Can Open Subsets of Real Numbers Form Countable Unions of Intervals?

  • Context: Graduate 
  • Thread starter Thread starter Bachelier
  • Start date Start date
  • Tags Tags
    Union
Click For Summary
SUMMARY

Open subsets of real numbers can indeed form countable unions of disjoint open intervals. This is established by demonstrating a bijection between the set of natural numbers and the open subset E, represented as E = disjoint U_(i in N) of (a_j, b_j) where j is in N and a_j, b_j are real numbers. The conclusion is that the cardinality of E is equal to that of the natural numbers, confirming that E is countable.

PREREQUISITES
  • Understanding of bijections and cardinality in set theory
  • Familiarity with open intervals in real analysis
  • Knowledge of the properties of natural numbers
  • Concept of disjoint unions in mathematics
NEXT STEPS
  • Study the concept of bijections in more depth
  • Explore the properties of open intervals in real analysis
  • Learn about countability and uncountability in set theory
  • Investigate the implications of the density of rational numbers in real numbers
USEFUL FOR

Mathematicians, students of real analysis, and anyone interested in set theory and the properties of open subsets of real numbers.

Bachelier
Messages
375
Reaction score
0
Question:
Given that any open subset E of the set of real numbers is a disjoint union of open intervals.
Is E a countable union of disj. opn intervls.

Answer:

Yes it is. to show this we need to find a Bijection from the set of natural numbers to E.

E = disjoint U_(i in N) of (a_j , b_j) with j in N and a_j , b_j in R

consider then g: N to E with f(n) = i

this is surely a bijection. Hence |E| = |N| hence E is countable.

?

thanks
 
Physics news on Phys.org
Hint: The rationals are dense in R.

Bachelier said:
Question:
Given that any open subset E of the set of real numbers is a disjoint union of open intervals.
Is E a countable union of disj. opn intervls.

Answer:

Yes it is. to show this we need to find a Bijection from the set of natural numbers to E.

E = disjoint U_(i in N) of (a_j , b_j) with j in N and a_j , b_j in R

consider then g: N to E with f(n) = i

this is surely a bijection. Hence |E| = |N| hence E is countable.

?



thanks

Is (0,1) countable?
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 5 ·
Replies
5
Views
11K