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- Jun 22, 2012

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I am currently reading Chapter 3: Continuous Functions on Intervals and am currently focused on Section 3.1 Limits and Continuity ... ...

I need some help in understanding the proof of Theorem 3.16 ...

Theorem 3.16 and its proof read as follows:

In the above proof by Andrew Browder we read the following:

" ... ... But \(\displaystyle f(b) \gt y\) implies (since \(\displaystyle f\) is continuous at \(\displaystyle b\)) that there exists \(\displaystyle \delta \gt 0\) such that \(\displaystyle f(t) \gt y\) for all \(\displaystyle t\) with \(\displaystyle b - \delta \lt t \leq b\). ... ... "

My question is as follows:

How do we demonstrate explicitly and rigorously that since \(\displaystyle f\) is continuous at \(\displaystyle b\) and \(\displaystyle f(b) \gt y \) therefore we have that there exists \(\displaystyle \delta \gt 0\) such that \(\displaystyle f(t) \gt y\) for all \(\displaystyle t\) with \(\displaystyle b - \delta \lt t \leq b\). ... ...

Help will be much appreciated ...

Peter

***NOTE***

The relevant definition of one-sided continuity for the above is as follows:

\(\displaystyle f\) is continuous from the left at \(\displaystyle b\) implies that for every \(\displaystyle \epsilon \gt 0\) there exists \(\displaystyle \delta \gt 0\) such that for all \(\displaystyle x \in [a, b]\) ...

we have that \(\displaystyle b - \delta \lt x \lt b \Longrightarrow \mid f(x) - f(b) \mid \lt \epsilon\)