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Find an open function \(f: \mathbb{R} \rightarrow \mathbb{R}\) that is not continuous. Hint: If \(x\)-int \((x)=0, \alpha_{1}, \alpha_{2}, \ldots\) is the binary expansion of the fractional part of \(x\), define \(g(x)=\lim \sup \left(n^{-1} \sum \alpha_{j} \mid j \leq n\right)\). Show that \(g(I)=[0,1]\) for every interval \(I\) in \(\mathbb{R}\). Take \(f=h \circ g\), where \(h\) is an arbitrary surjective function from \([0,1]\) to \(\mathbb{R}\).

Short Answer

Expert verified
Define \( f = h \circ g \) with \( g(x) = \lim \sup(n^{-1}\sum \alpha_j) \) and a surjective \( h: [0,1] \to \mathbb{R} \); f is open but not continuous.

Step by step solution

01

Understanding the Problem and Definitions

We need to find an open function \( f : \mathbb{R} \rightarrow \mathbb{R} \) that is not continuous. We start by analyzing the hint provided. Given any real number \( x \), we consider the binary expansion of its fractional part, \( \alpha_1, \alpha_2, \ldots \). The function \( g(x) = \lim \sup \left(n^{-1} \sum \alpha_j \mid j \leq n \right) \) is defined over \( \mathbb{R} \).
02

Properties of the Function g

The function \( g(x) \) calculates the lim sup of the average occurrence of 1's in the binary expansion of the fractional part of \( x \). The hint suggests that for any interval \( I \) in \( \mathbb{R} \), \( g(I) = [0,1] \). This means \( g \) maps every interval in \( \mathbb{R} \) surjectively (onto) \([0,1]\).
03

Defining Function f Using h

Choose a function \( h: [0,1] \rightarrow \mathbb{R} \) which is surjective. Such a function ensures that \( h([0,1]) = \mathbb{R} \). Now define the function \( f \) as \( f = h \circ g \). Since \( g(I) = [0,1] \) for any interval \( I \subset \mathbb{R} \), \( f \) is open, because \( f(x) = h(g(x)) \) covers all of \( \mathbb{R} \).
04

Proving f is Not Continuous

To show that \( f \) is not continuous, consider any point \( a \in \mathbb{R} \). Because \( g \) maps each interval in \( \mathbb{R} \) to the entire interval \([0,1]\), small changes in \( x \) around \( a \) can result in widely different outputs in \( h(x) \), since \( h \) is not constrained to be continuous in its mapping from \([0,1]\) to \( \mathbb{R} \). This property of \( h \circ g \) makes \( f \) non-continuous at every point.

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Key Concepts

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

Continuous Functions
A continuous function is a fundamental concept in mathematics, particularly in calculus and real analysis. A function \( f \) is considered continuous if, informally speaking, small changes in the input \( x \) result in small changes in the output \( f(x) \).
This means there is no abrupt jump or gap in the graph of the function.
Mathematically, a function \( f: \mathbb{R} \to \mathbb{R} \) is continuous at a point \( c \) if for every \( \varepsilon > 0 \), there exists a \( \delta > 0 \) such that whenever \( |x - c| < \delta \), it follows that \( |f(x) - f(c)| < \varepsilon \).
  • This definition essentially captures the idea that the function doesn’t "break" at \( c \).
  • Continuous functions at every point within an interval are sometimes referred to as being continuous on that interval.
In this exercise, we explore the opposite by looking for a function that is an open function but not continuous. Understanding continuous functions helps us see where the behavior of such a function differs.
Open Functions
Open functions in mathematics are an interesting counterpart to continuous functions.
Although less common in basic discussions, they play a significant role in advanced mathematical analysis.
An open function \( f: \mathbb{R} \to \mathbb{R} \) maps open sets in \( \mathbb{R} \) to open sets in \( \mathbb{R} \).
  • For example, if an interval \( (a, b) \) is mapped by an open function, its image will also be an open set in \( \mathbb{R} \).
  • Open functions can map a single interval onto the entirety of \([0, 1]\), creating very intriguing mappings.
In our exercise, we're tasked with finding an open function that is not continuous. This can be counterintuitive since many might assume open functions would naturally be continuous. However, the properties of open functions allow for more complex relationships than continuity alone.
Real Analysis
Real Analysis is a branch of mathematics dealing with real numbers and real-valued sequences and functions.
It provides the theoretical foundation for calculus and studies the behavior of real functions, which can range from simple to incredibly complex.
  • Key topics include limits, continuity, differentiation, integration, and sequences.
  • Real analysis probes deeply into the nuances of real-numbered functions, with a focus on their inherent properties and behavior.
This field is essential for both understanding the solutions to purely mathematical problems and for applications across disparate scientific disciplines.
In the context of the original exercise, real analysis provides the tools to explore functions that are open but not continuous, such as the construction of \( f = h \circ g \). Understanding real analysis concepts helps reveal why \( f \) behaves as it does, especially in non-intuitive cases.
Binary Expansion
Binary expansion is a way of representing numbers using powers of 2.
Each real number can be expressed in binary form, similar to how it's written in decimal form, but using only 0s and 1s.
  • The fractional part of a real number can be expanded in this form to analyze its properties.
  • This technique is fundamental in computer science, where binary numbers form the basis of computing.
In the problem, the binary expansion of the fractional part of \( x \) helps define the function \( g(x) \).
The binary digits (\( \alpha_1, \alpha_2, \ldots \)) are used to calculate the lim sup, which ultimately aids in the function construction. Understanding binary expansion is crucial as it provides insights into how and why \( g \) achieves its mapping from intervals in \( \mathbb{R} \) to \([0, 1]\), and how this contributes to the definition of open, non-continuous functions.

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Most popular questions from this chapter

Let \(X\) be a set and \(\mathscr{S}(X)\) the family of all subsets of \(X\). Show that the cardinality of the set \(\mathscr{S}(X)\) is strictly larger than that of \(X\), cf. E 1.1.6. Hint: If \(f: X \rightarrow \mathscr{S}(X)\) is a bijective function, set $$ A=\\{x \in X \mid x \notin f(x)\\} $$ and take \(y=f^{-1}(A)\). Either possibility \(y \in A\) or \(y \notin A\) will lead to a contradiction.

Let \((X, \tau)\) be a topological space and denote by \(C(X)\) the set of continuous functions from \(X\) to \(\mathbb{R} .\) Show that the following combinations of elements \(f\) and \(g\) in \(C(X)\) again produce elements in \(C(X)\) : \(\alpha f[\) if \(\alpha \in \mathbb{R}] ;|f| ; 1 / f[\) if \(0 \notin f(X)] ; f+g ; f g ; f \vee g ; f \wedge g .\)

Let \(\alpha\) be a positive, irrational number and show that the relation in \(\mathbb{Z} \times \mathbb{Z}\) given by $$ \left(x_{1}, x_{2}\right) \leq\left(y_{1}, y_{2}\right) \quad \text { if } \quad \alpha\left(y_{1}-x_{1}\right) \leq y_{2}-x_{2} $$ is a total order. Sketch the set $$ \left\\{\left(x_{1}, x_{2}\right) \in \mathbb{Z} \times \mathbb{Z} \quad \mid \quad\left(x_{1}, x_{2}\right) \geq(0,0)\right\\} . $$

(Connected spaces.) A topological space \((X, \tau)\) is connected if it cannot be decomposed as a union of two nonempty disjoint open sets. A subset of \(X\) is clopen if it is both open and closed. Show that \(X\) is connected iff \(\emptyset\) and \(X\) are the only clopen subsets. Let \(f: X \rightarrow Y\) be a surjective continuous map between topological spaces. Show that \(Y\) is connected if \(X\) is.

(The fundamental group.) Let \((X, \tau)\) be a nonempty arcwise connected (cf. E 1.4.14) topological space, and choose a base point \(x_{0}\) in \(X\). A loop in \(X\) is a continuous function (curve) \(f:[0,1] \rightarrow X\) such that \(f(0)=f(1)=x_{0}\). On the space \(L(X)\) of loops we define a composition \(f g\) (product) by $$ f g(t)=g(2 t), \quad 0 \leq t \leq \frac{1}{2} ; \quad f g(t)=f(2 t-1), \quad \frac{1}{2} \leq t \leq 1, $$ for \(f\) and \(g\) in \(L(X)\). We define homotopy of loops, written \(f \sim g\), if there is a continuous function \(F:[0,1] \times[0,1] \rightarrow X\) such that \(F(s, 0)=F(s, 1)=x_{0}\) for every \(s\) and \(F(0, t)=f(t), F(1, t)=g(t)\) for every \(t\). Show that the set \(\pi(X)\) of equivalence classes (under homotopy) of loops is a group under the product \(\pi(f) \pi(g)=\pi(f g)\), where \(\pi: L(X) \rightarrow \pi(X)\) is the quotient map. Hint: If \(F\) is a homotopy between the loops \(f_{1}\) and \(f_{2}\), and \(G\) is a homotopy between the loops \(g_{1}\) and \(g_{2}\), set $$ \begin{array}{cll} H(s, t)=F(s, 2 t) & \text { for } & 0 \leq s \leq 1, & 0 \leq t \leq \frac{1}{2} \\ (s, t)=G(s, 2 t-1) & \text { for } & 0 \leq s \leq 1, & \frac{1}{2} \leq t \leq 1 \end{array} $$ and check that \(H\) is a homotopy between \(f_{1} g_{1}\) and \(f_{2} g_{2}\). The product in \(\pi(X)\) is therefore well-defined. If \(f \in L(X)\), define \(f^{-1}\) in \(L(X)\) by \(f^{-1}(t)=f(1-t)\) and check that \(f^{-1} f \sim e\), where \(e(t)=x_{0}\) for all \(t\). The relevant homotopy is $$ \begin{array}{lll} F(s, t)=f(2 s t) & \text { for } & 0 \leq s \leq 1, & 0 \leq t \leq \frac{1}{2} \\ F(s, t)=f(2 s(1-t)) & \text { for } & 0 \leq s \leq 1, & \frac{1}{2} \leq t \leq 1 \end{array} $$ Similarly \(f f^{-1} \sim e, f e \sim e f \sim f\), so that \(\pi(e)\) is the identity in \(\pi(X)\). Given \(f, g, h\) in \(L(X)\) we have $$ \begin{aligned} &f(g h)(t)= \begin{cases}h(4 t) & \text { for } 0 \leq t \leq \frac{1}{4} \\ g(4 t-1) & \text { for } \frac{1}{4} \leq t \leq \frac{1}{2} \\ f(2 t-1) & \text { for } & \frac{1}{2} \leq t \leq 1\end{cases} \\ &(f g) h(t)= \begin{cases}h(2 t) & \text { for } 0 \leq t \leq \frac{1}{2} \\ g(4 t-2) & \text { for } \frac{1}{2} \leq t \leq \frac{3}{4} \\ f(4 t-3) & \text { for } \frac{3}{4} \leq t \leq 1\end{cases} \end{aligned} $$ To show that \(f(g h) \sim(f g) h\), use the homotopy $$ F(s, t)= \begin{cases}h\left(4 t(1+s)^{-1}\right) & \text { for } 4 t-1 \leq s \\\ g(4 t-s-1) & \text { for } 4 t-2 \leq s \leq 4 t-1 \\ f\left((4 t-s-2)(2-s)^{-1}\right) & \text { for } s \leq 4 t-2\end{cases} $$

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