Convergence of Sequences in Bounded Sets: Is the Limit Always the Supremum?

  • Context: Graduate 
  • Thread starter Thread starter semidevil
  • Start date Start date
Click For Summary
SUMMARY

The discussion centers on the convergence of sequences within bounded sets, specifically addressing the limit of a sequence derived from an infinite subset A of real numbers R, where R is bounded above. The supremum of A, denoted as u, is established as the limit of a sequence (Xn) belonging to A. The subsequence theorem is confirmed as applicable, asserting that if A converges to u, any sequence derived from A will also converge to u. The construction of a strictly increasing sequence that approaches the supremum is demonstrated through a methodical selection of elements from A.

PREREQUISITES
  • Understanding of real analysis concepts, specifically limits and supremum.
  • Familiarity with the subsequence theorem in the context of convergence.
  • Knowledge of constructing sequences from sets in mathematical analysis.
  • Basic proficiency in epsilon-delta definitions of limits.
NEXT STEPS
  • Study the subsequence theorem in detail to understand its implications on convergence.
  • Explore the properties of supremum and infimum in bounded sets.
  • Learn about constructing sequences that converge to specific limits in real analysis.
  • Investigate the epsilon-delta definition of limits for a deeper understanding of convergence.
USEFUL FOR

Mathematics students, educators, and researchers focusing on real analysis, particularly those interested in the properties of sequences and convergence within bounded sets.

semidevil
Messages
156
Reaction score
2
let A be an infinite subset of R and R is bounded above, and u:= sup A. show that there exist a sequence (Xn) with X(n) belongs to A, such that u = lim(Xn).


ok, so suppose that there does exist a sequence X(n) in A. We know that SupA = u. by the subsequence theorem, if A converges to u, then so will any sequence that belongs to it right? and by another theorem, the limit is the supremum...correct?

I don't know...maybe too easy? I feel I didn't cover everything
 
Physics news on Phys.org
semidevil said:
let A be an infinite subset of R and R is bounded above, and u:= sup A. show that there exist a sequence (Xn) with X(n) belongs to A, such that u = lim(Xn).


ok, so suppose that there does exist a sequence X(n) in A. We know that SupA = u. by the subsequence theorem, if A converges to u, then so will any sequence that belongs to it right?

Are you sure this is what the subsequence theorem says?
 
Let e>0, then there exists a(1) in A such that supA - e <= a(1) <=Sup(A)

Now let e be one half SupA -a(1)

Pick and a(2) in the range sup(A) - e to Sup A.

repeat and get a strictly increasing sequence tending to sup(A).
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K