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Chapter 12: Problem 8

Show that the intersection of any collection of closed sets in a metric space is closed.

Short Answer

Expert verified
Let \(\{C_i\}_{i \in I}\) be a collection of closed sets in a metric space M. Define \(C = \bigcap_{i \in I} C_i\) as the intersection of these sets. To prove that C is closed, we will show that it contains all its limit points. Let x be a limit point of C. Since C is a subset of each \(C_i\), x is also a limit point for each \(C_i\). As each \(C_i\) is closed, they all contain x. Hence, x is in their intersection, C. This shows that C contains all its limit points, and thus, C is a closed set.

Step by step solution

01

Assume a collection of closed sets

Assume we have a collection of closed sets in a metric space M: \(\{C_i\}_{i \in I}\), where I can represent any index set (finite or infinite).
02

Define the intersection of the closed sets

Let C be the intersection of all these sets, i.e., \(C = \bigcap_{i \in I} C_i\).
03

Prove that the intersection contains all its limit points

We will show that the set C is closed by proving that it contains all its limit points. Let x be a limit point of C. By definition, every open ball centered at x contains infinitely many points of C. Since x is a limit point for C, and C is a subset of each of the closed sets, x is also a limit point for each \(C_i\) in the collection. As each of these sets is closed, they all contain their limit points including x. Therefore, x is contained in every set \(C_i\).
04

Conclude that the intersection is closed

Since x is contained in every set \(C_i\), it follows that x is also in their intersection, C. As we have shown that C contains all its limit points, C is a closed set. Therefore, we have demonstrated that the intersection of any collection of closed sets in a metric space is closed.

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Key Concepts

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

Closed Sets
In the context of metric spaces, a closed set is one that contains all its limit points. But what does that mean? Imagine you have a set of points on a line or in a space, and no matter how closely you approach the boundary of this set, you'll never step outside it without crossing over a clear edge. That boundary comprises the limit points of the set. This complete inclusion of limit points ensures that the set is closed.
Closed sets are crucial because they have many applications in analysis and topology, such as in defining continuity and convergence. A helpful property is that the intersection of closed sets remains closed. This property is useful when working with complex sets made up of simpler ones.
Limit Points
A limit point of a set is a point where, if you draw an open ball (or a tiny neighborhood) around it, you will always find points from the set inside it. Even if the point itself isn't in the set, its presence influences the set, indicating a kind of 'edge' behavior.
Limit points help us determine whether sets are closed. For any set to be closed, it must contain all its limit points. Whether we have a familiar shape like a circle or an unusual collection of points in space, identifying these points is crucial for understanding the set's properties.
Intersection of Sets
The intersection of several sets is made up of just the elements they all share. Picture several overlapping circles; their intersection is the region where all circles overlap.
This idea transfers to closed sets. When you take the intersection of closed sets, the intersection stays closed. Why? Because all original closed sets already include their limit points, and any limit point of the intersection is necessarily a limit point of each of the original sets. Thus, the intersection inherently captures all these limit points, maintaining its closed nature.
Open Ball
An open ball in a metric space is like a circle around a point, but it can extend into higher dimensions like a bubble. It's defined by a center and a radius, containing all points closer to the center than the radius allows.
Open balls are important because they help define concepts like continuity and limit points. They allow us to test whether a set grabs onto points infinitely close to its edges (which would make those points limit points). This is essential when classifying sets as closed or open, and it aids in visualizing otherwise abstract concepts in metric spaces.

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Most popular questions from this chapter

Show that a function \(f:(M, d) \rightarrow\left(M^{\prime}, d^{\prime}\right)\) is continuous if and only if the inverse image of any open set is open, that is, if and only if \(f^{-1}(U)=\\{x \in M \mid f(x) \in U\\}\) is open in \(M\) whenever \(U\) is an open set in \(M^{\prime}\).

Show that a subset \(S\) of a metric space \(M\) is open if and only if \(S\) contains an open neighborhood of each of its points.

Let \(S \subseteq \ell^{\infty}\) be the subspace of all binary sequences (sequences of 0 's and 1 's). Describe the metric on \(S\).

Prove Minkowski's inequality $$ \left(\sum_{n=1}^{\infty}\left|x_{n}+y_{n}\right|^{p}\right)^{1 / p} \leq\left(\sum_{n=1}^{\infty}\left|x_{n}\right|^{p}\right)^{1 / p}+\left(\sum_{n=1}^{\infty}\left|y_{n}\right|^{p}\right)^{1 / p} $$ as follows: a) Prove it for \(p=1\) first. b) Assume \(p>1\). Show that $$ \left|x_{n}+y_{n}\right|^{p} \leq\left|x_{n}\right|\left|x_{n}+y_{n}\right|^{p-1}+\left|y_{n}\right|\left|x_{n}+y_{n}\right|^{p-1} $$ c) Sum this from \(n=1\) to \(k\) and apply H枚lder's inequality to each sum on the right, to get $$ \begin{aligned} &\sum_{n=1}^{k}\left|x_{n}+y_{n}\right|^{p} \\ &\quad \leq\left\\{\left(\sum_{n=1}^{k}\left|x_{n}\right|^{p}\right)^{1 / p}+\left(\sum_{n=1}^{k}\left|y_{n}\right|^{p}\right)^{1 / p}\right\\}\left(\sum_{n=1}^{k}\left|x_{n}+y_{n}\right|^{p}\right)^{1 / q} \end{aligned} $$ Divide both sides of this by the last factor on the right and let \(n \rightarrow \infty\) to deduce Minkowski's inequality.

a) Show that if \(\boldsymbol{x}=\left(x_{n}\right) \in \ell^{p}\), then \(\boldsymbol{x} \in \ell^{q}\) for all \(q>p\). b) Find a sequence \(\boldsymbol{x}=\left(x_{n}\right)\) that is in \(\ell^{p}\) for \(p>1\), but is not in \(\ell^{1}\).
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