Help with Proving X ⊂ P(N) - Cardinality of Continuum

  • Context: Graduate 
  • Thread starter Thread starter somebd
  • Start date Start date
  • Tags Tags
    Proof
Click For Summary
SUMMARY

The discussion focuses on proving that there exists a set X that is a subset of the power set P(N) with cardinality equal to the continuum. It establishes that for each element A in X, A is infinite and that any two distinct elements A and B in X have a finite intersection. The proof involves defining an almost disjoint family and utilizing the work of K. Kunen in "Set Theory, an Introduction to Independence Proofs." The construction of a bijective function from the countable set of finite subsets of natural numbers to demonstrate the desired properties is also outlined.

PREREQUISITES
  • Understanding of power sets, specifically P(N)
  • Familiarity with cardinality concepts, particularly the continuum
  • Knowledge of almost disjoint families in set theory
  • Basic comprehension of bijective functions and their properties
NEXT STEPS
  • Study "Set Theory, an Introduction to Independence Proofs" by K. Kunen for in-depth understanding
  • Explore the concept of almost disjoint families in set theory
  • Learn about cardinality and its implications in set theory
  • Investigate bijective functions and their applications in set theory proofs
USEFUL FOR

Mathematicians, set theorists, and students of advanced mathematics seeking to deepen their understanding of cardinality, power sets, and the properties of infinite sets.

somebd
Messages
5
Reaction score
0
Prove that exist X ⊂ P(N)

P(N) is a power set of natural numbers
1 cardinality of X is continuum
2 for each A (element of X) in X, A is infinite
3 for each A and B in X, A is not equal to B, A intersection with B is finite

Please, I have no idea how to solve this :o
 
Physics news on Phys.org
What you're searching for is called an almost disjoint family. A full treatise on this can be found in "Set theory, an introduction to independence proofs" by K. Kunen.

Here is a sketch of the proof:
If [tex]X\subseteq \mathbb{N}[/tex], define [tex]A_X=\{X\cap \{0,...,n\}~\vert~n\in \mathbb{N}\}[/tex]. If X is countable, then [tex]A_X[/tex] is countable. If [tex]X\neq Y[/tex], then [tex]A_X\cap A_Y[/tex] is finite.
Let [tex]\mathcal{A}=\{A_X~\vert~X\subseteq \mathbb{N}~\text{and X infinite}\}[/tex]. Then X has the cardinality of the continuum.
The problem is now, that it is not true that [tex]\mathcal{A}\subseteq P(\mathbb{N})[/tex]. To solve this, let [tex]I=\{A\subseteq \mathbb{N}~\vert~A~\text{finite}\}[/tex]. It is not that hard to see that I is countable. Let [tex]f[/tex] be a bijective function from I to [tex]\mathbb{N}[/tex]. Then [tex]\{f(A)~\vert~A\in\mathcal{A}\}[/tex] is the set you want...
 
Wow, thanks you very much for the fast response and solution idea :)
 

Similar threads

  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 18 ·
Replies
18
Views
4K
  • · Replies 29 ·
Replies
29
Views
6K
  • · Replies 5 ·
Replies
5
Views
2K