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Chapter 2: Problem 6

Es seien \(F: \mathbb{R} \rightarrow \mathbb{R}\) die Cantorsche Funktion und \(x, y \in C, x

Short Answer

Expert verified
This is a proof showing that given \(F: \mathbb{R} \rightarrow \mathbb{R}\) is the Cantor function and \(x, y \in C, x<y, F(x)=F(y)\), it can be concluded that there exist a \(n \geq 0\) and a \(j \in\left\{1, \ldots, 2^{n}\right\}\) such that \(x, y\left[=I_{n, j}\right.\). The proof utilizes the properties of the Cantor function, the Cantor set, and their interrelations.

Step by step solution

01

- Understanding The Cantor Function

It is important to note some properties of the Cantor function. The Cantor function \(F: \mathbb{R} \rightarrow \mathbb{R}\) is a special kind of function that is constant on the complement of the Cantor set. It is also monotonically increasing, and right-continuous everywhere.
02

- Assumptions

From the problem, we have \(x, y \in C\) and \(x<y\), but \(F(x)=F(y)\). We must show that there exist \(n \geq 0\) and \(j \in\left\{1, \ldots, 2^{n}\right\}\) such that \(x, y\left[=I_{n, j}\right.\). Here \(I_{n, j}\) denotes the j-th subinterval of the n-th level in the construction of Cantor set.
03

- Proof

Since \(F\) is constant on the complement of the Cantor's ternary set, and \(x, y \in C\), then the interval \([x, y]\) must be contained within the Cantor's ternary set. Moreover, since \(F(x) = F(y)\), it must mean that \(x, y\) are contained in the same sub-interval of the n-th level, denoted by \(I_{n,j}\), since \(F\)'s value increases strictly monotonically on the intervals \(I_{n,j}\).

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Key Concepts

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

Cantor Set
The Cantor set is a fascinating concept in mathematics, known for its unique construction and properties. The construction starts with the closed interval [0, 1]. The middle third of this interval is removed, leaving two subintervals: [0, 1/3] and [2/3, 1]. This process is repeated for each remaining interval, removing the middle third segment at every step. This iterative process ultimately results in the Cantor set.
  • It consists of all points that remain after infinitely many iterations of removing the middle thirds.
  • The Cantor set is uncountably infinite, but it has a total length of zero.
The Cantor set is a perfect set, meaning it's closed and contains no isolated points. It serves as a classic example of a set that is nowhere dense in real numbers.
Monotonicity
Monotonicity is a key attribute of the Cantor function. Essentially, a function is monotonic if it is either entirely non-increasing or non-decreasing. The Cantor function is monotonically increasing. This means:
  • As you move from left to right across its domain, the function value never decreases.
  • It can stay constant or increase but never drop back down.
For the Cantor function, this monotonicity is important because it helps to verify that if two points in the Cantor set have the same function value, then they must lie within the same subinterval, leading to the conclusion that there is no fluctuation between those points.
Right-continuity
The Cantor function exhibits right-continuity, an intriguing property of continuous functions. Right-continuity means the function behaves predictably at the limit from the right side, even if there might be jumps or discontinuities normally.
  • This means that for any point in the domain, approaching it from the right will yield the function's value at that point.
  • It ensures that even with the strange distribution of values in the Cantor function, no unexpected fluctuations occur from the right-hand side.
Thus, right-continuity complements monotonicity, adding predictability and further establishing the Cantor function as a peculiar yet structured transformation.
Subintervals in Cantor Set
The Cantor function's link to subintervals is crucial. During the construction of the Cantor set, subintervals like \(I_{n,j}\) are formed on each level, reducing and fragmenting continually.
  • While building the Cantor set, each level splits the previous subintervals into smaller segments.
  • These subintervals hold a significant role in understanding points with specific properties in the Cantor set, like equaling function values.
In context, if two points in the Cantor set, \(x\) and \(y\), have the same Cantor function value, they reside in the same \(I_{n,j}\). This reflects on how the function remains constant across those specific subintervals, reinforcing their structural importance in the set.

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Most popular questions from this chapter

Es seien \(F: \mathbb{R}^{p} \rightarrow \mathbb{R}\) wachsend und stetig und \(\eta_{F}\) das 盲u脽ere \(\mathrm{MaB}\) zu \(\mu_{F}\). a) Jede Hyperebene \(H=\left\\{x \in \mathbb{R}^{p}: x_{k}=\alpha\right\\}\) ist eine \(\eta_{F}\)-Nullmenge. b) Konstruieren Sie eine wachsende stetige Funktion \(F: \mathbb{R}^{2} \rightarrow \mathbb{R}\), zu welcher eine (zu keiner Koordinatenachse parallele) Gerade \(G\) existiert mit \(\eta_{F}(G)>0\).

Jede offene Teilmenge von \(\mathbb{R}\) ist disjunkte Vereinigung abz盲hlbar vieler offener Intervalle.

Konstruieren Sie eine Funktion \(f:[0,1] \rightarrow \mathbb{R}\), so da脽 die Menge \(D\) der Unstetigkeitsstellen von \(f\) das Lebesguesche MaB 0 hat und so da脽 f眉r jedes Teilintervall \(J \subset[0,1]\) mit \(\dot{J} \neq \emptyset\) der Durchschnitt \(J \cap D\) 眉berabz盲hlbar ist. (Hinweis: Es sei \(C_{1} \subset[0,1]\) das Cantorsche Diskontinuum. F眉r jedes der offenen Intervalle von \([0,1] \backslash C_{1}\) bilde man das entsprechende Cantorsche Diskontinuum; es sei \(C_{2}\) die Vereinigungsmenge dieser Diskontinua. Die induktive Fortsetzung dieser Konstruktion liefert eine Folge \(\left(C_{n}\right)_{n \geq 1}\) disjunkter Mengen. Es seien \(D:=\bigcup_{n=1}^{\infty} C_{n}\) und \(f(x):=2^{-n}\) f眉r \(x \in C_{n}(n \in \mathrm{N}), f(x):=0\) f眉r \(\left.x \in[0,1] \backslash D .\right)\)

Es seien \(\mu: \mathfrak{H} \rightarrow \overline{\mathbb{R}}\) und \(\nu: \mathfrak{K} \rightarrow \overline{\mathbb{R}}\) zwei Inhalte auf den Halbringen \(\mathfrak{H}, \mathfrak{K}\) 眉ber \(X\) bzw. \(Y\), und \(\rho: \mathfrak{H} * \mathfrak{K} \rightarrow \overline{\mathbb{R}}\) (s. Lemma I.5.3) sei definiert durch \(\rho(A \times B):=\mu(A) \cdot \nu(B) \quad(A \in\) \mathfrak{H} , ~ \(B \in \mathfrak{K}\) ); dabei wird das Produkt auf der rechten Seite definiert durch \(a \cdot \infty:=\infty\) f眉r \(0

Es sei \(X \neq \emptyset\) a) Es sei \(\mu: \mathfrak{P}(X) \rightarrow \mathbb{R}\) ein Inhalt mit \(\mu(X)=1\) und \(\mu(A) \in\\{0,1\\}\) f眉r alle \(A \subset X ; \mathfrak{U}:=\) \(\\{A \subset X: \mu(A)=1\\} .\) Zeigen Sie: (i) \(\emptyset \notin \mathfrak{U}\) (ii) \(A \in \mathscr{U}, A \subset B \subset X \Longrightarrow B \in \mathfrak{U} ;\) (iii) \(A, B \in \mathfrak{U} \Longrightarrow A \cap B \in \mathfrak{U}\) (iv) \(A \subset X \Longrightarrow A \in \mathfrak{U}\) oder \(A^{c} \in \mathfrak{U}\). (Eine Teilmenge \(\mathfrak{U} \neq \emptyset\) von \(\mathfrak{P}(X)\) mit den Eigenschaften (i), (ii), (iii) hei脽t ein Filter auf \(X\); gilt zus盲tzlich (iv), so hei脽t \(\mathfrak{U}\) ein Ultrafilter.) b) Ist \(\mathfrak{U}\) ein Ultrafilter auf \(X\), so ist \(\mu: \mathfrak{P}(X) \rightarrow \mathbb{R}, \mu(A):=1\) f眉r \(A \in \mathfrak{U}, \mu(A):=0\) f眉r \(A^{c} \in \mathfrak{U}\), ein Inhalt. \(\mu\) ist genau dann ein Ma脽, wenn f眉r jede Folge \(\left(A_{n}\right)_{n \geq 1}\) von Mengen aus U gilt \(\bigcap_{n=1}^{\infty} A_{n} \neq \emptyset\) c) Auf jeder unendlichen Menge gibt es einen nicht-trivialen Inhalt, der auf ganz \(\mathfrak{P}(X)\) definiert ist, nur die Werte 0 und 1 annimmt und auf allen endlichen Mengen verschwindet. (Hinweis: Zu jedem Filter \(\mathfrak{F}\) auf \(X\) gibt es einen Ultrafilter \(\mathfrak{U} \supset \mathfrak{F}\), s. z.B. SCHUBERT \([1], \mathrm{S}\). 49.)
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