Formal Proof of Uniform Continuity on a Closed Interval

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Homework Help Overview

The discussion revolves around proving the uniform continuity of a function on a closed interval, specifically addressing the implications of uniform continuity on subintervals. The problem is situated within the context of real analysis.

Discussion Character

  • Exploratory, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the original poster's rough proof idea, questioning whether the reasoning is correct and how to express it more formally. There is an emphasis on clarifying the definitions and conditions associated with delta and epsilon in the context of uniform continuity.

Discussion Status

Some participants have acknowledged the original poster's approach and suggested that it is a good starting point for a formal proof. There is an ongoing exploration of how to articulate the reasoning more clearly, with specific suggestions to elaborate on the conditions for delta.

Contextual Notes

Participants are considering the need for precise definitions and formal language in the proof, indicating a focus on rigor in mathematical writing. The discussion reflects the challenges of transitioning from informal reasoning to formal proof structure.

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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