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To prove the original statement, I invented (or discovered?) a function $f:\mathbb [a,b]\to[0,1]$ such that $f(x)=\frac1{b-a}x-\frac{a}{b-a}$ and proved that the function is bijective. Now I'm being asked to prove a similar relation, $(a,b)\sim(0,1)$. Would the same function, with similar ideas work in this case?

(I'm not sure if my tag is very adequate for this.)

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    Look at the graph of your function, or simply compute $f(a)$ and $f(b)$; you’ll see that the same function works. – 2017-01-08

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As your bijection $f \colon [a,b]\to [0,1]$ maps $a$ to $0$ and $b$ to $1$, its restriction to $(a,b)$, that is the map $f|_{(a,b)}$ is a bijection $(a,b) \to (0,1)$.

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    Hmm, is it really that simple? I can just change the codomain like that and not justify it? What if I wanted to restrict $f$ to $(a,\frac{a}2)$, I could just say $f|_{(a,\frac{a}2)}$ is a bijection $(a,\frac{a}2)\to (0,\frac12)$? – 2017-01-08
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    Yes, it is - you did an error in the calculation of the domain though, instead $\frac a2$, it's $\frac{a+b}2$, the midpoint - for every $I \subseteq (a,b)$, $f|_I$ is a bijection $I \to f(I)$. – 2017-01-08
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Consider $f(x) = \dfrac{b-x}{b-a}$ which is almost identical to the one you have, but it maps $(a,b)$ to $(0,1)$ in the opposite way.Another way to prove this is: the function $f(x) = \arctan \left(\dfrac{x-a}{b-a}\right)$ maps $(a,b)$ to $(0,\frac{\pi}{4})$ bijectively, and the map $g(x) = \dfrac{4x}{\pi}$ also maps $(0,\frac{\pi}{4})$ to $(0,1)$ bijectively. Hence the function $h(x) = (g\circ f)(x)$ maps $(a,b)$ to $(0,1)$ bijectively, proving that $(a,b) \sim (0,1)$

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    I used something like this to prove that $(0,1)\sim \mathbb R$, but thanks for the info! – 2017-01-08