I conjecture that if a function is continuous at a discreet point (say p), it is continuous at all points in some delta-neighborhood of p.
Am I correct? If yes - any pointers to prove this? If not - any counterexample.
For those who know the subject deeper - is this kind of formulation even important?
Thanks. But it sounds conter-intutive to me.
In layman's way, being continous at 'a' means 'very' close to 'a' value of the function varies 'very' little. So the same will apply to any point which is very very close to 'a' and thus I say it will be continous to at all points in some delta-neighbourhood of a.
I know I am talking an all vague language but it helps get a feel of definitons. Please excuse me if all this is not important/relevant.
Let be a rational number. Let's define the sequence by (note that this sequence converges to ). For all integer , is an irrationnal number hence . However so and is not continuous at . In other words, if is 'very large' then is 'very close to' but is 'not close to' . (this is true for all so it can be assumed that is 'very very close' to 0)
Using the same method it can be show that this result also holds if .