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1 vote
1 answer
55 views

Finiteness, finite sets and representing its elements.

A set $S$ is called finite if there exists a bijection from $S$ to $\{1,...,n\}$ for one $n \in \mathbf{N}$. It is then common to write its elements as $s_1,...,s_n$. I now wonder, why this is ...
MaxH's user avatar
  • 389
0 votes
3 answers
258 views

Can the natural numbers contain an element that is not representable by a number?

I read the following document: https://www.math.wustl.edu/~freiwald/310peanof.pdf . In this document, the author wants to formalize that natural numbers, that are informally thought of as a collection ...
MaxH's user avatar
  • 389
0 votes
2 answers
100 views

Directly proof $S$ is countable, where $S$ is set of function from $\{0, 1\}$ to $\mathbb{N}$

Suppose $S=\{f_1,f_2,f_3,f_4,f_5,........\}$ where $f_i$ is a function $f:\{0, 1\}\to\mathbb{N}.$ I have to prove $S$ is countable.Then need to prove direct one-to-one correspondence between $S$ and $\...
user avatar
2 votes
1 answer
106 views

Why Cantor diagonalization theorem is failed to prove $S$ is countable, Where $S$ is set of finite subset of $\mathbb{N}$?

I have an given set $S$ where $S=$ set of finite subsets of $\mathbb{N}.$ We need to prove $S$ is countably infinite. My approach: I need to prove there is one-to-one correspondence between $S$ and ${\...
user avatar
-1 votes
1 answer
173 views

Why bijection between $\mathbb{N^2}$ and $\mathbb{N}$ is not possible directly?

To understand bijection between $\mathbb{N^2}$ and $\mathbb{N}$ I found this pdf on internet. But have couple of confusion. N.B: Here we take $0 \in \mathbb{N}.$ Confusion:1 Why directly proof of ...
S. M.'s user avatar
  • 1
0 votes
0 answers
209 views

Are odd natural numbers an inductive set?

The definition of inductive set my textbook gave is: A set $T$ that is a subset of the integers is an inductive set provided that for each integer $k$, if $k$ is an element in the set $T$, then $k+1$ ...
Throw Away's user avatar
0 votes
0 answers
103 views

Why is the principle of induction for natural numbers not "self-evident"? [duplicate]

The principle of induction can be stated, in first-order logic, as follows. Let $S\subseteq\mathbb N$, and suppose that $0\in S$. $\forall n:n\in S\to n+1\in S$. Then, $S=\mathbb N$. Now, suppose ...
Joe's user avatar
  • 20.7k
-1 votes
1 answer
135 views

$X$ is infinite thus we have an injection from $\mathbb{N}$ to X [duplicate]

Hey guys I'm trying to prove the following: $X=\emptyset \lor$ There is a surjection $g: \mathbb{N} \rightarrow X \implies X$ is finite $\lor$ there is a Bijection from $\mathbb{N}$ to X I did case ...
Iwan5050's user avatar
  • 385
2 votes
0 answers
128 views

Statement of Well-ordering principle

The statement of well ordering principle appears in different mode - on subsets of natural numbers, or well-ordering of every (non-empty) set. For the question below, I am considering it w.r.t. non-...
Maths Rahul's user avatar
  • 3,047
1 vote
0 answers
515 views

Adding a Fixed Value to Each Element in a Set (How to Denote)

To denote a set such as, for example, the set of every natural number that is 3 greater than a multiple of 5, would $5\mathbb{N}+3$ be generally understood as $\{8,13,18,23,28,33,\dots\}$? If not, how ...
Mathgodpi's user avatar
3 votes
3 answers
187 views

Infinite natural numbers?

Only using the successor function $\nu$ and the other axioms, how do we guarantee that the "next" generated number is different from all the "previous" numbers (I am using ...
Roger Crook's user avatar
3 votes
1 answer
364 views

Showing that the natural numbers are totally ordered with respect to set membership

Working with the usual set theoretic construction of the natural numbers, denoted $\omega$ for now. I am trying to show that $\omega$ is totally ordered with respect to set membership, that is, $n<...
C Squared's user avatar
  • 3,612
0 votes
1 answer
85 views

In ZFC, do we use the set $\mathbb{N}$ in the definition of $\mathbb{N}$ recursively?

In ZFC set theory, we define the set of the natural numbers as follows: By the axiom of infinity, an inductive set exists. Let I be an inductive set. Then, $\mathbb{N}$ is defined as $\{ x\in I |\...
boyler's user avatar
  • 375
0 votes
0 answers
118 views

A Doubt about Well Ordering Principle and Principle of Mathematical Induction

I have had this lingering doubt in my mind for a very long time: One of the standard constructions of N starts by assuming the 5 Peano Axioms, proving that every non-zero is a successor and s(n) is ...
Tara Nanda's user avatar
0 votes
1 answer
33 views

Defining $A \in \mathcal{P}(\Bbb N \times\Bbb N)$ such that it is not any member of a countable subset $M \subseteq \mathcal{P}(\Bbb N \times \Bbb N)$

$ \newcommand{\N}{\mathbb{N}} \newcommand{\P}{\mathcal{P}(\N \times \N)} \newcommand{\set}[1]{\left\{ #1 \right\}} \newcommand{\i}{^{(i)}} \newcommand{\x}{^{[x]}} \newcommand{\y}{^{[y]}} \newcommand{\...
PrincessEev's user avatar
  • 45.9k

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