
#1
Apr1608, 10:59 AM

P: 2

1. The problem statement, all variables and given/known data
f(x) = {x^2 [tex]x \in Q[/tex] x^2 [tex]x \in R/Q [/tex] At what points is f continuous? 2. Relevant equations continuity: for every [tex]\epsilon > 0 [/tex] there exists [tex]\delta > 0 d(f(x),f(p)) < \epsilon[/tex] for all points [tex]x\inE[/tex] for which d(x,p) < [tex]\delta[/tex] 3. The attempt at a solution Alright my initial thought was that it would not be continuous at any point in Q, because for any two rationals there is an irrational between them (this is correct?), but then it would be continuous at all irrationals from a theorem (4.6 in Rudin) for [tex]p\in Q[/tex], lim(x> p) f(x) = p^2 [tex]\neq[/tex] p^2 = f(p) However, then this function is continuous at irrationals. For [tex]p\in R/Q[/tex], lim(x> p) f(x) = p^2 = f(p) is this reasoning sound ok? 



#2
Apr1608, 11:57 AM

Sci Advisor
HW Helper
PF Gold
P: 4,768

I didn't bother understanding what you tried (it's not very clear)... but you seem to be playing with the right idea, using the density of Q and R\Q to show failure of continuity. You should be able to use this reasoning to show that f is discontinuous everywhere but at one point.




#3
Apr1608, 01:22 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,882

You can say that f(x) is continuous at x_{0} if and only if lim f(x_{n})= f(x_{0}) for every sequence {x_{n}} converging to x_{0}. But you can't restrict that to rational numbers only or irrational numbers only.
Notice that the f(x_{n}) will be close to f(x_{0}) for both rational and irrational x_{n} if and only if x_{0}^{2}= x_{0}^{2}. For what x_{0} is that true? 


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