Exercise 2.3.4 (Open real intervals are homeomorphic)

Proof. We break down the proof into three cases: finite, semi-finite and infinite open sets. In each of the cases, we demonstrate that the interval in question is homeomorphic to the unit interval $(0,1)$.

• Finite intervals $(a,b)$.
  Consider the function

  $$f:(a,b)\to (0,1),x\mapsto \frac{x-a}{b-a},$$

  which shifts a point $x\in (a,b)$ in the direction of $0$ by subtracting $a$ and then scales to the unit interval by dividing by the length of the original interval $b-a$. It’s easy to see that $f$ is continuous as it is simply the identity function modified by constants. Similarly we deduce that the inverse

  $$f^{-1}:(0,1)\to (a,b),y\mapsto y(b-a)+a$$

  is continuous as well. It is easy to verify that $f$ is indeed bijective, i.e., $f\circ f^{-1}=f^{-1}\circ f=\mathrm{id}<!--\; FUNCTION\; APPLICATION\; -->$; thus, $f$ is a homeomorphism.

• Semi-finite intervals $(-\infty ,a)$ or $(a,+\infty )$.
  Consider the interval $(a,+\infty )$. The function defined by

  $$f:(a,+\infty )\to (0,1),x\mapsto \frac{(x-a)}{1+(x-a)}$$

  is a continuous mapping; its continuous inverse is given by

  $$f^{-1}:(0,1)\to (a,+\infty ),y\mapsto \frac{y+a(1-y)}{(1-y)}.$$

  The case for $(-\infty ,a)$ follows similarly.

• The infinite interval $(-\infty ,+\infty )$.
  Consider the inverse tangent function:

  $$f:(-\infty ,+\infty )\to \left(-\frac{\pi }{2},\frac{\pi }{2}\right),x\mapsto {\tan <!--\; FUNCTION\; APPLICATION\; -->}^{-1}(x)$$

  The inverse tangent function is known to be a continuous bijective function; thus, establishing a homomorphism between $ℝ$ and a finite open interval. Since finite open intervals are homeomorphic to $(0,1)$, so is $ℝ$.