Answer to Question #217081 in Differential Geometry | Topology for Prathibha Rose

Question

Prove that any two open intervals of the real.line are homeomorphic

Answer 1

Solution:

Proof:

Let's show that a bounded open interval (a, b) is homeomorphic to (0,1).

Consider the map f:(0,1) $\rightarrow$ (a, b) defined by f(t)=a+(b-a) t. It is continuous and bijective. It's inverse is continuous as well.

Now let's show that (0,1) is homeomorphic to (a, $\infty$ ). Consider g:(0,1) $\rightarrow(a, \infty)$ defined by

$g(t)=a+\tan (\pi t / 2)$

This is clearly bijective and continuous. Its inverse is continuous as well.

It's obvious that (-$\infty$, a) is homeomorphic to (-a, $\infty$).

Now we have that (-1,1) is homeomorphic to R using $h(t)=\tan (\pi t / 2)$

The property of being homeomorphic is transitive.

Hence, any two open intervals of the real line are homeomorphic.