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Consider the 'interleaving map' $f : (0,1]^2\rightarrow (0,1]$ which takes the decimal digits of its arguments and interleaves them, i.e.

$$\tag{1} (x,y)\equiv(0.a_1a_2a_3\ldots, 0.b_1b_2b_3\ldots) \,\mapsto\, f(x,y) := (0,a_1b_1a_2b_2a_3b_3\ldots)$$ for $x=\sum_{i=1}^\infty a_i\cdot 10^{-i}$ the decimal representation of $x$ (likewise for $y$) and $1=0.99999\ldots$

This map is e.g. featured in MJD's answer to this question, where it is attributed to Cantor, and also in [Klamke: Theory of Sets; Proof of Remark c) (p.31)] (as remarked in the comments to said answer).

It is clear that the map $f$ is injective, though I was wondering if it also holds that

  1. $f$ is Borel measurable (?), and
  2. $f$ maps Borel-sets to Borel-sets?
fsp-b
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1 Answers1

2

  1. It suffices to see that the inverse map $f$ sends elements of a basis of $(0,1]$ to a Borel set. Consider the collection of all open sets of the form $I_{a,n}=(a\cdot 10^{-n},(a+1)\cdot 10^{-n})$. Then the set of all $I_{a,n}$ forms a basis. You can see that the inverse image of $I_{a,n}$ under $f$ is Borel. For example, we can see $$f^{-1} (0.a_1b_1\cdots a_nb_n , 0.a_1b_1\cdots a_n(b_n+1)) = (0.a_1\cdots a_n,0.a_1\cdots (a_n+1))\times (0.b_1\cdots b_n,0.b_1\cdots (b_n+1))$$

    if $b_n<9$.

  2. We can see that $f$ is bijective. It suffices to show that $f$ maps $I_{a,n}\times (0,1]$ and $(0,1]\times I_{a,n}$ to a Borel set, by the same reason mentioned in 1. But we can prove it by direct evaluation (with some case division). For example, the direct image of $(0.a_1a_2\cdots a_n, 0.a_1\cdots (a_n+1))\times (0,1]$ ($a_n\neq 9$) under $f$ is the set of all reals of the form

    $$0.a_1\_a_2\_\cdots a_n\_\,\_\,\_\,\_\cdots,$$

    which is a finite union of intervals.

Hanul Jeon
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