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Chapter 9: Problem 3

Prove that if \(A\) is uncountable and \(A \subseteq B,\) then \(B\) is uncountable.

Short Answer

Expert verified
Assuming the opposite, i.e., that B is countable while A is uncountable, we define a one-to-one correspondence function \(f: B \to \mathbb{N}\). Then, we create a subset of natural numbers C such that \(C = \{f(x) | x \in A\}\). This results in a contradiction, as we now have A, an uncountable set, in one-to-one correspondence with the countable set C, which is against the definition of an uncountable set. Thus, B must be uncountable.

Step by step solution

01

Understanding the terms countable and uncountable sets

: A countable set is a set that can be put into a one-to-one correspondence with the set of natural numbers or any subset of natural numbers. In other words, it means that the elements of the countable set can be enumerated. Uncountable sets, on the other hand, are too large to be put into a one-to-one correspondence with the set of natural numbers.
02

Setting up the proof by contradiction

: In this step, we will assume the opposite of what we want to prove as true, i.e., we will assume that B is countable while A is uncountable. We will then try to prove that this leads to a contradiction, and this will, in turn, prove that B must be uncountable.
03

Defining a one-to-one correspondence for B

: Since B is a countable set (under our contradiction assumption), we can define a one-to-one correspondence function \(f: B \to \mathbb{N}\), which maps each element of B to the set of natural numbers.
04

Defining a subset of natural numbers

: Since A is a subset of B, using the function f, we can create a subset of natural numbers C, where \(C = \{f(x) | x \in A\}\). This means that for every element in A, there is a corresponding unique natural number in C.
05

Reaching a contradiction

: We have now shown that A, which is an uncountable set, can be put into a one-to-one correspondence with the countable set C (a subset of natural numbers). This leads to a contradiction since it goes against the definition of an uncountable set, which cannot be put into a one-to-one correspondence with the set of natural numbers or any of its subsets. Therefore, our initial assumption that B is countable while A is uncountable is contradictory, and hence, B must be uncountable.

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Key Concepts

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

Countable and Uncountable Sets
Understanding the difference between countable and uncountable sets is like distinguishing between things you can list out one by one, and those you simply can't because they are too vast. Countable sets are like your favorite playlist, where you can number the songs鈥攅ven if it takes a long time or continues indefinitely. Technically, these are sets that we can match up with the set of natural numbers, giving each element a distinct 'place' in line.

On the flip side, uncountable sets are like the grains of sand on a beach鈥攁ttempting to count each individually is a fool's errand. These sets are larger; so big, in fact, that not even the endless list of natural numbers can keep up!
Proof by Contradiction
When we use proof by contradiction, it's like catching a fish by draining the pond鈥攊t reveals the error by eliminating all other possibilities. In mathematical terms, we start by assuming the opposite of what we aim to prove and then demonstrate how this assumption logically leads to an impossibility or a contradiction.

If we end up contradicting a well-established fact or the very assumptions we started with, we prove that our original assumption must be incorrect. This method is a powerful tool in mathematics, often used when a direct proof is elusive.
One-to-One Correspondence
Imagine assigning every student in a class to their own unique locker. This scenario is akin to a one-to-one correspondence, a concept in mathematics where two sets can be paired off so that each element in one set is linked to one鈥攁nd only one鈥攅lement in the other set.

This concept is vital when comparing the sizes of sets. If you can establish such a correspondence between two sets, you show that they are, in essence, the same size. It is one of the most fundamental notions for understanding different types of infinity and is closely related to the concept of 'countability' of sets.
Natural Numbers
The natural numbers are the building blocks of basic arithmetic, like the counting numbers kids use to tally their candy on Halloween. These numbers start from 1 and go up (1, 2, 3, ...) indefinitely. They represent quantities or positions in a sequence and are a crucial part of our counting process.

In the grander scheme, they help us make sense of countable sets. Anything that can be paired up with the natural numbers in a one-to-one correspondence is countable. The natural numbers themselves, while infinitely many, are considered countable since they correspond exactly to their own set.

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Most popular questions from this chapter

State whether each of the following is true or false. (a) If a set \(A\) is countably infinite, then \(A\) is infinite. (b) If a set \(A\) is countably infinite, then \(A\) is countable. (c) If a set \(A\) is uncountable, then \(A\) is not countably infinite. (d) If \(A \approx \mathbb{N}_{k}\) for some \(k \in \mathbb{N},\) then \(A\) is not countable.

Prove that the function \(g: A \cup\\{x\\} \rightarrow \mathbb{N}_{k+1}\) in Lemma 9.4 is a surjection.

This exercise is a generalization of Exercise (8). Let \(m\) be a natural number, let \(A\) be a set, and assume that \(f: \mathbb{N}_{m} \rightarrow A\) is a surjection. Define \(g: A \rightarrow \mathbb{N}_{m}\) as follows: For each \(x \in A, g(x)=j,\) where \(j\) is the least natural number in \(f^{-1}(\\{x\\})\) Prove that \(f \circ g=I_{A},\) where \(I_{A}\) is the identity function on the set \(A,\) and prove that \(g\) is an injection.

Let \(A\) be a subset of some universal set \(U\). Prove that if \(x \in U\), then \(A \times\\{x\\} \approx A\)

The goal of this exercise is to use the Cantor-Schr枚der-Bernstein Theorem to prove that the cardinality of the closed interval [0,1] is \(c\). (a) Find an injection \(f:(0,1) \rightarrow[0,1]\). (b) Find an injection \(h:[0,1] \rightarrow(-1,2)\). (c) Use the fact that (-1,2)\(\approx(0,1)\) to prove that there exists an injection \(g:[0,1] \rightarrow(0,1) .\) (It is only necessary to prove that the injection \(g\) exists. It is not necessary to determine a specific formula for \(g(x) .)\) Note: Instead of doing Part (b) as stated, another approach is to find an injection \(k:[0,1] \rightarrow(0,1) .\) Then, it is possible to skip Part (c) and go directly to Part (d). (d) Use the Cantor-Schr枚der-Bernstein Theorem to conclude that [0,1]\(\approx(0,1)\) and hence that the cardinality of [0,1] is \(c .\)
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