I understand that I have to use the discontinuity criterion:Theorem: Let S ⊂ ℝ , let f : S → ℝ and let c ∈ S.Then f is discontinuous at c if and only if there exists a sequence { x n } in S such that lim ( x n ) = c, but the sequence { f ( x n ) } does not converge to f ( c ).I tried splitting up the function as f ( x ) = x 2 when x = p q , for p , q ∈ Z but I do not know how to define for irrational numbersI also don't know how to think of a sequence show the limit exists but it does not converge...

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I understand that I have to use the discontinuity criterion:Theorem: Let   S  ⊂      ℝ  , let   f  :  S  →      ℝ   and let   c  ∈  S.Then f is discontinuous at c if and only if there exists a sequence   {      x    n    } in S such that   lim  (      x    n    )  =  c, but the sequence   {  f  (      x    n    )  } does not converge to   f  (  c  ).I tried splitting up the function as   f  (  x  )  =      x    2   when   x  =      p    q  , for   p  ,  q  ∈      Z   but I do not know how to define for irrational numbersI also don&#039;t know how to think of a sequence show the limit exists but it does not converge...

Penelope Carson · Accepted Answer

If   x  ≠  0 is rational, then take       x    n    →  x with       x    n   irrational for all   n. Then:  0  =  f  (      x    n    )  →  0  ≠      x    2    =  f  (  x  )If   x  ≠  0 is irrational, then take       x    n    →  x with       x    n   rational for all   n. Then:      x    n    2    =  f  (      x    n    )  →      x    2    ≠  0  =  f  (  x  )In both cases,   f is discontinuous at   x.

I understand that I have to use the discontinuity criterion: Theorem: Let S ⊂ <mr

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