3a) $ \displaystyle f(\xi) \leq \eta $

Pick $ \displaystyle \epsilon = \eta - f(\xi) $ as you should know

f is continuous at ξ that means for any $ \displaystyle \epsilon >0 $ there exist $ \displaystyle \delta >0 $

such that whenever $ \displaystyle \mid x - \xi \mid < \delta $ we have $ \displaystyle \mid f(x) - f(\xi) \mid < \epsilon $

remember that epsilon>0 so if you choose a wrong epsilon the criteria for the continuity will be affected, back to our proof after we choose epsilon sub it

$ \displaystyle \mid f(x) - f(\xi) \mid < \eta - f(\xi) $

$ \displaystyle -\eta + f(\xi) < f(x) - f(\xi) < \eta - f(\xi) $

$ \displaystyle f(x) - f(\xi) < \eta - f(\xi) \Rightarrow f(x) < \eta $ for

$ \displaystyle \mid x - \xi \mid < \delta \Rightarrow \xi - \delta < x < \xi + \delta $

but $ \displaystyle \xi $ is an upper bound for S, $ \displaystyle \xi < \xi +\frac{\delta}{2} $

and

$ \displaystyle \xi + \frac{\delta}{2} \in S $ since

$ \displaystyle f\left(\xi + \frac{\delta}{2}\right) < \eta $ but $ \displaystyle \xi $ is an upper bound for S contradiction!! so $ \displaystyle \eta - f(\xi) $ is not positive so still two choices less than or equal zero

Lets see if $ \displaystyle f(\xi) > \eta $

f is continuous at $ \displaystyle \xi $ , now pick $ \displaystyle \epsilon = f(\xi) - \eta $

there exist delta

$ \displaystyle \mid f(x) - f(\xi) \mid < f(\xi) - \eta $ for $ \displaystyle \mid x - \xi \mid < \delta $

$ \displaystyle -f(\xi) + \eta < f(x) - f(\xi) $

$ \displaystyle \eta < f(x) $ for $ \displaystyle \xi - \delta < x < \xi + \delta $

$ \displaystyle \xi - \delta $ is in S the definition of the upper bound which means $ \displaystyle f \left(\xi - \delta \right) $ should be less than $ \displaystyle \eta $ not larger contradiction so our epsilon is not positive

This idea from Wiki see