Prove $|(0,1)| = |(a,b)|$

Question

I'm kinda confused how to prove it.

Information given : $a,b$ are real numbers such that $a<b$.

$(0,1)$ is interval (without $0$ and $1$) and $(a,b)$ is also interval without $a$ and $b$.

I had Idea of making one-to-one function as we do in $(0,1)$ to $(1,3)$ but I don't think that's the case, any ideas ?

Answer 1

I had Idea of making one-to-one function as we do in (0,1) to (1,3) but I don't think that's the case, any Ideas ?

Actually, that's exactly the case! I suggest you proceed as follows:

1. Write, rigorously and thoroughly, how you would prove that $|(0,1)|=|(1,3)|$
2. Try and follow how you prove (1), and see what happens of $(1,3)$ is, instead, $(a, 3)$ for some general $a<3$.
3. Try and follow how you prove (2), and see what happens if $(a, 3)$ is, instead, $(a,b)$ for some general $b>a$.

Answer 2

Alternatively, define $f:(0,1)\to (a,b)$ and $g:(a,b)\to (0,1)$ with $$f(x) = a(1-x) + bx \quad\mbox{and}\quad g(y) = \frac{y-a}{b-a}. $$ Verify that $f$ and $g$ are inverses of one another.

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More generally, one can readily obtain a bijection $(a,b) \to (c,d)$ from the above via composition $(a,b)\to (0,1) \to (c,d)$.

Further remark. Cantor-Bernstein is not necessary here. It is sufficient to give a bijection. The power of CBT appears when one has to prove $(a,b)$ and $(a,b]$ have same cardinality, for instance. It is sufficient to find injective maps $(a,b)\to (a,b]$ and $(a,b]\to (a,b)$. Giving an explicit bijection is not so straightforward (but not too difficult, either).