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I am looking for "classical" - simple yet striking - proofs that set of all real numbers is equivalent to the $(0, 1) \subset \mathbb{R}$ and $(0, 1] \subset \mathbb{R}$?

To my understanding, it is enough to show that two functions $f: (0, 1) \mapsto \mathbb{R}$ and $g: \mathbb{R} \mapsto (0, 1)$ are the inverses of each other. And the same for the $(0, 1]$. I assume there are well-known "classical" $f, g$, what are they?

Zazaeil
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    Did you really mean “rational numbers”? – José Carlos Santos Oct 29 '21 at 19:13
  • @JoséCarlosSantos typo, corrected. – Zazaeil Oct 29 '21 at 19:13
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    Well, $\tan$ is a bijection from $\left(-\frac\pi2,\frac\pi2\right)$ to $\mathbb R$. That's a pretty well known argument for showing that an open interval in $\mathbb R$ has the same cardinality as $\mathbb R$. – Rushabh Mehta Oct 29 '21 at 19:17
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    The classical argument for showing $(0,1)$ and $(0,1]$ have the same cardinality is the following bijection $f:(0,1]\to(0,1)$: $$f(x)=\begin{cases}2^{-k-1}&\exists k\in\mathbb N,;x=2^{-k}\x&\text{otherwise}\end{cases}$$ – Rushabh Mehta Oct 29 '21 at 19:19
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    What do you mean by equivalent? That they have the same cardinality? – Bonnaduck Oct 29 '21 at 19:37
  • @Bonnaduck yeah, equinumerous. – Zazaeil Oct 29 '21 at 19:38
  • $\arctan$ for $\Bbb{R} \cong (0, 1)$ then just a little bit of fiddling with a sequence of elements of $(0, 1)$ to get that $(0, 1) \cong (0, 1) \sqcup {1}$. This all seems routine rather than "striking" to me. – Rob Arthan Oct 29 '21 at 23:02

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Consider $f:(0,1)\to\mathbb R$ given by $f(x)=\cot(\pi x)$. It is easy to check that is a bijection.

Since $(0,1)\subset(0,1]\subset\mathbb R$, we have $|(0,1)|\leq|(0,1]|\leq|\mathbb R|$, so equality holds.

Bonnaduck
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