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This question already has an answer here:

Proving that $[0,1)$ and $(0,1)$ have the same cardinality (without assuming any previous knowledge) can be done easily using Cantor-Bernstein theorem.

However I'm wondering if someone can build an explicit bijection between these sets.

It's easy to build a bijection between $(0,1)$ and $\mathbb R$, so a bijection from $[0,1)$ to $\mathbb R$ will also fit the bill.

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marked as duplicate by Gabriel Romon, hardmath, Did, user1551, Cameron Buie Sep 7 '15 at 17:00

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Have a look at this question. $\endgroup$ – Krijn Sep 7 '15 at 15:43
  • $\begingroup$ @Krijn, that's exactly what I need. Too bad it didn't come up in the suggestions when I wrote my question. Closing as a duplicate. $\endgroup$ – Gabriel Romon Sep 7 '15 at 15:45
  • $\begingroup$ It comes up on the right of this page under Related, third entry. $\endgroup$ – André Nicolas Sep 7 '15 at 15:47
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    $\begingroup$ @AndréNicolas it should also appear here: imgur.com/5L9qViH $\endgroup$ – Gabriel Romon Sep 7 '15 at 15:50
  • $\begingroup$ I agree that it should have appeared there. Luckily, from now on, it wil. $\endgroup$ – Krijn Sep 7 '15 at 15:51
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Let us partition $(0,1)$ into a countable number of disjoint subsets of the form $[\frac{1}{n+1},\frac{1}{n})$ for $n=0,1,2,\ldots$.

These half-open intervals may then be positioned in reverse order to form a half-open interval of equal length. Whether this construction is sufficiently explicit is open to question, but it does allow the relocation of any $x\in (0,1)$ to $[0,1)$ to be computed in short order.

A more succinct construction is to define $f:[0,1) \to (0,1)$ by $f(0) = 1/2$, $f(1/n) = 1/(n+1)$ for integer $n \ge 2$, and $f(x) = x$ otherwise.

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  • $\begingroup$ Interesting idea. I wonder where I saw it already. Hey, wait a minute... $\endgroup$ – Did Sep 7 '15 at 16:02
  • $\begingroup$ @Did: I'm sure you will believe that I answered without look at your previous Answer, but I've converted my post to CW out of deference to your note. $\endgroup$ – hardmath Sep 7 '15 at 16:07

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