Are there sets that are not partitionable in certain ways?

  • Context: Graduate 
  • Thread starter Thread starter quasar987
  • Start date Start date
  • Tags Tags
    Sets
Click For Summary
SUMMARY

This discussion addresses the partitioning of sets, specifically the real numbers \(\mathbb{R}\) and the two-dimensional space \(\mathbb{R}^2\). It confirms that \(\mathbb{R}\) can be partitioned into singletons using the notation \(U_x = \{x\}\) for each \(x \in \mathbb{R}\), resulting in \(\cup_{x \in \mathbb{R}} U_x = \mathbb{R}\). Additionally, \(\mathbb{R}^2\) can be partitioned into lines of slope 2 with the sets defined as \(U_a = \{(x, y) | y = 2x + a\}\), leading to \(\cup_{a \in \mathbb{R}} U_a = \mathbb{R}^2\) and confirming that these sets are either equal or disjoint.

PREREQUISITES
  • Understanding of set theory and partitioning concepts
  • Familiarity with real numbers and their properties
  • Basic knowledge of Cartesian coordinates and linear equations
  • Ability to interpret mathematical notation and expressions
NEXT STEPS
  • Explore advanced topics in set theory, focusing on partitioning techniques
  • Study linear algebra concepts related to vector spaces and lines in \(\mathbb{R}^2\)
  • Learn about the implications of disjoint sets in mathematical proofs
  • Investigate the use of LaTeX for mathematical documentation and expressions
USEFUL FOR

Mathematicians, students of mathematics, and educators interested in set theory and its applications in higher mathematics.

quasar987
Science Advisor
Homework Helper
Gold Member
Messages
4,796
Reaction score
32
Are there sets that are not partitionable in certain ways? For exemple, can I partition [itex]\mathbb{R}[/itex] into a collection of singletons?

Can I partition [itex]\mathbb{R}^2[/itex] into a collection of lines of slope 2?

If so, how would you write each of those partitions?

Thx.
 
Physics news on Phys.org
For the first one: for every [tex]x \in \mathbb{R}[/tex] let [tex]U_x = \{x\}[/tex]. Then [tex]\cup_{x \in \mathbb{R}} U_x = \mathbb{R}[/tex] and obviously all [tex]U_x[/tex] are disjoint from one another, and are singletons.

For the second: for every [tex]a \in \mathbb{R}[/tex] let [tex]U_a = \{(x, y) | y = 2x + a\}[/tex] (I wish I knew LaTeX better).

Then take [tex](x_0, y_0) \in \mathbb{R}^2[/tex] and notice that [tex](x_0, y_0) \in U_{y_0 - 2x_0}[/tex]. Thus [tex]\cup_{a \in \mathbb{R}} U_a = \mathbb{R}^2[/tex]. A simple calculation will also reveal that [tex]U_a[/tex] and [tex]U_b[/tex] are either equal or disjoint (for any real a, b).
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K