A question on continuous function

[priyanka@]

can anybody please help me in solving the following question:

consider the function on [0,1] f(x)=1/q if x=p/q, p&q are non zero & p,q are positive integers,
p/q is in simplest form.
= 0 if x=0 or irrational

need to show the set of discontinuities of f(x) is the set of all non zero rationals in [0,1]

 

[losiu99]

Quick sketch:
Show that for each real number a, limit of d as x goes to a is 0. You can do that by showing that rationals with bounded numerator are nowhere dense. From this everything you need to prove follows trivially.
Good luck!

 

[priyanka@]

Thanks for the help..but I'm still not very clear with your answer. I was thinkng of doing this by using sequential continuity..like takng a random seq. <an> of non zero rationals converging to a non zero rational x. And then claiming <f(an)> not converging to f(x)..so don't know whether this approach is correct..

 

[losiu99]

Yes, it's correct. By taking any convergent sequence, we can show that limit of a function is zero at every point. It can be done from Cauchy definition of limit as well. Given a real number x, and arbitrary $$\epsilon$$, we show that in some neighbour of x rationals have denominators (I've made a mistake in previous post) large enough for function to have values less than $$\epsilon$$. Your approach (sequences) is essentially the same.

 

[DrRocket]

can anybody pls help me in solving d following question:

consider d function on [0,1] f(x)=1/q if x=p/q, p&q are non zero & p,q are +ve integers,
p/q is in simplest form.
= 0 if x=0 or irrational

need to show the set of discontinuities of f(x) is the set of all non zero rationals in [0,1]

This is not a particularly difficult question, but it most certainly sounds like a question from a class -- either homework or a test question.

So please explain the origin of the question and why providing you with an approach or an answer is consistent within the ethics of the course -- is not cheating.

When I took such classes, obtaining outside help on a question like this would have clearly been considered cheating.

 

[priyanka@]

this is neithr a homework nor a test question...m doing my msc. now...its a question of riemann integral dat i studied in b.sc...was going thru my notes...v actually need 2 show dis function is riemann integrable bt v cannot use d result"a bdd. functn for which d set of discontnuities has finfitely many limit points is riemann integrable." as dis isn't d case here..my prof. told me d set of continuities of dis fiunction bt i jst wantd 2 verify myself...was getng lil confused...so askd 4 help...jst 2 add 2 my knowledge...dats it..n nt 2 get gud marks or appreciation frm d prof...m nt into all dese thngs...

 

[DrRocket]

this is neithr a homework nor a test question...m doing my msc. now...its a question of riemann integral dat i studied in b.sc...was going thru my notes...v actually need 2 show dis function is riemann integrable bt v cannot use d result"a bdd. functn for which d set of discontnuities has finfitely many limit points is riemann integrable." as dis isn't d case here..my prof. told me d set of continuities of dis fiunction bt i jst wantd 2 verify myself...was getng lil confused...so askd 4 help...jst 2 add 2 my knowledge...dats it..n nt 2 get gud marks or appreciation frm d prof...m nt into all dese thngs...

It is Riemann integrable because the set of discontinuities has Lebesgue measure 0.

 

[priyanka@]

thnx..bt i knew dis function is rieman integrble...was stuck up sumwhere else...

 

[priyanka@]

Yes, it's correct. By taking any convergent sequence, we can show that limit of a function is zero at every point. It can be done from Cauchy definition of limit as well. Given a real number x, and arbitrary $$\epsilon$$, we show that in some neighbour of x rationals have denominators (I've made a mistake in previous post) large enough for function to have values less than $$\epsilon$$. Your approach (sequences) is essentially the same.

thank you for the help..i have got it..