A question on continuous function

  • Thread starter priyanka@
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  • #1
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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]
 
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Answers and Replies

  • #2
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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!
 
  • #3
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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 dont know whether this approach is correct..
 
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  • #4
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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 [tex]\epsilon[/tex], 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 [tex]\epsilon[/tex]. Your approach (sequences) is essentially the same.
 
  • #5
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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.
 
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  • #6
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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 isnt 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...
 
  • #7
112
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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 isnt 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.
 
  • #8
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thnx..bt i knew dis function is rieman integrble...was stuck up sumwhere else...
 
  • #9
5
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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 [tex]\epsilon[/tex], 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 [tex]\epsilon[/tex]. Your approach (sequences) is essentially the same.

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

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