Prove function continuous at only one point

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SUMMARY

The function defined as f(x) = x for rational x and f(x) = -x for irrational x is continuous only at the point 0. The proof involves demonstrating that for any point a ≠ 0, there exist sequences converging to a that yield different limits when applying the function f. Specifically, one sequence of rational numbers converging to a results in the limit a, while a sequence of irrational numbers converging to a results in -a. This discrepancy confirms that f is not continuous at any point other than 0.

PREREQUISITES
  • Understanding of limits and continuity in real analysis
  • Familiarity with rational and irrational numbers
  • Knowledge of sequence convergence
  • Basic principles of mathematical proofs
NEXT STEPS
  • Study the formal definition of continuity in real analysis
  • Learn about sequences and their convergence properties
  • Explore examples of piecewise functions and their continuity
  • Investigate the implications of the epsilon-delta definition of limits
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Students of mathematics, particularly those studying real analysis, and educators looking to clarify concepts of continuity and limits in piecewise functions.

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Homework Statement


Prove that the function defined as f(x)= x when x is rational and -x when x is irrational is only continuous at 0.


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The Attempt at a Solution


I have been looking at this website which proves this:
http://planetmath.org/encyclopedia/FunctionContinuousAtOnlyOnePoint.html

I don't understand how the writer proves that the function is not continuous when a[tex]\neq[/tex]0.
Why do they take xk= to something? How does this help them show that the limits equal a and -a.
How does showing that prove that f is not continuous at a?

I know those are a lot of question, but maybe an answer to even just the first ones would help me understand this proof.
 
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Do you agree that
"f is continuous at a, if for all sequences xk that converge to a, f(xk) converges to f(a)" ?

So then to show that f is not continuous at a, all you need is to find one sequence that does converge to a but for which the limit of the f(xk) is not f(a). (The negation of "for all sequences S, P(S)" is "there exists a sequence S such that not P(S)").

So what they do is take two different sequences which both have limit a, and show that when you apply f first and take the limit, you get two different results. Since f(a) cannot be equal to a and -a at the same time, at least one of this sequence forms a counter-example to the statement of continuity they start out with.
 

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