/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 Let \(T=\\{(i, j, k) \mid i, j, ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 8

Let \(T=\\{(i, j, k) \mid i, j, k \in \mathcal{N}\\} .\) Show that \(T\) is countable.

Short Answer

Expert verified
We can show that the set \(T = \{(i, j, k) \mid i, j, k \in \mathcal{N}\}\) is countable by constructing a bijection from \(T\) to the set of natural numbers \(\mathcal{N}\). We do this by creating a spiral walk through every possible triplet (i, j, k) in a 3-dimensional space where each axis is the set of natural numbers. At each level \(n\), we walk through all the triplets where \(i + j + k = n\), starting from \((n, 0, 0), (n-1, 1, 0), (n-2, 2, 0), \ldots\) and so on until \((0, 0, n)\). We then map each triplet to a unique natural number based on the order they are encountered on the walk. As this walk creates a surjective and injective function from \(T\) to \(\mathcal{N}\), we can conclude that \(T\) is countable.

Step by step solution

01

Understand the nature of the set T

The set \( T \) is composed of ordered triplets \((i, j, k)\) where \( i, j, k \) are natural numbers. If you think about it, these triplets can be thought of as points in a 3-dimensional space where each axis is the set of natural numbers.
02

Constructing the bijection function

To show that a set is countable, one approach is to construct a bijection from the set to the set of natural numbers \( \mathcal{N} \). A bijection means that each element in \( T \) maps to a unique element in \( \mathcal{N} \) and vice versa. For this problem, we can create a function \( f : T \rightarrow \mathcal{N} \) that takes an ordered triplet and outputs a natural number. We'll start by creating a spiral that "walks" through every possible triplet by going through the cube level by level. At each level \( n \), we'll walk through all the triplets \( (i,j,k) \) where \( i + j + k = n \), starting from \( (n, 0, 0), (n-1, 1, 0), (n-2, 2, 0), \ldots \) and so on until \( (0, 0, n) \). Once we've visited all the triplets at level \( n \), we'll go to the next level. Map each triplet to a unique natural number by simply ordering them as they are encountered on the walk.
03

Proving function is bijective

The function is surjective (onto) because for each natural number, there is an associated point in the walk. The function is injective (one-to-one) since for each point in the walk, there is a unique natural number that identifies it. So, by walk, we have shown that every possible triplet is covered and each triplet is associated with a unique natural number. Therefore, the set \( T \) is countable.
04

Concluding the proof

We have now shown a way to construct a bijection from the set \( T \) to the set of natural numbers \( \mathcal{N} \), hence \( T \) is countable. This concludes the proof.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Bijection
In mathematics, a bijection is a type of function that pairs every element of one set with a unique element of another set. This concept is crucial when determining if a set is countable. A bijection is both injective and surjective. This means that it matches every element in the first set to exactly one element in the second set, and every element in the second set is the image of exactly one element from the first set.

For example, to show that the set of 3-dimensional triplets \(i, j, k\) is countable, a bijection is created between this set and the natural numbers. During this bijection process, each 3D point maps to a specific natural number in a way that covers the entire set of triplets and all natural numbers, reflecting a bijective relationship.
Natural Numbers
Natural numbers, denoted by \(\), form the foundation of many mathematical concepts and are represented by \(0, 1, 2, 3, \, \text{and so on} \). They are countable, which means there is a bijection between the natural numbers and the elements of any countable set.

In the given problem, each triplet \(i, j, k\) from the set \(T\) is associated with a natural number. This association demonstrates that the triplets are countably infinite because there exists a systematic way (bijective function) to pair each triplet with a natural number. This method shows not only the countability of natural numbers but also builds a bridge to understanding the countability of more complex structures, like 3-dimensional spaces of points.
3-dimensional Space
3-dimensional space is a concept familiar in our physical world, defined by three axes: usually labeled x, y, and z. In the context of the problem, each axis corresponds to natural numbers, and every point in this space is denoted by a triplet \(i, j, k\).

In mathematics, dealing with spaces like these involves understanding their structure, such as how many points they include. The task in this context is to determine whether all these points form a countable set. By constructing a specific walk through this 3-dimensional space, visiting each triplet in a systematic way, we confirm that the space of natural-number-defined triplets is countable, aligning one intricate structure of space with the simplicity of natural numbers.
Injective and Surjective Functions
Understanding functions that are injective and surjective is essential to grasping bijections. An injective function ensures that different inputs map to different outputs, meaning no two distinct elements from the domain are assigned the same image in the codomain. Surjective functions ensure every element in the codomain is the image of at least one element from the domain.

In proving that \(T\) is countable, the function constructed is injective because each triplet is associated with one unique natural number. It is surjective because every natural number is matched with one triplet. Thus, this function construction process effectively covers all of \(T\), showing how each triplet can be transformed into a countable sequence of natural numbers. This ensures the function is bijective, as it holds both properties.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The following procedure decides \(A M B I G_{\mathrm{NFA}}\). Given an NFA \(N\), we design a DFA \(D\) that simulates \(N\) and accepts a string iff it is accepted by \(N\) along two different computational branches. Then we use a decider for \(E_{\mathrm{DFA}}\) to determine whether \(D\) accepts any strings. Our strategy for constructing \(D\) is similar to the NFA-to-DFA conversion in the proof of Theorem 1.39. We simulate \(N\) by keeping a pebble on each active state. We begin by putting a red pebble on the start state and on each state reachable from the start state along e transitions. We move, add, and remove pebbles in accordance with \(N\) 's transitions, preserving the color of the pebbles. Whenever two or more pebbles are moved to the same state, we replace its pebbles with a blue pebble. After reading the input, we accept if a blue pebble is on an accept state of \(N\) or if two different accept states of \(N\) have red pebbles on them. The DFA \(D\) has a state corresponding to each possible position of pebbles. For each state of \(N\), three possibilities occur: It can contain a red pebble, a blue pebble, or no pebble. Thus, if \(N\) has \(n\) states, \(D\) will have \(3^{n}\) states. Its start state, accept states, and transition function are defined to carry out the simulation.

Let \(B A L_{\mathrm{DFA}}=\\{\langle M\rangle \mid M\) is a DFA that accepts some string containing an equal number of os and 1s \\}. Show that \(B A L_{\mathrm{DFA}}\) is decidable. (Hint: Theorems about CFLs are helpful here.)

Let \(A=\\{\langle R, S\rangle \mid R\) and \(S\) are regular expressions and \(L(R) \subseteq L(S)\\} .\) Show that \(A\) is decidable.

Let \(\mathcal{B}\) be the set of all infinite sequences over \(\\{0,1\\} .\) Show that \(\mathcal{B}\) is uncountable using a proof by diagonalization.

The proof of Lemma 2.41 says that \((q, x)\) is a looping situation for a DPDA \(P\) if when \(P\) is started in state \(q\) with \(x \in \Gamma\) on the top of the stack, it never pops anything below \(x\) and it never reads an input symbol. Show that \(F\) is decidable, where \(F=\\{\langle P, q, x\rangle \mid(q, x)\) is a looping situation for \(P\\}\).
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Applied Mathematics

Read Explanation

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.