Answer to Question #217025 in Differential Geometry | Topology for Prathibha Rose

Question

Prove that the real line is a homeomorphic to the interval (0,1) with the subspace topology

Answer 1

Let us consider the function $f:(0;1)\to \mathbb{R}$ defined as $f(x)=\tan (-\frac{\pi}{2}+\pi x)$

This function is strictly increasing and thus is injective.

We know that $\lim_{x\to 0} f(x) = \lim_{t\to -\pi/2} \tan t = -\infty$ and $\lim_{x\to 1} f(x) = \lim_{t\to \pi/2} \tan t = +\infty$ and therefore it is surjective.

The function $x\to -\pi/2 +\pi x$ is continuous on $(0;1)$ and the function $\tan t$ is continuous on $(-\pi/2 ;\pi/2)$, so $f$ is continuous on $(0;1)$.

Finally, the inverse function defined as $g(y)= (\arctan(y)+\pi/2)/\pi$ is continuous on the real line $\mathbb{R}$. Therefore, the function $f$ realise a homeomorphism between $(0;1)$ and $\mathbb{R}$.