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91Ó°ÊÓ

Show directly, using a diagonal argument, that the half-open interval of reals, \(\\{x \in \mathbb{R}: 0

Short Answer

Expert verified
By contradiction and using the diagonal argument, it was demonstrated that the half-open interval (0,1) is not denumerable. This is because it is always possible to construct a new real number not already included in a supposed enumeration of this interval, which establishes a contradiction.

Step by step solution

01

Illustrate enumeration of the reals

Assume for contradiction that the half-open interval (0,1) is denumerable. This means there exists a listing of all of the real numbers in the interval (0,1)
02

Apply diagonal argument

Construct a new real number b in (0,1) as follows: for the nth decimal place of b, choose a number that is different from the nth decimal place of the nth number in the listing. This ensures b is clearly in (0,1), and is different from every number in our supposed enumeration.
03

Establish contradiction

Since b is different from every number in the supposed enumeration, it implies that the enumeration was not complete. This contradicts our assumption that such an enumeration exists. Therefore, the interval (0,1) is not denumerable.

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Key Concepts

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

Diagonal Argument
The diagonal argument is a powerful technique most famously used by mathematician Georg Cantor to show that certain sets are not countable, or 'denumerable'. The essence of the argument is to assume that you have listed all elements of a set and then, by constructing a new element that differs from each listed element in at least one aspect, you show that your list cannot be complete.

To envision this, imagine we have a list of real numbers between 0 and 1, written in their decimal expansion form. By changing the diagonal elements of this list—to be precise, changing the nth digit of the nth number—we can always create a number not on the list. The key here is the construction of a new number by ensuring that it differs from each of the listed numbers at some decimal place.

The diagonal argument is elegant because it doesn't rely on the size or specifics of the list, which makes it a universal refutation of any attempt to enumerate real numbers in an interval or set. It applies not just to the interval (0,1), but to any infinite set of real numbers.
Enumeration of Real Numbers
Enumeration, in the context of set theory, refers to a process by which elements of a set are listed or 'counted'. For a set to be denumerable, or countable, it must be possible to establish a one-to-one correspondence between the set's elements and the natural numbers (1, 2, 3, ...).

In practice, this means that if you can pair each element of your set with a distinct natural number without missing any element or natural number, your set is denumerable. For example, the set of all natural numbers is clearly denumerable since each element is it’s own pair. However, when applying such enumeration to the set of real numbers within an interval, as in our exercise, complications arise. Unlike natural numbers, real numbers cannot be listed in such a neat, simple pairing with natural numbers due to their uncountably infinite nature.

Apart from natural numbers, rational numbers too are denumerable, as there is a systematic way to list all fractions. However, the real numbers have 'gaps' between them, known as irrational numbers, which can't be expressed as fractions. Together, they form a continuum of numbers, making their enumeration impossible, which is a concept that often challenges our intuitive understanding of infinity.
Cantor's Theorem
Cantor's theorem is a fundamental result established by Georg Cantor in 1891, which states that for any given set, the set of all its subsets (its power set) has a strictly greater number of elements than the set itself. This applies even if the original set is infinite.

One of the most startling implications of Cantor's theorem is related to the real numbers. It implies that even though the set of natural numbers is infinite, the set of real numbers is a 'larger' kind of infinity, known as uncountable infinity. In the context of our exercise, Cantor's theorem indirectly supports the notion that real numbers cannot be enumerated, because it points to the existence of different sizes of infinity.

Cantor's diagonal argument, as applied in the exercise, actually originated from the proof of this theorem. It beautifully illustrates the concept that just being infinite does not imply that a set can be matched up with the natural numbers. These ideas laid the foundation for the entire field of set theory and our modern understanding of infinity.

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Most popular questions from this chapter

Determine the cardinality of each of the following sets: $$ \begin{aligned} &x_{1}=\left\\{f \in \mathbb{N}^{\mathbb{N}}:(\forall n \in \mathbb{N})(\forall p \in \mathbb{N})(n

Assume that the universe satisfies the axiom of choice. Let \(a\) and \(b\) be two infinite sets whose cardinalities are \(\lambda\) and \(\mu\), respectively. Assume \(\lambda>\mu\). Let \(g\) be an injective mapping from \(b\) into \(a\). The reader should determine the cardinalities Assume that the universe satisfies the axiom of choice. Let \(a\) and \(b\) be two infinite sets whose cardinalities are \(\lambda\) and \(\mu\), respectively. Assume \(\lambda>\mu\). Let \(g\) be an injective mapping from \(b\) into \(a\). The reader should determine the cardinalities of each of the following sets: $$ \begin{aligned} &y_{1}=\left\\{f \in b^{a}: \operatorname{card}(\vec{f}(a))=1\right\\} \\ &y_{2}=\left\\{f \in b^{a}:(\mathrm{V} x \in \wp(a))(\mathrm{card}(\bar{f}(x)) \leq 1)\right\\} \\ &y_{3}=\left\\{f \in b^{a}: \operatorname{card}\left(\bar{f}^{-1}(b)\right)=\lambda\right\\} \\ &y_{4}=\left\\{f \in b^{a}: \operatorname{card}(\bar{f}(a))=2\right\\} \\ &y_{5}=a-\bar{g}(b) ; \\ &y_{6}=\left\\{f \in b^{a}:(\forall y \in b)(f(g(y))=y)\right\\} \\ &y_{7}=\left\\{f \in b^{a}: \operatorname{card}(\bar{f}(a))=\mu\right\\} \end{aligned} $$ [Recall that if \(f \in b^{a}\), then \(\bar{f}\) and \(\bar{f}^{-1}\) respectively denote the direct image mapping induced by \(f\) from \(\wp(a)\) into \(\rho(b)\) and the inverse image mapping from \(\wp(b)\) into \(\wp(a)\).]

Let \(\Phi\) be a definable strictly increasing function from the class \(O n\) of ordinals into itself. We say that \(\Phi\) is continuous at a limit ordinal \(\alpha\) if \(\Phi(\alpha)=\) \(\sup _{\beta \in \alpha} \Phi(\beta) .\) Such a function \(\Phi\) is called continuous if it is continuous at all limit ordinals. An ordinal \(\alpha\) such that \(\Phi(\alpha)=\alpha\) is called a fixed point of \(\Phi\). (a) Show that every strictly increasing function \(\Phi\) from \(O n\) into \(O n\) has the following property: for every ordinal \(\alpha, \quad \Phi(\alpha) \geq \alpha\). (b) Show that if \(\Phi\) is a strictly increasing function from \(O n\) into \(O n\) that is continuous at every limit ordinal whose cofinality is \(\omega\), then for every ordinal \(\alpha, \Phi\) has a fixed point that is greater than \(\alpha\). (c) Show that if \(\Phi\) and \(\Psi\) are two strictly increasing functions from \(O n\) into \(O n\) that are continuous at every limit ordinal whose cofinality is \(\omega\), then for every ordinal \(\alpha, \Phi\) and \(\Psi\) have a common fixed point greater than \(\alpha\). (d) Suppose the universe satisfies the axiom of choice. Show that for every ordinal \(\alpha\), there exists an ordinal \(\beta>\alpha\) such that card \(\left(V_{\beta}\right)=\aleph_{\beta}=\beta\).

Suppose that the universe satisfies the axiom of choice. Let \(\mu\) be an infinite cardinal. By induction on the integers, we define a sequence of cardinal \(\left(\lambda_{n}\right)_{n \in \omega}\) by setting \- \(\lambda_{0}=\mu\); \- for every \(n \in \omega, \lambda_{n+1}=2^{\lambda_{n}} .\) Set \(\lambda=\sum_{n \in \omega} \lambda_{n} .\) (a) Show that \(\lambda^{\mu}=\mu^{2}=\lambda^{\lambda}=2^{\lambda}\). (b) Show that, for every cardinal \(\gamma\), $$ \begin{aligned} &\text { if } \aleph_{0} \leq \gamma \leq \lambda, \quad \text { then } \lambda^{\gamma_{0}}=\lambda^{\gamma}=\lambda^{\lambda} \\ &\text { if } \gamma \geq \lambda, \quad \text { then } \lambda^{\gamma}=2^{\gamma} \end{aligned} $$ (c) Show that there exist cardinals \(\alpha, \beta, \gamma\), and \(\delta\) such that $$ \alpha<\beta, \quad \gamma<\delta, \quad \text { and } \quad \alpha^{\gamma}=\beta^{\delta} . $$

Let \(x\) be a set and \(\Gamma(x)\) be the class of ordinals that are subpotent to \(x\). Show that \(\Gamma(x)\) is an ordinal, that it is the least ordinal not subpotent to \(x\), and that it is a cardinal. We call this Hartog's cardinality of \(x\). Characterize \(\Gamma(x)\) assuming that \(U\) satisfies the axiom of choice.

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