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Chapter 3: Problem 6

Give examples of incomplete metric spaces possessing complete subspaces.

Short Answer

Expert verified
Examples include the rationals \( \mathbb{Q} \) with finite decimal expansion as a complete subspace, and the open interval \((0, 1)\) with a closed subinterval like \([0.1, 0.9]\) as complete.

Step by step solution

01

Understand the Concept of Metric Spaces

A metric space is a set equipped with a function (called a metric) that defines a distance between any two elements in the set. The metric must satisfy certain conditions such as non-negativity, symmetry, and the triangle inequality. A space is complete if every Cauchy sequence (a sequence where terms become arbitrarily close to each other) in it has a limit that is also within the space.
02

Define an Incomplete Metric Space

An incomplete metric space is one in which there exists at least one Cauchy sequence that does not converge to a limit within the space. The prototypical example of an incomplete metric space is the set of rational numbers \( \mathbb{Q} \) with the standard metric \( d(x, y) = |x-y| \). There are Cauchy sequences of rational numbers whose limits are irrational, and thus not contained within \( \mathbb{Q} \).
03

Identify a Complete Subspace

To find a complete subspace within an incomplete metric space, consider the set of all rational numbers whose decimal representation has a finite number of digits. This set can be shown to be complete because any Cauchy sequence composed entirely of such numbers will converge to a number of the same type, since there is a limit on the possible number of digits due to finiteness.
04

Provide Another Example

Consider the open interval \((0, 1)\) under the standard metric \( d(x, y) = |x-y| \). This space is incomplete because there are sequences such as \( x_n = 1 - \frac{1}{n} \) that converge to 1, which is not in \((0, 1)\). However, the closed subinterval \([0.1, 0.9]\) is complete as every Cauchy sequence within this interval has a limit also within \([0.1, 0.9]\).

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Key Concepts

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

Cauchy sequence
A Cauchy sequence is a specific type of sequence in mathematics that has particularly interesting properties in metric spaces. Imagine you have a sequence of numbers, and as you proceed along the sequence, the numbers get increasingly closer to each other. This is the defining characteristic of a Cauchy sequence. The idea is that after a certain point in the sequence, the difference between any two numbers is arbitrarily small.

For a sequence \( \{x_n\} \) to be Cauchy in a given metric space with metric \( d \), it means that for every positive number \( \varepsilon \), however tiny, there is a certain stage in your sequence after which the distance \( d(x_m, x_n) < \varepsilon \) for all terms \( x_m \) and \( x_n \) that are beyond that stage.

This doesn't necessarily mean the sequence has a limit within the space; it just signifies the elements of the sequence are getting closer to each other. In complete metric spaces, all Cauchy sequences converge to a limit within the space, but in incomplete metric spaces, like the rational numbers, some Cauchy sequences can converge to values not in the space, such as irrational numbers.
complete subspace
A complete subspace is a fascinating construct, especially when considering an incomplete metric space. To visualize it, think about how in some metric spaces, even if not all sequences can find a limit within the space, specific subsets of that space still maintain completeness. These subsets are known as complete subspaces. They possess the property that any Cauchy sequence composed of elements within the subset has a limit that is also within the same subset.

For example, within the incomplete metric space of rational numbers, consider the subset of rational numbers that can be expressed with a finite number of digits. This subset is a complete subspace because any Cauchy sequence formed entirely from these numbers will have a limit that is another number of the same finite type. The finite representation restricts and bounds their convergence, ensuring the limit remains within the subset.

Complete subspaces are crucial when examining spaces that are not entirely complete, as they provide structures where solutions to Cauchy sequences are closed and conserved within the subsets themselves.
standard metric
The standard metric is one of the simplest and most intuitive ways to understand distances in mathematics. When discussing metric spaces, the standard metric, often denoted as \( d(x, y) = |x - y| \), is a way to measure the 'straight-line' distance between two points in a given space. It satisfies the basic conditions required of a metric: non-negativity (distances are always positive or zero), identity of indiscernibles (only identical points have zero distance), symmetry (the distance from \( x \) to \( y \) is the same as from \( y \) to \( x \)), and the triangle inequality (the distance from \( x \) to \( y \) together with \( y \) to \( z \) is at least \( x \) to \( z \)).

This metric is particularly significant as it is the foundational tool for defining and understanding Cauchy sequences, completeness, and convergence within metric spaces. In the standard metric space, such as the real numbers or rational numbers, it's used to assess how close sequences are to converging within the space, thereby revealing the complete or incomplete nature of those spaces.

Using the standard metric helps to easily identify both complete and incomplete spaces, and further helps in studying the behavior of sequences as they strive to reach limits within their sets.

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Most popular questions from this chapter

Prove that if \(\left(X, \rho^{\prime}\right)\) and \(\left(Y, \rho^{\prime \prime}\right)\) are metric spaces, then a metric \(\rho\) for the set \(X \times Y\) is obtained by setting, for \(x_{1}, x_{2} \in X\) and \(y_{1}, y_{2} \in Y\), (i) \(\rho\left(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)\right)=\max \left\\{\rho^{\prime}\left(x_{1}, x_{2}\right), \rho^{\prime \prime}\left(y_{1}, y_{2}\right)\right\\}\); or (ii) \(\rho\left(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)\right)=\sqrt{\rho^{\prime}\left(x_{1}, x_{2}\right)^{2}+\rho^{\prime \prime}\left(y_{1}, y_{2}\right)^{2}}\)

Metrize the extended real number system \(E^{*}\) by $$ \rho^{\prime}(x, y)=|f(x)-f(y)|, $$ where the function $$ f: E^{*} \underset{\text { onto }}{\longrightarrow}[-1,1] $$ is defined by \(f(x)=\frac{x}{1+|x|}\) if \(x\) is finite, \(f(-\infty)=-1,\) and \(f(+\infty)=1\) Compute \(\rho^{\prime}(0,+\infty), \rho^{\prime}(0,-\infty), \rho^{\prime}(-\infty,+\infty), \rho^{\prime}(0,1), \rho^{\prime}(1,2),\) and \(\rho^{\prime}(n,+\infty)\). Describe \(G_{0}(1), G_{+\infty}(1),\) and \(G_{-\infty}\left(\frac{1}{2}\right)\). Verify the metric axioms (also when infinities are involved).

Prove the additivity of the volume of intervals, namely, if \(A\) is subdivided, in any manner, into \(m\) mutually disjoint subintervals \(A_{1}, A_{2}, \ldots, A_{m}\) in \(E^{n}\), then $$ v A=\sum_{i=1}^{m} v A_{i} $$ (This is true also if some \(A_{i}\) contain common faces). [Proof outline: For \(m=2\), use Problem 8 Then by induction, suppose additivity holds for any number of intervals smaller than a certain \(m\) \((m>1) .\) Now let $$ A=\bigcup_{i=1}^{m} A_{i} \quad\left(A_{i} \text { disjoint }\right) $$ One of the \(A_{i}\) (say, \(\left.A_{1}=[\bar{a}, \bar{p}]\right)\) must have some edge-length smaller than the corresponding edge-length of \(A\left(\right.\) say \(\left., \ell_{1}\right) .\) Now cut all of \(A\) into $$ P=[\bar{a}, \bar{d}] \text { and } Q=A-P \text { (Figure 4) } $$ by the plane \(x_{1}=c\left(c=p_{1}\right)\) so that \(A_{1} \subseteq P\) while \(A_{2} \subseteq Q\). For simplicity, assume that the plane cuts each \(A_{i}\) into two subintervals \(A_{i}^{\prime}\) and \(A_{i}^{\prime \prime}\). (One of them may be empty.) Then $$ P=\bigcup_{i=1}^{m} A_{i}^{\prime} \text { and } Q=\bigcup_{i=1}^{m} A_{i}^{\prime \prime} $$ Actually, however, \(P\) and \(Q\) are split into fewer than \(m\) (nonempty) intervals since \(A_{1}^{\prime \prime}=\emptyset=A_{2}^{\prime}\) by construction. Thus, by our inductive assumption, $$ v P=\sum_{i=1}^{m} v A_{i}^{\prime} \text { and } v Q=\sum_{i=1}^{m} v A_{i}^{\prime \prime} $$ where \(v A_{1}^{\prime \prime}=0=v A_{2}^{\prime},\) and \(v A_{i}=v A_{i}^{\prime}+v A_{i}^{\prime \prime}\) by Problem \(8 .\) Complete the inductive proof by showing that $$ \left.v A=v P+v Q=\sum_{i=1}^{m} v A_{i} .\right] $$

Let \(f: E^{1} \rightarrow E^{1}\) be given by $$ f(x)=\frac{1}{x} \text { if } x \neq 0, \text { and } f(0)=0. $$ Show that \(f\) is bounded on an interval \([a, b]\) iff \(0 \notin[a, b] .\) Is \(f\) bounded on (0,1)\(?\)

Prove that \(\bar{A}\) is contained in any closed superset of \(A\) and is the intersection of all such supersets.
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