# another proof of the non-existence of a continuous function that switches the rational and the irrational numbers

Let $\mathrm{\pi \x9d\x95\x81}=\mathrm{\beta \x84\x9d}\beta \x88\x96\mathrm{\beta \x84\x9a}$ denote the irrationals.
There is no continuous function^{} $f:\mathrm{\beta \x84\x9d}\beta \x86\x92\mathrm{\beta \x84\x9d}$
such that $f\beta \x81\u2019(\mathrm{\beta \x84\x9a})\beta \x8a\x86\mathrm{\pi \x9d\x95\x81}$ and $f\beta \x81\u2019(\mathrm{\pi \x9d\x95\x81})\beta \x8a\x86\mathrm{\beta \x84\x9a}$.

Proof

Suppose $f$ is such a function. Since $\mathrm{\beta \x84\x9a}$ is countable^{}, $f\beta \x81\u2019(\mathrm{\beta \x84\x9a})$ and $f\beta \x81\u2019(\mathrm{\pi \x9d\x95\x81})$ are also countable. Therefore the image of $f$ is countable. If $f$ is not a constant function, then by the intermediate value theorem the image of $f$ contains a nonempty interval, so the image of $f$ is uncountable. We have just shown that this isnβt the case, so there must be some $c$ such that $f\beta \x81\u2019(x)=c$ for all $x\beta \x88\x88\mathrm{\beta \x84\x9d}$. Therefore $f\beta \x81\u2019(\mathrm{\beta \x84\x9a})=\{c\}\beta \x8a\x82\mathrm{\pi \x9d\x95\x81}$ and $f\beta \x81\u2019(\mathrm{\pi \x9d\x95\x81})=\{c\}\beta \x8a\x82\mathrm{\beta \x84\x9a}$. Obviously no number is both rational and irrational, so no such $f$ exists.

Title | another proof of the non-existence of a continuous function that switches the rational and the irrational numbers |
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Canonical name | AnotherProofOfTheNonexistenceOfAContinuousFunctionThatSwitchesTheRationalAndTheIrrationalNumbers |

Date of creation | 2013-03-22 16:23:54 |

Last modified on | 2013-03-22 16:23:54 |

Owner | neapol1s (9480) |

Last modified by | neapol1s (9480) |

Numerical id | 7 |

Author | neapol1s (9480) |

Entry type | Proof |

Classification | msc 54E52 |