(dis)prove this function does not have a limit at zero.

[Hodgey8806]

Homework Statement

The function g:ℝ→ℝ defined by "g(x) = 0 for x being rational and g(x) = 1 for x being irrational" does not have a limit at zero.

Homework Equations

I have to use the definition of the limit of a function at a point--I can't use the sequential criterion this time.

So the definition of the limit of a function at a point is: (assuming function and 0 being the cluster point)
A real number L is said to be a limit of f at c if, given any ε>0, there exists a δ>0 s.t. if x is in the domain, and 0<|x-c|<δ, then |f(x) - L|<ε.

Thus the negation of the definition is:
A real number L is NOT a limit of f at c if there exist ε>0 s.t. for all δ>0, x is in the domain, 0<|x-c|<δ, AND |f(x)-L|≥ε.

The Attempt at a Solution

I am going to prove this true by using the negation of the definition.

Proof:
Assume the limit of L ≤ 0. Choose ε=1/2.
By denseness of the irrationals, for all δ>0, there exists x in the irrationals s.t. 0<|x|<δ, but |f(x)-0|=|1-L|≥1 ≥ 1/2=ε. Thus L≤0 is not a limit of g.

Assume the limit of L ≥ 1. Choose ε=1/2.
By denseness of the rationals, for all δ>0, there exists x in the rationals s.t. 0<|x|<δ, but |f(x)-L|=|0-L|≥1 ≥ 1/2=ε. Thus L≥1 is not a limit of g.

Assume the limit of 0≤ L ≤ 1/2. Choose ε=1/2.
By denseness of the irrationals, for all δ>0, there exists x in the irrationals s.t. 0<|x|<δ, but |f(x)-L|=|1-L|≥1/2=ε. Thus 0≤ L ≤ 1/2 is not a limit of g.

Assume the limit of 1/2< L ≤ 1. Choose ε=1/2.
By denseness of the rationals, for all δ>0, there exists x in the rationals s.t. 0<|x|<δ, but |f(x)-L|=|0-L|>1/2=ε. Thus 1/2< L ≤ 1 is not a limit of g.

Thus, we have proved for ever L in the reals that L cannot be a limit of this function.
Q.E.D

 

[Dick]

Way too many cases, don't you think? Use the triangle inequality. If r is rational and i is irrational then |f(r)-f(i)|=|f(r)-L+L-f(i)|<=|f(r)-L|+|f(i)-L|.

 

[Hodgey8806]

Hmm, I don't quite see where this is going though. I understand the triangle inequality, but I don't understand its usefulness here. Do I need to show the epsilon as well? I'm assuming there is a way to "compact" what I did into this inequality, but I'm having trouble seeing that. Thank you very much :)

 

[Hodgey8806]

Oh ok! I think I see it now! We have
1=|f(r)-f(i)|=|f(r)-L+L-f(i)|<=|f(r)-L|+|f(i)-L|<=ε+ε=2ε by our definition for all ε>0
Taking ε=1/4, we are given:
1<=1/2, which is our contradiction. Thus, g does not have a limit at 0.
Q.E.D.

How is that?

 

[Dick]

Oh ok! I think I see it now! We have
1=|f(r)-f(i)|=|f(r)-L+L-f(i)|<=|f(r)-L|+|f(i)-L|<=ε+ε=2ε by our definition for all ε>0
Taking ε=1/4, we are given:
1<=1/2, which is our contradiction. Thus, g does not have a limit at 0.
Q.E.D.

How is that?

Right. So the value of L doesn't really matter. No need to break into cases.

 

[Hodgey8806]

Oh ok, great! Thank you so much! This is genius and, for me at least, it is elegant. Haha, my teacher will be very impressed I think.

 

[Dick]

Oh ok, great! Thank you so much! This is genius and, for me at least, it is elegant. Haha, my teacher will be very impressed I think.

The thinking, of course, is just that 0 and 1 can't possibly both be contained in an interval whose length is less than 1. The triangle inequality is just the formal way of saying that.

 

[Hodgey8806]

I see that now, and this application I think will be very useful for future problems. I have seen similar proofs even in our text that use the triangle inequality in such a way. But it was always just a little "in the dark" for me. However, I think this will very much help me see how to do a problem like this much more elegantly!