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Suppose we have a countable family of function graphs (each function is $\mathbb R\to\mathbb R$, not necessary continuous). Obviously, they cannot cover the whole plane $\mathbb R^2$, because they cannot even cover every of uncountably many points on a single vertical line.

But suppose now we are allowed to rotate each graph from the family by arbitrary angle around an arbitrary point of the plane (the total number of graphs is still countable). Is it possible to cover the whole plane $\mathbb R^2$ in this case?

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    $\begingroup$ math.stackexchange.com/questions/35606/… $\endgroup$
    – user61527
    Jun 17, 2014 at 3:59
  • $\begingroup$ Why not start with a line thru the origin and rotate it by every angle from $0$ to $2\pi$. Is that what you're trying to get at? $\endgroup$
    – user99680
    Jun 17, 2014 at 4:32
  • $\begingroup$ @user99680 But there are a countable family of functions, and there are uncountable angles to turn through, so I don't think this will work. $\endgroup$
    – user105475
    Jun 17, 2014 at 4:52
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    $\begingroup$ @user99680 Each graph can have a unique shape and is not necessary a line — a function can be discontinuous at every point, and its graph can even be dense in $\mathbb R^2$ $\endgroup$ Jun 17, 2014 at 5:12
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    $\begingroup$ This is a really cool question. $\endgroup$ Jun 20, 2014 at 22:07

2 Answers 2

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You can construct just one function $f: R \rightarrow R$ such that countably many rotations of the graph of $f$ cover the plane. This is due to R. O. Davies. See theorem 31 and the remark below it in Arnie's notes which has many other interesting results (without proofs) about set theory of plane.

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The idea below does not work: I wrote "empty interior" when I meant "nowhere dense". But graphs do not need to be nowhere dense.


We cannot cover the plane this way. The reason is that the graph of any $f:\mathbb R\to\mathbb R$ has empty interior (as a subset of the plane), since every vertical line meets $f$ in at most one point. Having empty interior is not affected by rotations. The Baire category theorem gives us that $\mathbb R^2$ is not a countable union of sets with empty interior, and this concludes the proof.

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    $\begingroup$ IIRC, Baire category theorem gives us that $\mathbb R^2$ is not a countable union of closed sets with empty interior. A function graph is not necessary a closed set, e.g. the graph of Dirichlet function. $\endgroup$ Jun 17, 2014 at 14:52
  • $\begingroup$ Ah, I made a mistake (somehow mixed up nowhere dense with empty interior). $\endgroup$ Jun 17, 2014 at 15:22
  • $\begingroup$ I believe Sierpinski has proved this in 1933 under CH. Some information about this problem can be found in Roy Osborne Davies, Covering the plane with denumerably many curves, Journal of the London Mathematical Society (1) 38 (1963), 433-438 and in John Carson Simms, Sierpinski's theorem, Simon Stevin 65 #1-2 (March-June 1991), 69-163. However, I don't have access to these papers right now or the time to look into this. I also believe John C. Morgan's book Point Set Theory has a lot of information on this topic. $\endgroup$ Jun 20, 2014 at 21:02
  • $\begingroup$ Another quote for the Sierpinski construction is in "Application of point set theory in real analysis" by A. B Kharazishvili p.188: assuming CH there is a single function that can be rotated or translated $\omega$ times and thus cover the plane. $\endgroup$
    – Eran
    Jun 20, 2014 at 21:42

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