Is This Sequence Compact? Analysis of {1, 1/2, 2/3, 3/4, 4/5...}

  • Thread starter Thread starter cragar
  • Start date Start date
  • Tags Tags
    Compact Set
Click For Summary
SUMMARY

The sequence {1, 1/2, 2/3, 3/4, 4/5...} is compact as it contains its cluster point, which is 1. This conclusion is supported by the theorem stating that any closed and bounded subset of the real numbers is compact. The proof involves demonstrating that for any open cover of the set, there exists a point within the cover that satisfies the conditions of compactness. Specifically, the sequence converges to 1, ensuring that all but a finite number of terms are contained within any open set that includes 1.

PREREQUISITES
  • Understanding of compactness in topology
  • Familiarity with open covers and cluster points
  • Knowledge of real number properties
  • Basic concepts of sequences and convergence
NEXT STEPS
  • Study the definition and properties of compact sets in topology
  • Learn about the Heine-Borel theorem regarding closed and bounded subsets
  • Explore examples of sequences and their convergence behavior
  • Investigate open covers and their applications in proving compactness
USEFUL FOR

Mathematics students, particularly those studying real analysis and topology, as well as educators seeking to deepen their understanding of compactness and convergence in sequences.

cragar
Messages
2,546
Reaction score
3

Homework Statement


{1,1/2,2/3,3/4,4/5...} is this set compact.

The Attempt at a Solution


I think this set is compact because it contains its cluster point which is 1.
is this correct?
 
Physics news on Phys.org
That is correct.
 
You could prove that by appealing to "any closed and bounded subset of the real numbers is compact" or directly from the definition:
Let [itex]\{U_\alpha\}[/itex], for [itex]\alpha[itex]in some index set, be an open cover for this set. Since 1 is in the set, 1 is in at least one of those- relabel the sets, if necessary, so that set is called [itex]U_1[/itex]. Since [itex]U_1[/itex] is open, there exist a point p, in that set, and a number [itex]\delta> 0[/itex] such that [itex]N_\delta(p)= \{x | |x- p|<\delta\}[/itex] is in [itex]U_1[/itex]. It then follows that if [itex]n> 1/\delta[/itex] [itex]|1- (n-1)/n|= 1/n< \delta[/itex] so that all except a finite number of the fractions in the sequence are in [itex]U_1[/itex]. Every number in the sequence, [itex](n-1)/n[/itex] for n less than that might be in as separate set in the original sequence but there are only a finite number of them.[/itex][/itex]
 

Similar threads

  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 9 ·
Replies
9
Views
7K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K