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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

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    $\begingroup$ looks good at first glance! $\endgroup$
    – Alberto Takase
    Feb 14, 2019 at 13:39
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    $\begingroup$ 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). $\endgroup$
    – Jakobian
    Feb 14, 2019 at 13:41
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    $\begingroup$ I would also add an assumption on the interval to be nonempty, otherwise $\emptyset =(0,0)$ makes trouble. $\endgroup$
    – Felix
    Feb 14, 2019 at 13:42
  • $\begingroup$ @Jakobian Thank you. I just forgot that. $\endgroup$
    – Heptapod
    Feb 14, 2019 at 13:48

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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.

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