Open Real Intervals are Homeomorphic

From ProofWiki
Jump to navigation Jump to search


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

Let $I_1 := \openint a b$ and $I_2 := \openint c d$ be non-empty open real intervals.

Then $I_1$ and $I_2$ are homeomorphic.


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:

$\forall x \in I_1: \map f x = c + \dfrac {\paren {d - c} \paren {x - a} } {b - a}$

Then after some algebra:

$\forall x \in I_2: \map {f^{-1} } x = a + \dfrac {\paren {b - a} \paren {x - c} } {d - c}$

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

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.

Hence the result by definition of homeomorphic.