Question #217081

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


1
Expert's answer
2021-07-29T11:47:03-0400

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) (a,)\rightarrow(a, \infty) defined by

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

The property of being homeomorphic is transitive.

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


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