Function continuous at irrationas and integers

Click For Summary

Discussion Overview

The discussion revolves around the existence of a real-valued function that is continuous at all irrational numbers and integers, but discontinuous everywhere else. Participants explore the definition and properties of such a function, examining its continuity and the implications of its construction.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant proposes a function defined as f(x) = 0 for irrationals and integers, and f(x) = 1/q for rational numbers p/q that are not integers, suggesting this function could be continuous at irrationals and integers.
  • Another participant agrees with the initial argument but raises concerns about the continuity of the proposed function, particularly questioning its definition on rationals.
  • A different participant challenges the clarity of the function's definition, suggesting that a requirement for p and q to be relatively prime might be necessary to avoid ambiguity.
  • Concerns are expressed about the continuity of the function at irrationals, with a suggestion that any delta neighborhood will include both rationals and irrationals, complicating the continuity argument.
  • One participant asserts that the original proposer has the right idea but emphasizes the need for the condition that p and q are expressed in lowest terms.
  • The original proposer acknowledges the oversight regarding the specification of rational numbers in lowest terms, indicating a refinement of their initial definition.

Areas of Agreement / Disagreement

Participants express differing views on the continuity of the proposed function, with some supporting the initial idea and others raising significant concerns. There is no consensus on the validity of the function's continuity or its definition.

Contextual Notes

Limitations include the lack of a precise definition of continuity being used by participants, and unresolved questions about the behavior of the function in neighborhoods of irrational numbers.

zolit
Messages
6
Reaction score
0
I have to either give an example or show that no such function exists:

A real valued function f(x) continuous at all irrationals and at all the
integers, but discontinuous everywhere else.

I think such function exists and I would define it as follows:

f(x) = 0 if x is an irrational or an integer
1/q if x is rational (p/q) but not an integer.

This way the proof is similar to the case for continuous for irrationas, discontinuous everywhere else. Each rational can be viewed as a limit of a sequence of irrationals, right? So those rationals that have the same value as that sequence (integers in my case) will be continuous, those that don't won't be.
Does this sound right?
 
Physics news on Phys.org
Your argument seems very right to me.
 
It is not clear to me that your function as defined is continuous anywhere,

I am having troubles with your function on the rationals, it is not clear that it is even a function.

[tex]f ( \frac 1 2 ) <> f( \frac 2 4 )[/tex]

Seems like you would need to throw in a requirement that P and q must be relatively prime. But then, I think you will be able to find a rational (y) in every neighborhood of an irrational (x) such that f(y) is not in the corresponding neighborhood of f(x). so your function on the irrationals is not continuous.

I know that I have not expressed this precisely, but since I do not know what definition of continuity you are working with, I have tried to make it general. Do a bit of analysis along the lines I am describing, see if it holds. I have not done a detailed analysis and may be wrong, that is your job! :smile:
 
Also think about it like this. Any delta neighborhood is going to include both rationals and irrationals. Therefore for any x_0, you can find an x such that |f(x) - f(x_0)| (is not less than) epsilon.

I really think there is a way to do this though, just think about how you can satisfy your definitions.
 
zolit does, in fact, have the right function. (once he adds the condition that p and q are relatively prime)
 
Yeah, that's what I forgot to specify: all rational non-integers p/q expressed in the lowest order. this takes cares of 2/4, etc.
Thanks for all input.
 

Similar threads

Undergrad Uniform convergence of family of functions with continuous index
  • · Replies 3 ·
Replies
3
Views
3K
Undergrad Continuity of Mean Value Theorem
  • · Replies 12 ·
Replies
12
Views
3K
Undergrad Should the function f to be continuous for applying MVTI or not?
  • · Replies 4 ·
Replies
4
Views
2K
High School Can the continuity of functions be defined in the field of rational numbers?
  • · Replies 5 ·
Replies
5
Views
3K
Undergrad Doubt about theorem in Calculus on Manifolds
  • · Replies 9 ·
Replies
9
Views
4K
Undergrad Largest subset in which a function is continuous
  • · Replies 1 ·
Replies
1
Views
2K
Undergrad A counterexample to "the integral of the limit is the limit of the integral"
  • · Replies 1 ·
Replies
1
Views
1K
Graduate Continuity proved by differentiation
  • · Replies 5 ·
Replies
5
Views
2K
Undergrad Riemann Integral: Does g(x)=f(x) Almost Everywhere?
  • · Replies 28 ·
Replies
28
Views
8K
Graduate Spivak Thomae's Function proof explanation
  • · Replies 35 ·
2
Replies
35
Views
8K