functions - Explicit bijection between $[0,1)$ and $(0,1)$

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.

Answer

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.