Have I proved this obvious fact correctly? (Real Analysis)

Click For Summary
SUMMARY

The discussion centers on the Riemann–Stieltjes integral as presented in "baby Rudin," specifically the claim that for any ε>0, a positive η can be chosen such that (α(b)-α(a))η<ε. The user proposes a contradiction by assuming no such η exists, leading to a real number r that violates the arbitrary nature of ε. Another participant counters this by suggesting that for given η and ε, choosing x greater than ε/η disproves the initial claim. This highlights the importance of understanding the conditions under which the Riemann–Stieltjes integral operates.

PREREQUISITES
  • Understanding of Riemann–Stieltjes integrals
  • Familiarity with concepts of limits and ε-δ definitions in real analysis
  • Basic knowledge of proofs by contradiction
  • Proficiency in manipulating inequalities involving real numbers
NEXT STEPS
  • Study the properties of Riemann–Stieltjes integrals in detail
  • Explore ε-δ definitions and their applications in real analysis
  • Learn about proofs by contradiction and their effectiveness in mathematical arguments
  • Investigate counterexamples in real analysis to strengthen understanding of concepts
USEFUL FOR

Students of real analysis, mathematicians exploring integration techniques, and anyone seeking to deepen their understanding of the Riemann–Stieltjes integral and its implications.

Arian.D
Messages
101
Reaction score
0

Homework Statement


It's not a HW problem. I was reading baby Rudin, in chapter 6 when it talks about Riemann–Stieltjes integral, it claims that given ε>0, we could choose η>0 such that (α(b)-α(a))η<ε. I wonder why it is true. I proposed this question to myself:

Suppose that ε>0 is an arbitrarily given number. Is there a positive real number η such that for every real x we have: |x|η<ε?

The Attempt at a Solution



My solution is this: Suppose that such an η>0 doesn't exist. Therefore there exists a real number r such that for any η>0 we have: |r|η ≥ ε. Since η is now arbitrary, let's set η=1. we'll have 0<ε<|r|, but this restricts ε and that is in contradiction with the hypothesis that ε is arbitrarily given. (for example we could set ε=|r|+1 and that fails for sure).
 
Physics news on Phys.org
Arian.D said:

Homework Statement


It's not a HW problem. I was reading baby Rudin, in chapter 6 when it talks about Riemann–Stieltjes integral, it claims that given ε>0, we could choose η>0 such that (α(b)-α(a))η<ε. I wonder why it is true. I proposed this question to myself:

Suppose that ε>0 is an arbitrarily given number. Is there a positive real number η such that for every real x we have: |x|η<ε?

The Attempt at a Solution



My solution is this: Suppose that such an η>0 doesn't exist. Therefore there exists a real number r such that for any η>0 we have: |r|η ≥ ε.

I don't think that follows. Just do this: Given ##\eta## and ##\epsilon##, pick ##x>\frac \epsilon \eta##, which shows the statement is false.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 28 ·
Replies
28
Views
4K
  • · Replies 14 ·
Replies
14
Views
5K
  • · Replies 10 ·
Replies
10
Views
3K
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 2 ·
Replies
2
Views
4K
Replies
3
Views
2K