21i 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?
15 Answers
21
1I 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 .
21according 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?
21secondly 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......????
21i strongly think that there is no such function....thinking for a proof:)
341Your 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)
341In 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
66@theprophet
The discontinuity of the Dirichlet function are of the second kind; they are not jump discontinuity.
1I 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 : (
341@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)
21@ sir
can u pls give me the prove of the fact that if f has removable discontinuities everywhere
it is not always discontinuous...
thank you
62To be true without disturbing the underlying wave of discussion here... there is no need to know such a function...
You can search for Weierstrass on the net if you want to know more...
A very good hint for such functions would be Fractals
Or even Snow Flakes (which seem to be non differentiable everywhere!) (Even though we have to reduce the domain to make them differentiable..)
62@Prophet Sir,
A simple justification
The discontinuties are not jump discontinuities becasue in any neighbourhoood of a rational number we will have infinitely many rationals and irrationals. So by removing a discontinuity, we are not going to fulfil the epsilon delta condition..
66@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.
341I 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.