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Chapter 1: Problem 48

Prove that every subset of a countable set is countable.

Short Answer

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In conclusion, we have three cases to consider: 1. If the countable set is finite, a bijection exists between the subset and the first few natural numbers up to the size of the subset, making the subset countable. 2. If the countable set is infinite and the subset is finite, a bijection exists between the finite subset and the first few natural numbers, making the subset countable. 3. If both the countable set and its subset are infinite, we can enumerate their elements and define a bijection between the subset and the natural numbers, making the subset countable. Thus, in all cases, every subset of a countable set is countable.

Step by step solution

01

Case 1: Countable set is finite

If the countable set is finite (has a finite number of elements), then it has a finite number of subsets. This means that we can create a bijection between the subset and the first few natural numbers up to the size of the subset. Since the natural numbers are countable, the finite subset is also countable.
02

Case 2: Countable set is infinite and the subset is finite

In this case, we can create a bijection between the finite subset and the first few natural numbers, just like in the first case. Since the natural numbers are countable, the finite subset is countable as well.
03

Case 3: Countable set and the subset are both infinite

Given that both the countable set (A) and its subset (B) are infinite, we can enumerate the elements of the countable set A as \(a_1, a_2, a_3, ...\). For each element in the subset B, we will denote it as \(b_i\), where \(b_i\) is an element of A, and i is the index of the element in B. Now we will define a bijection between the natural numbers (1, 2, 3, ...) and the subset B. Since B is an infinite subset of A, we can list its elements according to the order that they appear in the enumeration of A. For example, the first element of B will be the element with the lowest index i such that \(a_i \in B\). We can continue this process for all elements in B, creating a bijection between the natural numbers and B. Since the natural numbers are countable, B is also countable. To conclude, we have shown that every subset of a countable set is countable, whether the countable set and its subset are finite or infinite.

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Key Concepts

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

Finite Subsets
A finite subset contains a specific, limited number of elements. Regardless of how large or small, these elements can be represented completely as they are limited, not infinite. Picture a bag of marbles: once you count them all, that's it, as you have reached the end of the collection. In mathematics, when we mention finite subsets, we mean the set can be paired with a portion of the natural numbers, such as 1 to 5, if the subset has five elements. This pairing, or mapping, is called a *bijection* and allows for each element in the subset to be matched with exactly one natural number. Such a mapping demonstrates that finite subsets are indeed countable. The main takeaway is that finite means limited and graspable, making finite subsets always countable, no matter the starting or parent countable set.
Infinite Subsets
An infinite subset, as opposed to a finite one, extends indefinitely. It could contain elements that do not stop, much like counting natural numbers one after another forever. When we say a set is infinite, it still remains countable if we can establish a systematic way to list all its elements. The intriguing part about infinite subsets of countable sets is that they don't run out of elements either; however, they can still be listed in a sequence. We rely on an ordered, eternal recounting, similar to naming each star in an infinite sky one by one. Through this endless yet structured listing, we can prove that infinite subsets are countable, as each element in these subsets can be mapped in a one-to-one manner with natural numbers, ensuring all elements are accounted for.
Bijection
A bijection is like a perfect matchmaker. It ensures that for every element in one set, there is exactly one partner in another set. This concept is quite useful in proving countability because if every element in a subset can be matched with a unique natural number, the subset can indeed be deemed countable. Think of it as arranging a flawless seating chart: each seat has precisely one guest assigned. In mathematics, creating a bijection between two sets establishes a definitive alignment, whereby each member of one set pairs with one and only one member of another set, with none left out or doubled. For subsets of countable sets, demonstrating a bijection confirms they are also countable, whether they are finite or infinite.
Natural Numbers
Natural numbers are the foundation of counting. Beginning from 1 and progressing upwards indefinitely (such as 1, 2, 3, ...), they serve as a basic set of elements in mathematics. Their endless nature allows them to mesh seamlessly with both finite and infinite subsets. Because natural numbers themselves are infinitely countable, they provide a standard measure against which any countable set or subset can be compared or linked through a bijection. This comparison is key: if a subset can be matched completely with a section of natural numbers, that subset is countable. Even for infinite subsets, aligning them in a one-to-one fashion with natural numbers reveals their countability. In simple terms, natural numbers are like an infinite, ever-ready yardstick for measuring any set that seeks to be countable.

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Most popular questions from this chapter

Let \(m, n \in \mathbb{Z}\). Show that \(m\) and \(n\) are relatively prime if and only if \(\operatorname{GCD}(m, n)=1\). Also show that \(m\) and \(n\) are relatively prime if and only if \(u m+v n=1\) for some \(u, v \in \mathbb{Z} .\) Is it true that if a positive integer \(d\) satisfies \(u m+v n=d\) for some \(u, v \in \mathbb{Z}\), then \(d=\operatorname{GCD}(m, n) ?\)

Prove that every rational number \(r\) can be written as \(r=\frac{p}{q}\), where \(p, q \in \mathbb{Z}, q>0\) and \(p, q\) are relatively prime, and moreover the integers \(p\) and \(q\) are uniquely determined by \(r .\)

If \(m, n, n^{\prime} \in \mathbb{Z}\) are such that \(m\) and \(n\) are relatively prime and \(m \mid n n^{\prime}\), then show that \(m \mid n^{\prime}\). Deduce that if \(p\) is a prime (which means that \(p\) is an integer \(>1\) and the only positive integers that divide \(p\) are 1 and \(p\) ) and if \(p\) divides a product of two integers, then it divides one of them. (Hint: Exercise 38.)

The Fundamental Theorem of Algebra states that if \(p(x)\) is a polynomial in \(\mathbb{C}[x]\) of positive degree, then \(p(x)\) has at least one root in \(\mathbb{C} .\) (i) Assuming the Fundamental Theorem of Algebra, show that if \(p(x)\) is a polynomial in \(\mathbb{C}[x]\) of positive degree \(n \in \mathbb{N}\), then we can write $$ p(x)=c\left(x-\alpha_{1}\right) \cdots\left(x-\alpha_{n}\right) $$ where \(c\) is the leading coefficient of \(p(x)\) and \(\alpha_{1}, \ldots, \alpha_{n}\) are (not necessarily distinct) complex numbers. (ii) Show that if \(p(x) \in \mathbb{R}[x]\) and if a complex number \(\alpha=a+i b\) is a root of \(p(x)\), then its conjugate \(\bar{\alpha}:=a-i b\) is also a root of \(p(x)\). (iii) Show that the Fundamental Theorem of Algebra, as stated above, and the Real Fundamental Theorem of Algebra, as stated in this chapter, are equivalent to each other, that is, assuming one of them, we can deduce the other.

Let \(\mathbb{C}[x, y]\) denote the set of all polynomials in two variables \(x\) and \(y\) with coefficients in \(\mathbb{C}\). Elements of \(\mathbb{C}[x, y]\) look like \(P(x, y)=\sum c_{i, j} x^{i} y^{j}\), where \(i\) and \(j\) vary over finite sets of nonnegative integers, and \(c_{i, j} \in \mathbb{C}\). If \(P(x, y)\) is not the zero polynomial, that is, if some \(c_{i, j}\) is nonzero, then the (total) degree of \(P(x, y)\) is defined be \(\max \left\\{i+j: c_{i, j} \neq 0\right\\}\). We say that \(P(x, y)\) is homogeneous of degree \(m\) if each term has degree \(m\), that is, \(i+j=m\) whenever \(c_{i, j} \neq 0\). As in the case of polynomials in one variable, we can substitute real or complex numbers for the variables \(x\) and \(y . \mathrm{A}\) pair \((\alpha, \beta)\), where \(\alpha, \beta \in \mathbb{C}\), is called a root of \(P(x, y)\) if \(P(\alpha, \beta)=0\). (i) Show that there are nonzero polynomials in \(\mathbb{C}[x, y]\) with infinitely many roots. Show, however, that there is no nonzero polynomial \(P(x, y) \in \mathbb{C}[x, y]\) such that \(P(\alpha, \beta)=0\) for all \(\alpha \in D\) and \(\beta \in E\) where both \(D\) and \(E\) are infinite subsets of \(\mathbb{C}\). (ii) Show that if \(P(x, y)\) is a homogeneous polynomial of positive degree \(m\), then \(P(x, y)\) factors as a product of homogeneous polynomials of degree 1 , that is, $$ P(x, y)=\prod_{i=1}^{m}\left(\alpha_{i} x+\beta_{i} y\right) \quad \text { for some } \alpha_{i}, \beta_{i} \in \mathbb{C}, 1 \leq i \leq m $$ Deduce that the pair \(\left(\beta_{i},-\alpha_{i}\right)\) is a root of \(P(x, y)\) for \(i=1, \ldots, m\) and up to proportionality, these are the only roots of \(P(x, y)\), that is, if \((\alpha, \beta)\) is a root of \(P(x, y)\), then \((\alpha, \beta)=\left(\lambda \beta_{i},-\lambda \alpha_{i}\right)\) for some \(\lambda \in \mathbb{C}\) and \(i \in\\{1, \ldots, m\\} .\) (Hint: Consider \(P(x, 1)\) or \(P(y, 1)\), and use Exercise 69.)
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