How to construct a bijection from $[0,1]$ to $[0,1] \times [0,1]$?

Question

I'm trying to find a bijective map $f$ such that

$$ f : [0,1] \mapsto [0,1] \times [0,1] $$

I've succeeded in constructing one in the 1D case : $[0,1] \mapsto [0,1]$, but I don't know how to approach in 2D case as the above.

Is there anyone to help me out?

https://math.stackexchange.com/questions/183361/examples-of-bijective-map-from-mathbbr3-rightarrow-mathbbr?lq=1 — Y.Guo, Sep 08 '18 at 03:20
@taritgoswami Hilbert space-filling curves are not bijections. — Angina Seng, Sep 08 '18 at 07:41

Rushabh Mehta · Answer 1 · 2018-09-08T03:11:03.307

2

Here's a good one. Suppose that the decimal expansion of the input to $f$ is $0.abcdefgh...$. Let $f$ map to $(0.aceg...,0.bdfh...)$.

Injectivity: Suppose two tuples $(a,b),(c,d)$ of real numbers in $[0,1]^2$ are distinct. WLOG assume $a\neq c$. Let $j$ be the first decimal position where $a$ and $c$ differ. Then, the input that produces $(a,b)$ will differ from the input that produces $(c,d)$ at decimal position $2j-1$. If we applied this process to $b$ and $d$, it would be at position $2j$.

Surjectivity: Let $(a,b)\in[0,1]^2$. I can take one decimal place from each number, and interweave them to produce a new number, which can be done for any $a$ and $b$.

These are not formal proofs, but should give you a starting point.

edited Sep 08 '18 at 03:11

answered Sep 08 '18 at 02:58

Rushabh Mehta

13,663

1

The accepted answer that Y.Guo links to explains why this approach doesn’t (exactly) work. There’s an issue of the map not being well-defined on some rationals. – Santana Afton Sep 08 '18 at 03:37
Ah, I see now. The whole $.9999...=1$ issue. Oops. – Rushabh Mehta Sep 08 '18 at 14:40

How to construct a bijection from $[0,1]$ to $[0,1] \times [0,1]$?

1 Answers1