This is Lay's exercise $8.4.b$. Prove that any two open intervals are
equinumerous.
Is my proof correct? And even if it is how can I make it better? Is there a better alternative?
Consider two open intervals $(0,1)$ and $(m,n)$ for any $m,n\in\Bbb R$ such that there exists a function $f$ such that $f:(0,1)\longrightarrow (m,n)$. Let $f(x)=(n-m)x+m$. Then we can see that $f$ is a bijection between $(0,1)$ and $(m,n)$. Now consider another open interval $(p,q)$ such that there exists a bijective function $g$ such that $g:(0,1)\rightarrow (p,q)$. From the bijective functions $f$ and $g$ we get the bijective function $h$ such that $h=g\circ f^{-1}:(m,n)\longrightarrow (p,q)$. Since $(m,n)$ and $(p,q)$ were arbitrary we can conclude that any two open intervals are equinumerous.
I think my proof is correct but since the statement seems to be a general one how could it be proved so simply?
Thank you in advance
Best Answer
Just an observation. You may take it as an answer.
Define, $f:(a,b)\to (c,d)$ by, $$f(x)=c+\frac{d-c}{b-a}(x-a)$$ for all $x\in(a,b)$
This is basically an open bijective continuous map i.e. Homeomorphism between any two open intervals.