# Thread: Continuous Functions on Intervals ... B&S Theorem 5.3.2 ....

1. 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:

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 $\displaystyle x$ ... ?

Peter

2. Originally Posted by Peter
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 $x$.

Originally Posted by Opalg
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