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Around 1998, I encountered a (forgotten) reference to a particularly strange space-filling curve.

Consider a foliation as a collection of continuous nonintersecting curves that start at $(0,0)$ and end at $(1,1)$ and collectively fill the unit square, such as the graphs of functions $f_t(x) = x^t$ where $t \ge 0$. Supposedly there exists a continuous curve G that starts at $(1,0)$, ends at $(0,1)$, fills the unit square, and crosses each $f_t$ curve only once.

This initially sounds even more impossible than the Cantor curve. But intuitively a space-filling curve could trace back and forth over the $f_t$ curves and only cross at the corners $(0,0)$ and $(1,1)$. Can someone please explain a construction of such a space-filling curve?

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  • $\begingroup$ For $n > 0$ take the family of curves $\{x^a\}_{a \in S(n)}$ in $[0,1]^2$ where $S(n) := \{n^{-1},(n-1)^{-1},\dots,1,\dots,n-1,n\}$. The union $\gamma_n$ of these curves can be modified within balls of radius $\epsilon$ about the origin and $(1,1)$ to form a single curve $\gamma_{n,\epsilon}$ from the origin to $(1,1)$. I think (not quite sure) that an appropriate limit gives what you're looking for. $\endgroup$
    – Steve Huntsman
    Commented Aug 27, 2010 at 23:14
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    $\begingroup$ What exactly is your definition of "crossing"? $\endgroup$
    – fedja
    Commented Aug 27, 2010 at 23:33
  • $\begingroup$ @Steve - The curve $x^{-n}$ only intersects $[0,1]^2$ at the point $(1,1)$. $\endgroup$
    – Simon Rose
    Commented Aug 27, 2010 at 23:33
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    $\begingroup$ However, a more major concern is that you won't hit anything between, say, $x$ and $x^2$ with that process, so this can't be a space-filling curve. $\endgroup$
    – Simon Rose
    Commented Aug 27, 2010 at 23:36
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    $\begingroup$ Somthing strange here: if the curve intersects $\{ (t, t^2) \}$ in only one point, it's not filling the square, right? $\endgroup$
    – Jeff Strom
    Commented Aug 28, 2010 at 15:50

2 Answers 2

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The space filling curve you are looking for does not exist.

Assume by contradiction that such a space filling curve $\gamma:I\rightarrow [0,1]^2$ exists. Since $\gamma$ intersects each curve $f_t\subset [0,1]^2$ only once, the preimage $\gamma^{-1}(f_t)$ is either a point or an interval. The curve $\gamma$ being space-filling, that preimage can't be a point. It is therefore an interval and, in particular, of positive measure.

Letting $t$ vary, we have constructed an uncountable family of disjoint subsets of $[0,1]$, all of whom have positive measure: contradiction!

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  • $\begingroup$ Thanks, this is convincing. But it begs another question. What was I misremembering? MathOverflow doesn't have a category for mind-reading questions, but is there an interesting space-filling curve associated with a foliation of the unit square? $\endgroup$
    – Steven Heston
    Commented Aug 28, 2010 at 21:41
  • $\begingroup$ Can't help you. $\endgroup$
    – André Henriques
    Commented Aug 29, 2010 at 2:56
  • $\begingroup$ This arguement may be true that a "curve" in the sense of a contunuous line or function but what about a "curve" that is a dust as Steve mentioned, a Cantor or Appolonian (Spelling?) set. $\endgroup$
    – user34569
    Commented May 31, 2013 at 12:13
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André's answer is correct, but what you really are looking for (I think) is this example, due to Katok, but explained (beautifully) by Milnor.

There is a family of disjoint smooth real-analytic curves $\Gamma_{\beta}$ that fill the unit square $I^2$, and a subset $E \subset I^2$ of full measure, so that each curve $\Gamma_{\beta}$ intersects $E$ in at most one point.

This means that $E$ can be constructed by choosing one point for each parameter $\beta$, so in order not to contradict Fubini, the dependence of $\Gamma_{\beta}$ on $\beta$ cannot be continuous.

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