# injective map between real numbers is a homeomorphism

Lemma. Assume that $I$ is an open interval and $f:I\to \mathbb{R}$ is an injective^{}, continuous map^{}. Then $f(I)\subseteq \mathbb{R}$ is an open subset.

Proof. Since $f$ is injective, then of course $f$ is monotonic^{}. Without loss of generality, we may assume that $f$ is increasing. Let $y=f(x)\in f(I)$. Since $I$ is open, then there are $\alpha ,\beta \in I$ such that $$. Therefore $$ and (because continuous functions^{} are Darboux functions) for any ${y}^{\prime}\in (f(\alpha ),f(\beta ))$ there exists ${x}^{\prime}\in I$ such that $f({x}^{\prime})={y}^{\prime}$. This shows that $(f(\alpha ),f(\beta ))$ is an open neighbourhood of $y$ contained in $f(I)$ and therefore (since $y$ was arbitrary) $f(I)$ is open. $\mathrm{\square}$

Proposition^{}. Assume that $I$ is an open interval and $f:I\to \mathbb{R}$ is an injective, continuous map. Then $f$ is a homeomorphism onto image.

Proof. Of course, it is enough to show that $f$ is an open map. But if $U\subseteq I$ is open, then there are disjoint, open intervals ${I}_{\alpha}$ such that

$$U=\bigcup _{\alpha}{I}_{\alpha}.$$ |

Therefore we obtain continuous, injective maps ${f}_{\alpha}:{I}_{\alpha}\to \mathbb{R}$ which are restrictions^{} of $f$ to ${I}_{\alpha}$. By lemma we have that ${f}_{\alpha}({I}_{\alpha})$ is open and therefore

$$f(U)=f\left(\bigcup _{\alpha}{I}_{\alpha}\right)=\bigcup _{\alpha}f({I}_{\alpha})=\bigcup _{\alpha}{f}_{\alpha}({I}_{\alpha})$$ |

is open. This shows that $f$ is a homeomorphism onto image. $\mathrm{\square}$

Title | injective map between real numbers is a homeomorphism |
---|---|

Canonical name | InjectiveMapBetweenRealNumbersIsAHomeomorphism |

Date of creation | 2013-03-22 18:53:58 |

Last modified on | 2013-03-22 18:53:58 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 4 |

Author | joking (16130) |

Entry type | Theorem |

Classification | msc 54C05 |