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I am asked the following question: (I write $\{a,b\}$ for points in $\mathbb{R}^2$)

Let $\{0,0\},\{1,1\} \notin K \subset [0,1]^2$ such that the projections of $K$ onto the $x$-axis and the $y$-axis are (1 dimensional) lebesgue null-sets. Is there a curve $\gamma : [0,1] \longrightarrow [0,1]^2\backslash K$ such that $\gamma(0)=\{0,0\}, \gamma(1)=\{1,1\}$ and $\ell(\gamma)\leq2$?

My idea was to consider a family of disjunct curves to do a dimension argument, but I stumbled across the cantor-set and therefore saw that $K$ doesn't have to be countable, might there even be a counter example?

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@joriki: That is a possible solution, but I want to find such a curve for arbitrary $K$. –  Listing Nov 1 '11 at 9:18
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I think you misunderstood my first comment. It's unusual and thus confusing to write "every" for two. Inserting "coordinate" has reduced the room for misunderstanding, but still one wonders briefly whether there are some other coordinate axes involved that would warrant the use of "every". –  joriki Nov 1 '11 at 9:28
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OK, but simply changing "every" to "each" would have done the job :-) –  joriki Nov 1 '11 at 9:38
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@Olivier: I was trying to think along similar lines. We could consider pairs of coverings of $\pi_x$ and $\pi_y$ (defined as in my answer) with countable collections of closed intervals and look at the set of values of $\alpha$ blocked by their products. This is again a countable collection of closed intervals, but I can't see how to deduce that the infimum of the sum of their lengths is zero from the zero infima for $\pi_x$ and $\pi_y$. In fact after trying for a while to deduce something like that my impression is that perhaps such an $\alpha$ isn't in fact guaranteed to exist. –  joriki Nov 1 '11 at 17:12
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@Olivier: Yes -- in fact that in a sense is the entire problem, because once you can somehow escape from the end points even just a tiny bit, you've necessarily crossed a line that you can use to get all the way to the other side. But of course escaping by a tiny bit is exactly as difficult as getting to the other end point -- and the solution is to try to get towards the end points instead of trying to get away... –  joriki Nov 1 '11 at 17:26

1 Answer 1

up vote 3 down vote accepted

I believe you can construct such a curve $\gamma$ for all such sets $K$ by zig-zagging in axis-parallel lines. Let $\pi_x$ and $\pi_y$ be the projections of $K$ onto the $x$ and $y$ axis, respectively. Pick some starting point $(x_0,y_0)$ with $y_0\notin\pi_y$, and pick some $x_1\notin\pi_x$ with $x_1\le x_0/2$. Add the segment from $(x_1,y_0)$ to $(x_0,y_0)$ to the curve (say, by mapping $[\frac14,\frac12]$ to it). Then pick some $y_1\notin\pi_y$ with $y_1\le y_0/2$ and add the segment from $(x_1,y_1)$ to $(x_1,y_0)$ to the curve (say, by mapping $[\frac18,\frac14]$ to it). Repeating this procedure indefinitely constructs a continuous curve $\gamma$ that can be continuously extended to $\gamma(0)=(0,0)$, since the $x_i$ and $y_i$ converge to $0$. We can do the same thing in the other direction towards $(1,1)$. The length of the curve is $2$ as required.

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@Listing: Sorry, this was simply wrong. –  joriki Nov 1 '11 at 9:47
    
@Listing: I've replaced the answer with a more sensible one -- I hope I didn't make a basic mistake again? –  joriki Nov 1 '11 at 16:19
    
Very nice answer! –  Olivier Bégassat Nov 1 '11 at 16:36
    
This is awesome, I can find no mistake in the proof :-) –  Listing Nov 1 '11 at 17:09

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