Prove or Disprove: There exists $f: \mathbb{R} \rightarrow \mathbb{R}$ that is onto and one-to-one but not continuous.

Question

There exists $f: \mathbb{R} \rightarrow \mathbb{R}$ that is onto and one-to-one but not continuous.

I believe this statement is wrong, because values exist in the function's image that will not be "used" due to the function not being continuous. And since the function is one to one, no other number in the domain will be able to use up that specific y value in the range/image.

Am I correct? What could I use for a more concrete proof?

Duplicate: http://math.stackexchange.com/q/1780909/339790 – Rodrigo de Azevedo May 16 '16 at 15:21 — Rodrigo de Azevedo, May 16 '16 at 15:21

score 3 · Answer 1 · answered May 16 '16 at 15:17

3

Take $f: \mathbb R \to \mathbb R$ defined by $f(x) = x$ if $x$ is rational, and $f(x) = - x$ if $x$ is irrational. This is one-to-one and onto but continuous nowhere (except at $x = 0$).

answered May 16 '16 at 15:17

User8128

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So this proves the statement is correct. Is there a concrete way to prove that the statement is true though ? – May 16 '16 at 15:21
@RonaldB: An example is a concrete proof. – carmichael561 May 16 '16 at 15:22

Prove or Disprove: There exists $f: \mathbb{R} \rightarrow \mathbb{R}$ that is onto and one-to-one but not continuous.

1 Answers1