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

[Math Amateur]

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 \(\displaystyle I\) is closed and the elements of \(\displaystyle X'\) belong to \(\displaystyle I\), it follows from Theorem 3.2.6 that \(\displaystyle x \in I\). Then \(\displaystyle f\) is continuous at \(\displaystyle x\) ... ... "Can someone please explain exactly why/how we can conclude that \(\displaystyle f\) is continuous at \(\displaystyle x\) ... ?Peter

 

[Opalg]

In the above text from Bartle and Sherbert we read the following:

" Since \(\displaystyle I\) is closed and the elements of \(\displaystyle X'\) belong to \(\displaystyle I\), it follows from Theorem 3.2.6 that \(\displaystyle x \in I\). Then \(\displaystyle f\) is continuous at \(\displaystyle x\) ... ... "Can someone please explain exactly why/how we can conclude that \(\displaystyle f\) is continuous at \(\displaystyle x\) ... ?

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 $x$.

 

[Math Amateur]

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