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Chapter 13: Problem 5

Show that \(Z=\\{f \in M[a, b]:\) the range of \(f\) is nowhere dense \(\\}\) is a residual subset of the space \(M[a, b]\) of bounded functions on \([a, b]\).

Short Answer

Expert verified
The set \(Z\) is residual in \(M[a, b]\) because its complement is a countable union of nowhere dense sets, making it of the first category.

Step by step solution

01

Define M[a, b] and residual sets

The space \(M[a, b]\) consists of all bounded functions defined on the interval \([a, b]\). For a subset of a topological space \(X\), it is called residual if it is a countable intersection of open dense sets, or equivalently, its complement is of the first category (a countable union of nowhere dense sets).
02

Understand nowhere dense set

A set \(S\) is called nowhere dense in a space if its closure has empty interior. Hence, if the range of a function \(f\) is nowhere dense in \(\mathbb{R}\), this means the closure of the range of \(f\) has empty interior in \(\mathbb{R}\).
03

Space of understudy

Consider a function \(f\) in \(M[a, b]\). We are tasked to show that the set \(Z\), consisting of functions with a nowhere dense range, is residual in \(M[a, b]\). This means showing that the complement of \(Z\) is of the first category.
04

Establish the range non-density

We first consider the complement of \(Z\), the set of functions whose ranges are not nowhere dense. For a function to have a range that is not nowhere dense, the closure of its range must have non-empty interior. This implies the existence of an interval fully contained in the range of \(f\).
05

Showing the complement is of the first category

Now consider intervals \((q - \epsilon, q + \epsilon)\) in \(\mathbb{R}\), centered at rational points \(q\) and of rational radius \(\epsilon\). Functions whose ranges intersect any such interval in a non-trivial way (contain an interval) form a set that is nowhere dense in \(M[a, b]\). Thus, the complement of \(Z\) is the union of sets, each nowhere dense, since it requires having a range with a non-empty interior.
06

Conclude Z is residual

Since the complement of \(Z\) is the countable union of nowhere dense sets, it is of the first category. Therefore, by definition, \(Z\) is a residual set in \(M[a, b]\).

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Key Concepts

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

Residual Set
A residual set in a topological space is quite an interesting notion. It refers to a subset that is significant in the sense of density. Specifically, a set is residual if its complement is of the first category. What does that mean? It means that the complement is a countable union of nowhere dense sets.

In our context with bounded functions, we have a space denoted as \(M[a, b]\). It includes all functions that are bounded on the interval \([a, b]\). The set of interest, \(Z\), includes functions whose range is nowhere dense. To show that \(Z\) is residual, we demonstrate that its complement consists of functions with a range that is not nowhere dense, i.e., the range closure has non-empty interior.

Thus, \(Z\) can be considered residual due to its complement forming a countable union of nowhere dense sets, thereby satisfying the criteria of a residual set.
Bounded Functions
Bounded functions are functions that do not go to infinity within their range. More formally, a function \(f\) defined on an interval \([a, b]\) is bounded if there is a real number \(M\) such that for all \(x\) in \( [a, b] \), \(|f(x)| \leq M\). This concept is crucial as it provides a boundary within which the function operates.

In the scenario of our study, the space \(M[a, b]\) is comprised of these bounded functions. This boundedness assures us that the functions do not exhibit wild behavior like shooting off to infinity, making them more manageable to work with.

Working with bounded functions also implies that their range is contained within certain limits, and this characteristic plays a part in discussing whether their ranges can be nowhere dense.
Nowhere Dense Sets
To understand a nowhere dense set, it helps to recall that if a set's closure has an empty interior, it is considered nowhere dense. In simpler terms, nowhere dense sets are spread out and sparse. They don't "stick" anywhere because they don't occupy any 'chunk' of space.

In the context of functions, if the range of a function is nowhere dense, it implies that although the range covers some real numbers, it doesn't cluster densely around any particular interval. This idea is integral when proving that the set \(Z\) is residual.

Consider the notion of being nowhere dense in \(M[a, b]\): the set of all bounded functions is considered majorly by examining those that intersect with certain types of intervals, particularly targeting function ranges that fail to demonstrate dense clustering. Understanding this concept provides insight into how the complement of our set \(Z\) becomes composed of nowhere dense subsets.

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Most popular questions from this chapter

Let \((X, d)\) be a metric space and let \(M(X)\) denote the set of all bounded real-valued functions on \(X\) furnished with the sup metric. Choose a fixed \(x_{0} \in X\) and define a mapping \(h: X \rightarrow M(X)\) by writing \(h(x)\) for the function $$ (h(x))(y)=d(x, y)-d\left(y, x_{0}\right) \quad(y \in X) $$ (a) Show that each \(h(x)\) is a bounded function on \(X\) for each \(x \in X\). (b) Show that \(h\) is an isometry of \(X\) to a subspace of \(M(X)\). (c) Show that every metric space is isometric to a subspace of some complete function space (i.e., a complete metric space of real-valued functions defined on a set and furnished with the sup metric).

Let \(c_{Q}\) denote the set of all sequences of real numbers converging to a rational number, that is, $$ c_{Q}=\left\\{x \in c: \lim _{i \rightarrow \infty} x_{i} \in \mathbb{Q}\right\\} $$ Is \(c_{Q}\) nowhere dense in \(c ?\) Is \(c_{Q}\) of the first category in \(c ?\)

Prove the following more general version of the Arzel脿-Ascoli theorem: Let \(K\) be a closed subset of the metric space \(C(X)\), where \(X\) is a compact metric space and \(C(X)\) denotes the space of continuous real-valued functions on \(X\) furnished with the supremum metric. Then \(K\) is compact in \(C(X)\) if and only if \(K\) is uniformly bounded and equicontinuous.

Use the Baire category theorem to show that the set $$ \\{f \in \mathcal{C}[0,1]:|f(x)| \leq 1\\} $$ in \(\mathcal{C}[0,1]\) cannot be covered by a sequence of compact sets. (This is also done, using the Arzel脿-Ascoli theorem, in Exercise \(13.12 .57 .\) )

Use the Arzel脿-Ascoli theorem to show that the set $$ \\{f \in \mathcal{C}[0,1]:|f(x)| \leq 1\\} $$ in \(\mathcal{C}[0,1]\) cannot be covered by a sequence of compact sets. (This is also done, using the Baire category theorem, in Exercise 13.13.9.)
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