Showing discontinuity at infinitely many points

Click For Summary

Homework Help Overview

The problem involves a function f: ℝ → ℝ that takes each of its values exactly twice. The objective is to demonstrate that f is discontinuous at infinitely many points. The context suggests a focus on properties of continuous functions and their implications.

Discussion Character

  • Conceptual clarification, Assumption checking, Exploratory

Approaches and Questions Raised

  • Participants discuss the implications of the function's behavior, particularly in relation to continuity and the values it attains. There is an exploration of specific examples, such as f(x) = |x|, to illustrate potential discontinuities. Questions arise regarding the necessity of showing discontinuity at infinitely many points versus just at some points.

Discussion Status

The discussion is ongoing, with participants offering observations and questioning the validity of certain approaches. Some guidance has been provided regarding the implications of continuity and the mapping of values, but no consensus has been reached on a definitive method or conclusion.

Contextual Notes

Participants note the importance of the function mapping all of ℝ to ℝ, and there is mention of specific counterexamples that could influence the understanding of the problem. The original poster expresses significant difficulty in progressing with the problem.

delta_epsilon
Messages
2
Reaction score
0

Homework Statement



Suppose f: ℝ → ℝ takes on each of its values exactly twice; that is, for each y in ℝ, the set {x: y = f(x)} has either 0 or 2 elements. Show that f is discontinuous at infinitely many points.

Homework Equations



I don't know if this is relevant, but in the prior text to this problem, this was perhaps the only one that may be relevant:

If I is an interval in ℝ and if f: I → ℝ is a nonconstant continuous function, then f(I) is an interval. In particular, if a,b in I with f(a) ≠f(b), then f assumes every value between f(a) and f(b)

The Attempt at a Solution



I've tried going through definitions, theorems, etc. for several hours of the course of 3 days and I am still totally stumped. Ugh. Please, any help is much appreciated.
 
Physics news on Phys.org
I'd like to note how close f=|x| comes to breaking this!

So, I can get that the function isn't continuous, and perhaps this will help and perhaps it won't. Suppose it were continuous, and let y be in the range of f. Then consider I=[x_1,x_2] where f(x_1)=f(x_2)=y. Now, on I f attains a max, say m=f(x_3). Now, since f is continuous, f attains every value between y and m at least twice, once in [x_1,x_3] and once in [x_3,x_2]. Also, it can't attain m again on this interval or else you break the hypotheses, as values will be attained more than twice. Thus it attains m again somewhere outside the interval, and you should get that you attain values three times.

If this observation hasn't been made, perhaps you can tweek it to win the problem, though I'm not positive you can or cannot.
 
The fact that this function maps all of R to all of R is crucial. Otherwise, just take f(x)= |x| for x not 0, f undefined at 0 as a counter example.
 
HallsofIvy said:
The fact that this function maps all of R to all of R is crucial. Otherwise, just take f(x)= |x| for x not 0, f undefined at 0 as a counter example.

So are you saying blinktx411 solution is incorrect?
 
He's shown that the curve must be discontinuous but not that it must be discontinuous at infinitely many points.
 
Last edited:

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 39 ·
2
Replies
39
Views
6K
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
6K
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K