Consider the question:
Show that any two open intervals of $\mathbb{R}$ are isomorphic.
Definition. Given two well-ordered sets $(A, <_A)$, $(B, <_B)$, we say that $A$ and $B$ are isomorphic if there is a bijective $f: A \to B$ such that
- for all $a,b \in A$ such that $a<_A b$, then $f(a)<_B f(b)$.
my attempt:
Note that, for a function to be order-preserving, it suffices for it to be monotonically increasing. Since $\mathbb{R} \approx \;\;]0,1[\;\;$ (are isomorphic). We have the cases for $I \subseteq\mathbb{R}$:
$I = \;\;]a,b[\;\;$ with $a<b$. Let $f: \;\;]0,1[\;\;\to \;\;]a,b[\;\;$ given by $$f(x) = a+(b-a)x$$ which is incresing for $a<b$.
$I = \;\;]a, \infty[\;\;$. Let $f$ be given by $$f(x) = a+\tan\left(\frac{\pi x}{2} \right)$$ which is increasing in the interval $\;\;]0,1[\;\;$.
$I = \;\;]-\infty,b[\;\;$, Here is where I'm having trouble, I (for some reason) cannot think of an increasing bijective function from $\;\;]0,1[\;\;$ to $\;\;]-\infty, b[\;\;$.
Since we can establish a order-preseving function from any two open subsets, we can find an order-preserving composite function from one open interval to another.