# Prove that any two open intervals are equinumerous.

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?

• You don't take $f$ just to be any function from $(0, 1)$ to $(m, n)$, you want to show that such function exists that is a bijection, and you do by saying that $f(x) = (n-m)x+m$ ($x$ is important here). Feb 14, 2019 at 13:41
• I would also add an assumption on the interval to be nonempty, otherwise $\emptyset =(0,0)$ makes trouble. Feb 14, 2019 at 13:42
Define, $$f:(a,b)\to (c,d)$$ by, $$f(x)=c+\frac{d-c}{b-a}(x-a)$$ for all $$x\in(a,b)$$