Prove that $[0, \infty)$ is not order isomorphic to $(\frac{1}{2}, \infty)$. Prove that $[0, \infty)$ is not order isomorphic to $(\frac{1}{2}, \infty)$.
My attempt: Let $f: [0, \infty)\rightarrow (\frac{1}{2}, \infty)$ is an order isomorphism.
Let $a= f(0) \in (\frac{1}{2}, \infty)$. Now choose $b$ such that $(\frac{1}{2}<b<a$. Now  I can  not proceed further.
How to construct a contradiction?
 A: There is just not a minimum element in $(\frac{1}{2},\infty)$. Since $f(0)$ must be the least element of $(\frac{1}{2},\infty)$, let's say $f(0)=b$. Suppose $b \in (\frac{1}{2},\infty)$ is the minimum of the interval. By definition of open set, you have that $(b-\delta , b + \delta) \subseteq (\frac{1}{2},\infty)$ with $\delta > 0$ sufficiently small, then $b- \frac{\delta}{2} \in (\frac{1}{2},\infty)$ but this is a contradiction (because we have found an element in $(\frac{1}{2},\infty)$ smaller than $b=f(0)$). So they cannot be order isomorphic
A: Let $b_1 = f^{-1}(b)$. As $b \lt a$ and $f$ is an order isomorphism, you need to have $b_1 = f^{-1}(b) \lt f^{-1}(a) = 0$. A contradiction.
A: An order isomorphism between two ordered sets must preserve any property describable using only the order properties.
One of these ordered sets has a minimum element and the other does not, so they can't be order isomorphic. There's no need to explain why any particular possible order isomorphism isn't one.
A: An order isomorphism $f\colon X\to Y$ is a bijection such that, for all $a,b\in X$, it holds
$$
a<b \iff f(a)<f(b)
$$
which is equivalent to saying that $f$ is increasing and the inverse map is increasing as well.
Requiring that $f$ is bijective and increasing is not sufficient.
A simple counterexample is the set $X=\{1,2\}$ with the relation $\{(1,1),(2,2)\}$ and $Y=\{1,2\}$ with the relation $\{(1,1),(1,2),(2,2)\}$. The identity map $f\colon X\to Y$ satisfies “for all $a,b\in X$, if $a<b$ then $f(a)<f(b)$”, but the inverse map doesn't satisfy the similar condition.
In your case, set $a=f(0)$ and choose $b$ such that $b<a$ (which surely exists). Then $f^{-1}(b)<f^{-1}(a)$ by definition of order isomorphism and this is a contradiction, because $f^{-1}(b)<0$ is impossible.
