Can anyone check my proof involving least-upper-bounds?

  • Thread starter Thread starter Eclair_de_XII
  • Start date Start date
  • Tags Tags
    Proof
Click For Summary
SUMMARY

The discussion revolves around proving that if \( x = \sup(S) \), then for every \( \epsilon > 0 \), there exists an element \( a \in S \) such that \( x - \epsilon < a \leq x \). Participants analyze two cases: when \( x \) is an element of \( S \) and when it is not. The proof requires understanding the definition of the least upper bound, which states that \( x \) is greater than or equal to all elements in \( S \) and is less than or equal to any upper bound of \( S \). The conclusion emphasizes that \( x - \epsilon \) cannot be an upper bound for \( S \), ensuring the existence of the required element \( a \) in the interval \( (x - \epsilon, x) \).

PREREQUISITES
  • Understanding of the concept of supremum (least upper bound) in real analysis.
  • Familiarity with the properties of real numbers and intervals.
  • Knowledge of mathematical proof techniques, particularly in analysis.
  • Ability to manipulate inequalities involving real numbers.
NEXT STEPS
  • Study the definition and properties of supremum in detail.
  • Learn about the least-upper-bound property and its implications in real analysis.
  • Explore examples of sets with different types of upper bounds.
  • Practice constructing proofs involving supremum and infimum in various contexts.
USEFUL FOR

Students of mathematics, particularly those studying real analysis, as well as educators and anyone interested in understanding the concept of least upper bounds and their applications in proofs.

  • #31
So, ##x-\epsilon## is not an upper bound for ##S## because there exists an ##a∈S## such that ##a>x-\epsilon##.

...Oh, so because ##x-\epsilon## is not an upper bound for ##S##, then it is implied that there must exist an element ##a∈S## greater than ##x-\epsilon##?
 
Physics news on Phys.org
  • #32
Eclair_de_XII said:
So, ##x-\epsilon## is not an upper bound for ##S## because there exists an ##a∈S## such that ##a>x-\epsilon##.

...Oh, so because ##x-\epsilon## is not an upper bound for ##S##, then it is implied that there must exist an element ##a∈S## greater than ##x-\epsilon##?

Yes!
 
  • #33
Thanks for bearing with me and helping me finish my homework. I'm kind of apprehensive about how I'll handle the rest of the class, to be honest...
 
  • Like
Likes   Reactions: PeroK
  • #34
Eclair_de_XII said:
So, ##x-\epsilon## is not an upper bound for ##S## because there exists an ##a∈S## such that ##a>x-\epsilon##.

...Oh, so because ##x-\epsilon## is not an upper bound for ##S##, then it is implied that there must exist an element ##a∈S## greater than ##x-\epsilon##?
I like the fact that you italicized the word "must" above. That's the first inkling you have given that you actually get it. Good for you. That's what this forum is here for.
 

Similar threads

  • · Replies 25 ·
Replies
25
Views
3K
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
32
Views
3K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K