injective map between real numbers is a homeomorphism

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