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Chapter 13: Problem 10

Let \((X, d)\) be a metric space. (a) Prove that \(X\) and \(\emptyset\) are both open and closed. (b) Prove that a finite union of closed sets is closed and a finite intersection of open sets is open.

Short Answer

Expert verified
In metric spaces, both \(X\) and \(\emptyset\) are open and closed. A finite union of closed sets is closed, and a finite intersection of open sets is open.

Step by step solution

01

Understanding Open and Closed Sets

In a metric space (X, d), a subset U of X is called open if for every x in U , there exists an \( \epsilon > 0 \) such that \( B_\epsilon(x) \subseteq U \) . A subset F of X is called closed if its complement \( X \setminus F \) is open.
02

Proving X is Open and Closed

Since X is the entire space, for any point x \in X and any \( \epsilon > 0 \), the open ball \( B_\epsilon(x) \) will always be contained in X . Thus, X is open. To show that X is closed, consider its complement, \( X \setminus X = \emptyset \), which is open (by part (a)). Hence, X is also closed.
03

Proving Empty Set is Open and Closed

The empty set \( \emptyset \) has no elements, so the condition that every point x \in \emptyset has an \( \epsilon -neighborhood inside \emptyset \) is vacuously true, making \emptyset open. Its complement is X, which is open, hence \emptyset is also closed.
04

Proving Finite Union of Closed Sets is Closed

Consider closed sets \( F_1, F_2, \ldots, F_n \subseteq X \). These are closed, meaning \( X \setminus F_i \) is open for each i. The union \( \bigcup_{i=1}^n F_i \) complements to \( X \setminus (\bigcup_{i=1}^n F_i) = \bigcap_{i=1}^n (X \setminus F_i) \), a finite intersection of open sets, which is open. Thus, \( \bigcup_{i=1}^n F_i \) is closed.
05

Proving Finite Intersection of Open Sets is Open

Let open sets \( U_1, U_2, \ldots, U_n \subseteq X \) be given. Each point \( x \in \bigcap_{i=1}^n U_i \) has an \( \epsilon_i \) ball within \( U_i \). Choosing \( \epsilon = \min(\epsilon_1, \ldots, \epsilon_n) \) guarantees that \( B_\epsilon(x) \subseteq U_i \) for each i, so \( B_\epsilon(x) \subseteq \bigcap_{i=1}^n U_i \). Thus, the intersection is open.

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Key Concepts

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

Open Sets
In the context of metric spaces, open sets are fundamental. A subset \( U \) of a metric space \((X, d)\) is considered open if, for every point \( x \) in \( U \), there is some distance \( \epsilon > 0 \) such that the open ball \( B_\epsilon(x) \), which includes all points within that \( \epsilon \) radius of \( x \), is entirely contained within \( U \).
The idea is that, around each point in an open set, you can "wiggle around" or move a little without leaving the set.
This concept helps in defining boundaries and understanding how spaces behave.
  • Example: A circle without its boundary in a plane is open because you can move a bit in any direction around any point without leaving the circle.
  • An entire metric space \( X \) is open because any open ball centered in \( X \) with any radius stays within \( X \).
  • The empty set \( \emptyset \) is trivially open because there are no points to challenge this definition.
Closed Sets
Closed sets complement the concept of open sets in metric spaces. A set \( F \) within \( X \) is called closed if its complement \( X \setminus F \) is open. Essentially, open sets have no boundaries, while closed sets include their boundaries.
This means that every limit point of \( F \), or point where sequences can converge to, must be contained within \( F \).
  • Example: A filled-in circle (with the boundary) is closed because any point that approaches the boundary is also contained within the circle.
  • The empty set \( \emptyset \) is closed, owing to its complement, the whole space \( X \), being open.
  • The whole space \( X \) itself is closed because its complement, the empty set, is open.
Topology in Metric Spaces
Topology in metric spaces involves studying geometric properties preserved under continuous transformations. It focuses on whether things like shape, size, or distance change. The terms open and closed sets form the basis of this study.
These concepts help define 'closeness' of points and continuity of functions in space.
  • The entire collection of open sets in a space forms a topology, giving the space its topological structure.
  • Taking unions and intersections of these sets helps understand the fabric of the topology.
  • Understanding topology helps grasp more complex topics, like convergence, continuity, and compactness, all critical in advanced mathematical analysis.
Finite Unions and Intersections
Finite unions and intersections are important operations in set theory and topology. They unveil how sets combine or overlap.
  • **Finite Union of Closed Sets**: If you take several closed sets in a metric space and form a union of these sets, the resulting set is also closed. This property stems from the fact that the complement of the union is the intersection of their complements, each of which is open.
  • **Finite Intersection of Open Sets**: Conversely, if you take several open sets and intersect them, the resulting set remains open. This holds because the intersection behaves similarly to taking the most restrictive open ball contained within all of them.
  • These properties ensure the flexibility to build larger structures in a way that preserves their topological nature, aiding in constructing more complex topological spaces.

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Most popular questions from this chapter

State precisely and prove a theorem that asserts under which conditions the composition \(f \circ g\) of two continuous functions is continuous.

Show that a metric space \((X, d)\) is compact if and only if for every family \(\mathcal{F}\) of closed subsets of \(X\) for which $$ \bigcap_{F \in \mathcal{F}} F=\emptyset $$ there must be a finite collection \(F_{1}, F_{2}, \ldots, F_{m}\) of sets in \(\mathcal{F}\) so that $$ \bigcap_{i=1}^{m} F_{i}=\emptyset $$

Let \((X, d)\) be a metric space and let \(A\) be a nonempty subset of \(X\). Define \(f: X \rightarrow \mathbb{R}\) by $$ f(x)=\operatorname{dist}(x, A)=\inf \\{d(x, y): y \in A\\} $$ (a) Show that \(|f(x)-f(y)| \leq d(x, y)\) for all \(x, y \in X\). (b) Show that \(f\) defines a continuous real-valued function on \(X\). (c) Show that \(\\{x \in X: f(x)=0\\}=\bar{A}\). (d) Show that \(\\{x \in X: f(x)>0\\}=\operatorname{int}(X \backslash A)\). (e) Show that, unless \(X\) contains only a single point, there exists a continuous real-valued function defined on \(X\) that is not constant. (f) If \(E \subset X\) is closed and \(x_{0} \notin E\), show that there is a continuous real-valued function \(g\) on \(X\) so that \(g\left(x_{0}\right)=1\) and \(g(x)=0\) for all \(x \in E\) (g) If \(E\) and \(F\) are disjoint closed subsets of \(X\), show that there is a continuous real-valued function \(g\) on \(X\) so that \(g(x)=1\) for all \(x \in F\) and \(g(x)=0\) for all \(x \in E\). (h) If \(E\) and \(F\) are disjoint closed subsets of \(X\), show that there are disjoint open sets \(G_{1}\) and \(G_{2}\) so that \(E \subset G_{1}\) and \(F \subset G_{2}\). (i) In the special case where \(X\) is the real line with the usual metric and \(K\) denotes the Cantor ternary set, sketch the graph of the function \(f(x)=\operatorname{dist}(x, K)\) (j) Give an example of a metric space, a point \(x_{0}\), and a set \(A \subset X\) so that \(\operatorname{dist}\left(x_{0}, A\right)=1\) but so that \(d\left(x, x_{0}\right) \neq 1\) for every \(x \in \bar{A}\).

Show that if \(A_{1}, \ldots, A_{n}\) are nowhere dense subsets of a metric space \(X\), then \(A_{1} \cup \cdots \cup A_{n}\) is also nowhere dense in \(X\).

Show that every finite subset of a metric space is compact.
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