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Chapter 2: Problem 10

Give an example of two uncountable sets \(A\) and \(B\) such that \(A-B\) is a) finite. b) countably infinite. c) uncountable.

Short Answer

Expert verified
a) \( \{0, 1, 2, ..., n\} \) is finite. b) \( \mathbb{Q} \) is countably infinite. c) \( [0, 0.5) \) is uncountable.

Step by step solution

01

Understand uncountable sets

An uncountable set is a set that has more elements than the set of natural numbers, meaning its cardinality is greater than \( \aleph_0 \). Examples of uncountable sets are the set of real numbers \( \mathbb{R} \) and the set of all points in a plane.
02

Example with A - B is finite

Let \( A = \mathbb{R} \) (the set of all real numbers) and \( B = \mathbb{R} - \{0, 1, 2, 3, ..., n\} \) where \( n \) is any natural number. Then \( A - B = \{0, 1, 2, 3, ..., n\} \), a finite set.
03

Example with A - B is countably infinite

Let \( A = \mathbb{R} \) (the set of all real numbers) and \( B = \mathbb{R} - \mathbb{Q} \) where \( \mathbb{Q} \) is the set of all rational numbers. Then \( A - B = \mathbb{Q} \), a countably infinite set since the rationals are countably infinite.
04

Example with A - B is uncountable

Let \( A = [0, 1] \) (the set of real numbers between 0 and 1 inclusive) and \( B = [0.5, 1] \) (the set of real numbers between 0.5 and 1 inclusive). Then \( A - B = [0, 0.5) \), an uncountable set since any continuous subset of the real numbers is uncountable.

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Key Concepts

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

Finite Sets
A finite set is a set that has a limited number of elements. This means we can count all the elements and the counting will end at some point. For example, the set \(\)\{1, 2, 3, 4, 5\}\(\) is a finite set since it has only 5 elements. Finite sets are quite intuitive because we often work with numbers or objects that can be easily counted. When we talk about the difference between two sets being finite, like in our example where \( A - B = \{0, 1, 2, 3, ..., n\} \), it means that there are only a finite number of elements in that difference. This makes it easy to understand and visualize.
Countably Infinite Sets
A countably infinite set is a set that has the same cardinality (size) as the set of natural numbers \(\mathbb{N}\). This means that you can list or enumerate all the elements of the set, even though the list will never end. An example of a countably infinite set is the set of rational numbers \(\mathbb{Q}\). Although it may seem surprising, rational numbers can be arranged in such a way that we can count them one by one.

In our context, if set \( B \) is the set of real numbers excluding the rational numbers, and set \( A \) is the set of all real numbers, then \( A - B = \mathbb{Q} \), which is countably infinite. Even though there are an infinite number of rational numbers, their infinity is 'countable'.
Uncountable Infinite Sets
Uncountable infinite sets are sets whose size is so large that they cannot be counted or listed out in a sequence that would match the natural numbers. The most famous example is the set of all real numbers \(\mathbb{R}\). This set cannot be put into one-to-one correspondence with \(\mathbb{N}\) because there are simply 'too many' real numbers.

When we talk about an uncountable set in the context of our example, we look at the interval \([0, 0.5)\). Although this interval seems smaller, it still contains an uncountable number of real numbers because any continuous subset of \(\mathbb{R}\) is uncountable. This is why \([0, 0.5)\) is uncountable, even though it is only a part of \(\mathbb{R}\).

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Most popular questions from this chapter

Show that \((0,1)\) and \(R\) have the same cardinality by a) showing that \(f(x)=\frac{2 x-1}{2 x(1-x)}\) is a bijection from \((0,1)\) to \(\mathbf{R} .\) b) using the Schr枚der-Bernstein theorem.

Show that \(\left\lfloor x+\frac{1}{2}\right\rfloor\) is the closest integer to the number \(x\) , except when \(x\) is midway between two integers, when it is the larger of these two integers.

In this exercise, we prove the Schr枚der-Bernstein theorem. Suppose that \(A\) and \(B\) are sets where \(|A| \leq|B|\) and \(|B| \leq|A| .\) This means that there are injections \(f : A \rightarrow B\) and \(g : B \rightarrow A\) . To prove the theorem, we must show that there is a bijection \(h : A \rightarrow B,\) implying that \(|A|=|B|\) To build \(h,\) we construct the chain of an element \(a \in A .\) This chain contains the elements \(a, f(a), g(f(a)), f(g(f(a))), g(f(g(f(a)))), \ldots\) It also may contain more elements that precede \(a,\) extending the chain backwards. So, if there is a \(b \in B\) with \(g(b)=a\) , then \(b\) will be the term of the chain just before \(a .\) Because \(g\) may not be a surjection, there may not be any such \(b,\) so that \(a\) is the first element of the chain. If such a \(b\) exists, because \(g\) is an injection, it is the unique element of \(B\) mapped by \(g\) to \(a ;\) we denote it by \(g^{-1}(a) .\) (Note that this defines \(g^{-1}\) as a partial function from \(B\) to \(A .\) . We extend the chain backwards as long as possible in the same way, adding \(f^{-1}\left(g^{-1}(a)\right), g^{-1}\left(f^{-1}\left(g^{-1}\left(g^{-1}(a)\right)\right), \ldots \text { To construct }\right.\) the proof, complete these five parts. a) Show that every element in \(A\) or in \(B\) belongs to exactly one chain. b) Show that there are four types of chains: chains thaform a loop, that is, carrying them forward from every element in the chain will eventually return to this element (type 1\()\) , chains that go backwards without stopping (type 2\()\) , chains that go backwards and end nin the set \(A\) (type \(3 ),\) and chains that go backwards and end in the set \(B\) (type 4\()\) . c) We now define a function \(h : A \rightarrow B .\) We set \(h(a)=\) \(f(a)\) when \(a\) belongs to a chain of type \(1,2,\) or \(3 .\) Show that we can define \(h(a)\) when \(a\) is in a chain of type \(4,\) by taking \(h(a)=g^{-1}(a)\) . In parts \((\mathrm{d})\) and (e), we show that this function is a bijection from \(A\) to \(B,\) proving the theorem. d) Show that \(h\) is one-to-one. (You can consider chains of types \(1,2,\) and 3 together, but chains of type 4 separately.) e) Show that \(h\) is onto. (You need to consider chains of types \(1,2,\) and 3 together, but chains of type 4 separately.)

Find \(\bigcup_{i=1}^{\infty} A_{i}\) and \(\bigcap_{i=1}^{\infty} A_{i}\) if for every positive integer \(i\) a) \(A_{i}=\\{-i,-i+1, \ldots,-1,0,1, \ldots, i-1, i\\}\) b) \(A_{i}=\\{-i, i\\}\) c) \(A_{i}^{\prime}=[-i, i],\) that is, the set of real numbers \(x\) with \(-i \leq x \leq i .\) d) \(A_{i}=[i, \infty),\) that is, the set of real numbers \(x\) with \(x \geq i .\)

Find a recurrence relation for the balance \(B(k)\) owed at the end of \(k\) months on a loan of \(\$ 5000\) at a rate of 7\(\%\) if a payment of \(\$ 100\) is made each month. [Hint: Express \(B(k)\) in terms of \(B(k-1) ;\) the monthly interest is \((0.07 / 12) B(k-1) . ]\)
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