Formal Proof of Uniform Continuity on a Closed Interval

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SUMMARY

The discussion focuses on proving that if a function f is uniformly continuous on the intervals [a,b] and [a,c], then it is also uniformly continuous on the interval [a,c]. The proof approach involves selecting an epsilon greater than 0 and identifying delta_1 for [a,b] and delta_2 for [b,c]. The conclusion is that the minimum of delta_1 and delta_2 serves as the delta for the interval [a,c], establishing uniform continuity. Participants emphasize the need for clarity and formality in the proof presentation.

PREREQUISITES
  • Understanding of uniform continuity
  • Familiarity with epsilon-delta definitions
  • Knowledge of closed intervals in real analysis
  • Ability to construct formal mathematical proofs
NEXT STEPS
  • Study the formal definition of uniform continuity in detail
  • Learn how to construct epsilon-delta proofs
  • Explore examples of functions that are uniformly continuous on closed intervals
  • Investigate the implications of uniform continuity in real analysis
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Students of real analysis, mathematicians focusing on continuity concepts, and anyone interested in formal proof construction in mathematical contexts.

MathSquareRoo
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Homework Statement



Prove that if f is uniformly continuous on [a,b] and on [a,c] implies that f is uniformly continuous on [a,c].

Homework Equations




The Attempt at a Solution



This is my rough idea for a proof, can someone help be say this more formally? Is my thinking even correct?

Let epsilon > 0.
Then there is some delta_1 for [a,b] and some delta_2 for [b,c].
Then the minimum of delta_1 and delta_2 is the delta we want for [a,c].
 
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MathSquareRoo said:

Homework Statement



Prove that if f is uniformly continuous on [a,b] and on [a,c] implies that f is uniformly continuous on [a,c].

Homework Equations




The Attempt at a Solution



This is my rough idea for a proof, can someone help be say this more formally? Is my thinking even correct?

Let epsilon > 0.
Then there is some delta_1 for [a,b] and some delta_2 for [b,c].
Then the minimum of delta_1 and delta_2 is the delta we want for [a,c].

Sure, that's the idea. It's not that hard to fill that out to a formal proof.
 
Thanks. What changes should I make to make it more formal?
 
MathSquareRoo said:
Thanks. What changes should I make to make it more formal?

Just fill in some words. "there is some delta_1 for [a,b]" doesn't mean much. There is some delta_1 for [a,b] such that what? I know what you mean, but spell it out.
 

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