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Chapter 1: Problem 9

Unstetigkeitsstellen. Geben Sie Beispiele von Funktionen \(f: \mathbb{R} \rightarrow \mathbb{R}\) an, die an folgenden Stellen stetig: sind (a) nirgends, (b) auf der Menge aller irrationalen Zahlen \(\mathbb{R} \backslash \mathbb{Q}\) und sonst nicht, (c) an der Stelle 0 und sonst nicht. (d) Gibt es eine Funktion, die auf \(\mathbb{Q}\) stetig ist und sonst nicht?

Short Answer

Expert verified
(a) Dirichlet function, (b) Identity on irrationals, (c) Zero map at zero, (d) No such function.

Step by step solution

01

Understanding Discontinuities

Discontinuities in a function occur where the function is not smooth or jumps and are key points where it is not continuous. Continuity requires a function's limit to approach its function value at a point.
02

Example for Part (a): Nowhere Continuous Function

A classic example of a nowhere continuous function is the Dirichlet function defined as: \[ f(x) = \begin{cases} 1, & \text{if } x \in \mathbb{Q} \ 0, & \text{if } x \in \mathbb{R} \setminus \mathbb{Q} \end{cases} \]This function is not continuous at any point because at every point there's a mixture of rational and irrational numbers, leading to an infinite oscillation.
03

Example for Part (b): Continuous on Irrational Numbers

Consider the function defined as: \[ g(x) = \begin{cases} x, & \text{if } x \in \mathbb{R} \setminus \mathbb{Q} \ 0, & \text{if } x \in \mathbb{Q} \end{cases} \]This function is continuous at all irrational numbers because for irrational points, it follows the identity map. It is discontinuous at all rational numbers since the values jump between 0 and the actual value of the irrational point.
04

Example for Part (c): Continuous Only at Zero

Consider the function defined as:\[ h(x) = \begin{cases} 0, & \text{if } x = 0 \ 1, & \text{if } x eq 0 \end{cases} \]This function is only continuous at zero because the limit and the function value at zero are both zero. However, for any other point, the function value is constantly 1, causing discontinuity because the limit at any non-zero point would not be 1.
05

Evaluating Part (d): Continuity on Rational Numbers Only

There is no function that is continuous on the set of rational numbers (\(\mathbb{Q}\)) and nowhere else. Continuity, by its nature, cannot isolate itself to just rationals; the topological structure of \(\mathbb{R}\) involving dense interleaving of \(\mathbb{Q}\) and \(\mathbb{R} \setminus \mathbb{Q}\) prevents this.

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Key Concepts

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

Discontinuous Functions
Discontinuous functions are those in which the graph does not form an unbroken curve. Discontinuities occur where the function jumps, has gaps, or sharp turns. Continuity requires both that a function's limit exists at a point and matches the function's value. When these conditions are not met, the function is discontinuous. Key aspects of discontinuities include:
  • Removable Discontinuity: A gap in the graph at a point that can be "filled" by defining or redefining the function's value.
  • Jump Discontinuity: Occurs when the function value jumps abruptly from one point to another, causing a big jump or leap in the graph.
  • Infinite Discontinuity: Triggered when the function values increase or decrease without bound near a point.
When studying functions, understanding these types is crucial, especially in advanced mathematics where they often arise.
Dirichlet Function
The Dirichlet function is a classic example used to illustrate discontinuity. This peculiar function is defined as follows:\[ f(x) = \begin{cases} 1, & \text{if } x \in \mathbb{Q} \ 0, & \text{if } x \in \mathbb{R} \setminus \mathbb{Q} \end{cases} \]In essence, it assigns values based on whether a number is rational or irrational:
  • If a number is rational (i.e., can be expressed as a fraction \(\frac{a}{b}\) where \(a\) and \(b\) are integers and \(b eq 0\)), the function outputs 1.
  • If a number is irrational (cannot be expressed as a simple fraction), the output is 0.
The Dirichlet function is nowhere continuous. At every real number, the nearby rationals and irrationals cause the function to oscillate wildly without settling, illustrating a complete lack of smoothness and continuity.
Rational and Irrational Numbers
Rational and irrational numbers are fundamental in understanding discontinuities. Rational numbers (\(\mathbb{Q}\)) include any number that can be expressed as a fraction, such as 1/2 or 3. These numbers have either terminating or repeating decimal expansions. In contrast, irrational numbers maintain non-repeating, non-terminating decimal expansions. Pi (\(\pi\)) and the square root of 2 are examples of irrational numbers.
A key aspect of these numbers is their density in the number line:
  • Rational numbers are densely packed between any two real numbers.
  • Irrational numbers fill gaps between rationals, ensuring every segment on the number line is filled.
Thus, when understanding functions like the Dirichlet function, the interruption and oscillation between rational and irrational numbers cause significant discontinuities.
Topological Structure of Real Numbers
The topological structure of real numbers describes their arrangement and properties on the real number line, focusing on closeness and continuity. This structure is vital in analyzing how functions behave when rational and irrational numbers interplay.
Key characteristics include:
  • Density: Both rational and irrational numbers are densely intertwined; no matter how small the distance, they are closely packed.
  • Completeness: Every limit point of a sequence of real numbers actually lies on the real number line, adhering to no gaps.
  • Interleaving: Rational and irrational numbers alternatingly fill every section on the number line, contributing to the Dirichlet function's discontinuity.
This structure makes any attempt to have a function continuous solely on one type of number (such as rationals only) impossible, due to their dense intermingling. Understanding these characteristics helps in grasping why specific functions exhibit certain discontinuities.

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Most popular questions from this chapter

R盲nder. Bestimmen Sie die R盲nder der folgenden Teilmengen des \(\mathbb{R}^{2}\) : (a) \(M=\\{(x, y) \mid x>0\) und \(y \neq 0\\}\) (b) \(M=\left\\{(x, y) \mid 0 \leqslant x^{2}-y^{2}<1\right\\}\)

Verklebung stetiger Funktionen. Sei \(X=A \cup B\), und \(A, B\) seien abgeschlossen. Sei \(f: A \rightarrow Y\) und \(g: B \rightarrow Y\) stetig und \(f(x)=g(x)\) f眉r alle \(x\) im Durchschnitt \(A \cap B\). Zeigen Sie, dass dann die Abbildung \(h: X \rightarrow Y\) mit \(h(x)=f(x)\) f眉r \(x \in A\) und \(h(x)=g(x)\) f眉r \(x \in B\) wohldefiniert und stetig ist.

Mengenlehre. Sei \(f: X \rightarrow X^{\prime}\) eine Abbildung zwischen den Mengen \(X\) und \(X^{\prime} .\) Untersuchen Sie, ob die folgenden Aussagen immer gelten; und wenn nicht, so geben Sie m枚glichst einfache Bedingungen f眉r \(f\) an, die notwendig und hinreichend sind. (a) \(f(A \cup B)=f(A) \cup f(B)\) (a') \(\quad f^{-1}\left(A^{\prime} \cup B^{\prime}\right)=f^{-1}\left(A^{\prime}\right) \cup f^{-1}\left(B^{\prime}\right)\) (b) \(f(A \cap B)=f(A) \cap f(B)\) (b') \(\quad f^{-1}\left(A^{\prime} \cap B^{\prime}\right)=f^{-1}\left(A^{\prime}\right) \cap f^{-1}\left(B^{\prime}\right)\) (c) \(f(X \backslash A)=X^{\prime} \backslash f(A)\) (c') \(\quad f^{-1}\left(X^{\prime} \backslash A^{\prime}\right)=X \backslash f^{-1}\left(A^{\prime}\right)\) (d) \(\quad f^{-1}(f(A)) \subseteq A\) (d') \(\quad f\left(f^{-1}\left(A^{\prime}\right)\right) \subseteq A^{\prime}\) (e) \(f^{-1}(f(A)) \supseteq A\) (e') \(\quad f\left(f^{-1}\left(A^{\prime}\right)\right) \supseteq A^{\prime}\) jeweils f眉r alle \(A, B \subseteq X\) bzw. alle \(A^{\prime}, B^{\prime} \subseteq X^{\prime}\)

Das kleine Einspluseins. Gibt es eine Abbildung \(f: \mathbb{R} \rightarrow \mathbb{R}\), welche den, Bedingungen \(f(x+y)=f(x)+f(y)\) und \(f(x \cdot y)=f(x) \cdot f(y)\) f眉r alle reellen Zahlen \(x\) und \(y\) gen眉gt, und die nicht stetig ist?

Dreimal ist keinmal. Sei \(\mathcal{T}\) eine Topologie auf der Menge \(X=\\{1,2,3\\}\). Dann ist die Hom枚omorphismengruppe von \((X, \mathcal{J})\) eine Untergruppe der symmetrischen Gruppe mit \(3 !=6\) Elementen. Man zeige: Es gibt keine Topologie \(\mathcal{T}\) deren Hom?omorphismengruppe genau drei Elemente hat. Gibt es 眉berhaupt einen topologischen Raum, dessen Hom枚omorphismengruppe genau drei Elemente hat?
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