# 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 $\alpha. Therefore $f(\alpha) and (because continuous functions are Darboux functions) for any $y^{\prime}\in\big{(}f(\alpha),f(\beta)\big{)}$ there exists $x^{\prime}\in I$ such that $f(x^{\prime})=y^{\prime}$. This shows that $\big{(}f(\alpha),f(\beta)\big{)}$ is an open neighbourhood of $y$ contained in $f(I)$ and therefore (since $y$ was arbitrary) $f(I)$ is open. $\square$

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

 $f(U)=f\bigg{(}\bigcup_{\alpha}I_{\alpha}\bigg{)}=\bigcup_{\alpha}f(I_{\alpha})% =\bigcup_{\alpha}f_{\alpha}(I_{\alpha})$

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

Title injective map between real numbers is a homeomorphism InjectiveMapBetweenRealNumbersIsAHomeomorphism 2013-03-22 18:53:58 2013-03-22 18:53:58 joking (16130) joking (16130) 4 joking (16130) Theorem msc 54C05