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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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    I do not see any connection between $K$ and $\gamma$ in your question. Have I overlooked something? Also, $\ell(\gamma)$ denotes the length of the curve, correct?\\What is also a little unusual is that you denote ordered pairs by $\{0,0\}$, this notation is more frequently used for sets. I guess it might help the others if you explained your notation.2011-11-01
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    In the case of two things, the more usual expression is "each axis". With "every axis", it sounds like you might mean not just the coordinate axes but arbitrary axes.2011-11-01
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    Thank you two, I fixed both problems.2011-11-01
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    What precludes $K=\emptyset$? and $\gamma$ the diagonal?2011-11-01
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    @joriki: That is a possible solution, but I want to find such a curve for arbitrary $K$.2011-11-01
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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".2011-11-01
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    I see, I changed it to be more clear now.2011-11-01
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    OK, but simply changing "every" to "each" would have done the job :-)2011-11-01
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    I bet you you can take a curve of the form $t\in[0,1]\mapsto t^{\alpha}$ for some $\alpha>0$... though I have no idea how to prove it2011-11-01
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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.2011-11-01
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    @joriki I thought similarly (maybe not ^^) using two open covers of length $<\epsilon$ via outer measure. One additional problem is that one of the end points $(0,0)$ or $(1,1)$ might be irrevocably concealed by a series of thin rectangles of fixed length that accumulate to a side of the square if $0$ or $1$ is in $\pi_x$ or $\pi_y$.2011-11-01
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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...2011-11-01

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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.2011-11-01
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    @Listing: I've replaced the answer with a more sensible one -- I hope I didn't make a basic mistake again?2011-11-01
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    Very nice answer!2011-11-01
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    This is awesome, I can find no mistake in the proof :-)2011-11-01