Open Real Interval is Homeomorphic to Real Number Line

Theorem

Let $\R$ be the real number line with the Euclidean topology.

Let $I := \openint a b$ be a non-empty open real interval.

Then $I$ and $\R$ are homeomorphic.

Proof 1

By definition of open real interval, for $I$ to be non-empty it must be the case that $a < b$.

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

Thus $a - b \ne 0$.

Let $I' := \openint {-1} 1$ denote the open real interval from $-1$ to $1$.

From Open Real Intervals are Homeomorphic, $I$ and $I'$ are homeomorphic.

Consider the real function $f: I' \to \R$ defined as:

$\forall x \in I': \map f x = \dfrac x {1 - \size x}$

Then after some algebra:

$\forall x \in \R: \map {f^{-1} } x = \dfrac x {1 + \size x}$

Both of these are defined, as $\size x < 1$.

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

Thus by definition $I'$ and $\R$ are homeomorphic.

From Homeomorphism Relation is Equivalence it follows that $I$ and $\R$ are homeomorphic.

$\blacksquare$

Proof 2

Let $I := \openint {-\dfrac \pi 2} {\dfrac \pi 2}$ denote the open real interval from $-\dfrac \pi 2$ to $\dfrac \pi 2$.

Consider the real function $f: I \to \R$ defined as:

$\forall x \in I: \map f x = \tan x$

Then we have:

$\forall x \in \R: \map {f^{-1} } x = \arctan x$

From Homeomorphism Relation is Equivalence it follows that $I$ and $\R$ are homeomorphic.

Then by Open Real Intervals are Homeomorphic, $I$ is homeomorphic to every other open real interval.

$\blacksquare$