limit point of \(P\) belongs to \(P\). Suppose \(p' \in \overline{P}\), the closure of \(P\). For any neighborhood \(U\) of \(p'\), there exists a \(p \in P\) such that \(p \in U\). Since \(p\) is a condensation point of \(E\), this implies that \(U\) contains uncountably many points of \(E\). Since \(U\) is arbitrary, this means \(p'\) is a condensation point of \(E\), so \(p' \in P\). Therefore, \(P\) is closed.

Relating \(W^{c}\) and \(P\)

Now, we want to show that \(P=W^{c}\). Let \(x \in W^{c}\), which means \(x\) does not belong to any \(V_n\) for which \(E \cap V_n\) is at most countable. This means that for any \(V_n\) containing \(x\), \(E \cap V_n\) is uncountable. As \(\{V_n\}\) is a base, for any neighborhood \(U\) of \(x\), there exists \(V_n\) such that \(x \in V_n \subset U\). Since \(E \cap V_n\) is uncountable, it follows that every neighborhood of \(x\) contains uncountably many points of \(E\). Thus, \(x\) is a condensation point of \(E\), so \(x \in P\). This shows that \(W^{c} \subset P\). Conversely, let \(x \in P\). We want to show that \(x \in W^{c}\). Since \(x\) is a condensation point of \(E\), every neighborhood of \(x\) contains uncountably many points of \(E\). So, for any \(V_n\) containing \(x\), we must have that \(E \cap V_n\) is uncountable, otherwise, \(x\) would belong to a union of some \(V_n\) for which \(E \cap V_n\) is at most countable, which would contradict that \(x \in P\). Therefore, \(x \in W^{c}\), and it follows that \(P \subset W^{c}\). Combining the two inclusions, we have shown that \(P=W^{c}\).

Intersection of \(P^{\epsilon}\) and \(E\) is at Most Countable

Finally, we want to show that \(P^{\epsilon} \cap E\) is at most countable. Note that \(W = \bigcup_{n=1}^{\infty} V_n'\), where \(V_n' = V_n\) if \(E \cap V_n\) is at most countable, and \(V_n'= \emptyset\) otherwise. Since each \(V_n'\) is open, it follows that \(W\) is open, and as \(P=W^{c}\), then \(P^{\epsilon} \subset W\). Thus, \(P^{\epsilon} \cap E \subset W \cap E\). Since \(W \cap E\) is the union of at most countable sets intersected with \(E\), we have \(W \cap E = \bigcup_{n=1}^{\infty} (E \cap V_n')\), which is also at most countable. Therefore, \(P^{\epsilon} \cap E\) is at most countable, completing the proof.

Key concepts

Metric Space

A metric space is a foundational concept in mathematics, particularly in the field of topology. It includes a set of objects and a function known as the 'metric' or 'distance function' that defines the distance between any two objects within that set. The distance function must satisfy several criteria: it must be non-negative, symmetric (the distance from object A to object B is the same as from B to A), and adhere to the triangle inequality (the distance from A to B via C is always longer than or equal to the direct distance from A to B).

Mathematically, a metric space is a pair \( (X, d) \) where \( X \) is a set and \( d: X \times X \rightarrow \mathbb{R} \) with \( d(x,y) \) fulfilling the conditions mentioned. In the context of textbook exercises, understanding metric spaces is crucial for analyzing concepts like convergence, continuity, and in this particular case, the nature of condensation points – which helps us to explore set behaviors in multi-dimensional real spaces.

Perfect Set

A perfect set in the context of metric spaces is one that is closed and every point within it is a limit point, also known as a condensation point for certain cases. A closed set is one that contains all of its limit points, meaning no points 'escape' the set as we take limits. But for a set to be perfect, it must also be so rich in points that each point is the limit of other points within the set itself.

This is an intriguing property because it suggests that the set has no isolated elements. In terms of the textbook exercise, the set \( P \) of condensation points is perfect because it has been shown that every limit point of \( P \) actually belongs to \( P \) and that the set contains no isolated points, fulfilling the criteria for perfection.

Uncountable Set

An uncountable set is a set too large to be counted by natural numbers – this implies that it has more elements than the set of all natural numbers, \( \mathbb{N} \). The classic example of an uncountable set is the real numbers, \( \mathbb{R} \).

It diverges from countable sets, which might be infinite but can be mapped to natural numbers, like the integers or rational numbers. The concept of uncountability is central to the exercise at hand; neighborhoods of condensation points in \( E \) are required to contain uncountably many points of \( E \) itself, indicating a dense and infinitely complex structure within any region of the set.

Countable Base

A countable base for a topology is a set of open sets such that every open set in the topology can be represented as a union of sets from this collection. Importantly, this base is countable, meaning its size is the same as the set of natural numbers, even if the topology itself is on an uncountable set.

In the given exercise, the countable base \( \{V_n\} \) of \( \mathbb{R}^{k} \) is used strategically to partition the space and analyze the nature of the set \( E \) and its condensation points. By understanding which parts of \( E \) are countable or uncountable within the frames of this base, we reveal the properties of the set \( P \) and its relationship to \( E \) in the context of the problem – a sophisticated yet elegant intersection of these foundational mathematical concepts.