FUNCTION

i think there is no such function

bcz if we r to draw the graph of such function ,we cant help leaving the paper blank..

wat u ppl think?

I think there indeed exists such a function , namely ," Dirichlet ' s Function ". You can check the details about it on Wikipedia . I ' ll just give its definition -
f ( x ) = 1 ............if " x " [ = pq ] is a rational number ,
= 0 ................. if " x " is an irrational number .

according to your definition i think it can be continuous in irrational points..

because i think an irrational point is surrounded by many other irrational points ( am i wrong here? )

who are infinitely close to each other. so if zero is given for all of them surely continuity will occur ..

what do u say?

secondly consider a point which is infinitely close to a rational point (one side) and an irrational point (other side) , then we cant say that it has removable discontinuities......????

i strongly think that there is no such function....thinking for a proof:)

Your hunch is right, because the points of removable discontinuities is a countable set (this is broadly because to every point of discontinuity you can map a rational number and the rational numbers are countable)

Ricky's example doesnt count as the discontinuities are all jump discontinuities in this case. Make that f(x) = 1/q when x=p/q in its lowest terms and you have removable discontinuities at every point of discontinuity i.e. the rational numbers(This is known as Thomae's function)

In fact, there is a discussion that may be relevant to you:

http://math.stackexchange.com/questions/3777/is-there-a-function-with-a-removable-discontinuity-at-every-point

@theprophet
The discontinuity of the Dirichlet function are of the second kind; they are not jump discontinuity.

I am very sorry , I did not notice the part " removable discontinuities " in the original post , I just thought it was " discontinuities " only . One again , Sorry : (

@kaymant you are right. they arent jump discontinuities either, because even the limit does not exist anywhere.

(Because as I learnt later even the jump discontinuities form a countable set)

@Nishant,
What exactly that justification means? And what does "..... removing a discontinuity, we are not going to fulfil the epsilon delta condition..." exactly means? As has been already said, the discontinuities are not removable.

I was out of town so i couldnt reply.

I was only dimly aware of this fact, as I had come across a proof in case of jump discontinuities for monotonic functions, where the proof is as i have indicated (the reference is from a book by Zorich, a russian author)

The same proof is extended for other functions too, but here there are concepts of σ-algebra of sets, again of which i have read very little in trying to self-learn concepts like measure theory.

I could of course throw stuff at you from sources on the internet, but I would rather wait till i learn it in the normal course in an organic manner. Till then i wouldnt venture to explain the proof.

Also, maybe you could save your interest in these matters for later, because they involve concepts way beyond even an undergraduate math student's grasp.