Open Real Intervals are Homeomorphic

Theorem
Consider the real numbers $\R$ as a metric space under the Euclidean metric.

Let $I_1 := \left({a \,.\,.\, b}\right)$ and $I_2 := \left({c \,.\,.\, d}\right)$ be non-empty open real intervals.

Then $I_1$ and $I_2$ are topologically equivalent.

Proof
By definition of open real interval, for $I_1$ and $I_2$ to be non-empty it must be the case that $a < b$ and $c < d$.

In particular it is noted that $a \ne b$ and $c \ne d$.

Thus $a - b \ne 0$ and $c - d \ne 0$.

Consider the real function $f: I_1 \to I_2$ defined as:
 * $\forall x \in I_1: f \left({x}\right) = c + \dfrac {\left({d - c}\right) \left({x - a}\right)} {b - a}$

Then after some algebra:
 * $\forall x \in I_2: f^{-1} \left({x}\right) = a + \dfrac {\left({b - a}\right) \left({x - c}\right)} {d - c}$

Both of these are defined as $a - b \ne 0$ and $c - d \ne 0$.

By the Combination Theorem for Continuous Functions, both $f$ and $f^{-1}$ are continuous on the open real intervals on which they are defined.

Hence the result by definition of topologically equivalent.