Continuity of Dirichlet looking function

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The function f(x) is continuous only at x = 0, as approaching any other point c with rational numbers yields c, while approaching with irrational numbers results in a limit of 0. The concept of density is crucial, as both rational and irrational numbers are densely packed in the real numbers, allowing sequences from both sets to approach any real number. The discussion clarifies that a sequence of irrationals can be constructed to converge to any rational number, demonstrating the limits involved. The proof of density can be technical, but examples provided illustrate the concept effectively at a second-year level. Overall, the function is not continuous at any point other than zero due to differing limits from rational and irrational approaches.
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Homework Statement



Where is the function f(x) continuous?

f(x) =
x, if x is rational

0, if x is irrational

Homework Equations





The Attempt at a Solution



Is this correct?: I approach some c =/= 0, 1st through x's that are rational
and prove there is the limit c, and then approach through x's that are irrational and prove that the limit now cannot be c, that now I can conclude that the limit at c does not exist, and hence the function is not continuous at any c=/=0?

If no, why, and in what other way must I solve it then? If yes, please try to explain as rigourously as you can why this can be done. Remind you though, don't get technical above 2nd year in which I am. Thank you.
 
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You have it right. If you approach through irrationals, the limit is always zero. Which is not equal to c unless c=0. There's not really much to get technical about beyond the statement that both the rationals and irrationals are dense in R.
 
Well you just did! Sorry, but by dense you mean that: for any pair of irrationals there is a rational between them, and the same the other way? This is what I vaguely remember I read some day on the web. How is it that dense implies you can do that? Thanks again.
 
A set S is dense if for any x you can find an s in S that is as close to x as you want. Which would let you say for any x you can find a sequence of rationals approaching it and a sequence of irrationals. Do you think you are expected to prove that?
 
No way! But how is it that the rationals and irrational are proven to be dense?
 
just basically showing that the limit as s in S approaches x is x itself, and that's it?
 
The proof depends on how you define an irrational number and can be sort of 'technical'. But try this, if x is rational, then {x+sqrt(2)/n} is a sequence of irrationals approaching x. If you think of an irrational in terms of it's decimal expansion then for example sqrt(2) is the limit of the sequence 1,1.4,1.41,1.414,... Each term in the series adds one more decimal place to the expansion and each term in the series is rational. How's that for 2nd year level?
 
Sweet! :) Even my neighbor can get that. I'll go there now and show it to him (joking). Thanks so much.
 

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