Continuous Functions on Intervals .... B&S Theorem 5.3.2 ....

Click For Summary
SUMMARY

The discussion centers on Theorem 5.3.2 from "Introduction to Real Analysis" (Fourth Edition) by Robert G. Bartle and Donald R. Sherbert, specifically regarding the continuity of a function \( f \) on a closed interval \( I \). Participants clarify that since \( I \) is closed and the point \( x \) belongs to \( I \) as established by Theorem 3.2.6, it follows that \( f \) is continuous at \( x \). The definition of continuity at a point within the interval confirms this conclusion. The exchange emphasizes the importance of understanding the definitions and theorems in real analysis.

PREREQUISITES
  • Understanding of real analysis concepts, specifically continuity and closed intervals.
  • Familiarity with Theorem 3.2.6 from "Introduction to Real Analysis".
  • Knowledge of the definitions of continuous functions.
  • Basic mathematical proof techniques used in analysis.
NEXT STEPS
  • Review Theorem 3.2.6 in "Introduction to Real Analysis" for a deeper understanding of closed intervals.
  • Study the definition of continuity in the context of real-valued functions.
  • Examine additional examples of continuous functions on closed intervals.
  • Explore related theorems in real analysis that discuss continuity and limits.
USEFUL FOR

Students and educators in mathematics, particularly those studying real analysis, as well as anyone seeking to strengthen their understanding of continuity in mathematical functions.

Math Amateur
Gold Member
MHB
Messages
3,920
Reaction score
48
I am reading "Introduction to Real Analysis" (Fourth Edition) by Robert G Bartle and Donald R Sherbert ...

I am focused on Chapter 5: Continuous Functions ...

I need help in fully understanding an aspect of the proof of Theorem 5.3.2 ...Theorem 5.3.2 and its proof ... ... reads as follows:View attachment 7277In the above text from Bartle and Sherbert we read the following:

" Since $$I$$ is closed and the elements of $$X'$$ belong to $$I$$, it follows from Theorem 3.2.6 that $$x \in I$$. Then $$f$$ is continuous at $$x$$ ... ... "Can someone please explain exactly why/how we can conclude that $$f$$ is continuous at $$x$$ ... ?Peter
 
Physics news on Phys.org
Peter said:
In the above text from Bartle and Sherbert we read the following:

" Since $$I$$ is closed and the elements of $$X'$$ belong to $$I$$, it follows from Theorem 3.2.6 that $$x \in I$$. Then $$f$$ is continuous at $$x$$ ... ... "Can someone please explain exactly why/how we can conclude that $$f$$ is continuous at $$x$$ ... ?
The theorem contains the hypothesis that $f$ is continuous on $I$. By definition, that means that $f$ is continuous at each point of $I$. The proof has already shown that $x\in I$. So it follows that $f$ is continuous at $x$.
 
Opalg said:
The theorem contains the hypothesis that $f$ is continuous on $I$. By definition, that means that $f$ is continuous at each point of $I$. The proof has already shown that $x\in I$. So it follows that $f$ is continuous at $x$.
Oh ... careless of me not to notice that ...!

Thanks Opalg ...

Peter
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
2K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K