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

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The discussion centers on understanding the proof of Theorem 5.3.2 from "Introduction to Real Analysis" by Bartle and Sherbert, specifically regarding the continuity of a function at a point within a closed interval. It is clarified that since the interval is closed and the point belongs to it, the function must be continuous at that point due to the theorem's hypothesis. The definition of continuity states that if a function is continuous on an interval, it is continuous at every point within that interval. The confusion was resolved by recognizing that the proof already established the point's inclusion in the interval. Ultimately, the continuity of the function at that point follows directly from the theorem's conditions.
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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
 
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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
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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