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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?

TauMu
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    http://math.stackexchange.com/questions/35606/lebesgue-measure-of-the-graph-of-a-function –  Jun 17 '14 at 03:59
  • 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? – user99680 Jun 17 '14 at 04:32
  • @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. –  Jun 17 '14 at 04:52
  • @user99680 That would be an uncountable family of lines, but the question says the total number of graphs must be countable. – Vladimir Reshetnikov Jun 17 '14 at 05:00
  • @Bhoot: I was not clear on what the OP was getting at; I guess s/he wants a way of generating an uncountable set with infinitely-many operations? – user99680 Jun 17 '14 at 05:02
  • @Vladimir Reshetnikov: I guess I did not understand the question too well. – user99680 Jun 17 '14 at 05:03
  • Maybe we can look at this from the perspective of group theory: an uncountable group cannot be finitely-generated; can it be countably-infinitely generated? Here rotation would be the group operation, of course. – user99680 Jun 17 '14 at 05:04
  • @user99680 To generate an uncountable set is not a problem — each graph itself is an uncountable set of points. The interesting question is to how to cover the whole plane (if it is possible at all)? – Vladimir Reshetnikov Jun 17 '14 at 05:06
  • @Vladimir Reshetnikov: I meant to generate the whole plane (meaning all the lines in the plane, say all thru the origin)by applying rotations to the countably-infinite family of lines. But it becomes more complicated if we have to consider many different angles of rotation. – user99680 Jun 17 '14 at 05:07
  • So we know an uncountable group cannot be finitely-generated; can it be countably-infinitely-generated, considering the countably-many lines as generators and given rotation by a certain angle as the group operation? – user99680 Jun 17 '14 at 05:12
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    @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 $\mathbb R^2$ – Vladimir Reshetnikov Jun 17 '14 at 05:12
  • @Vladimir Reshetnikov: I understand, but I was trying to fix an approach for testing and see where it leads. I know we can use space-filling curves, etc. – user99680 Jun 17 '14 at 05:13
  • Graphs of maps from $\mathbb R$ to itself have empty interior. Rotating them does not change this fact. Now use the Baire category theorem. – Andrés E. Caicedo Jun 17 '14 at 05:39
  • @Andres:Caicedo: Are you excluding space-filling curves? – user99680 Jun 17 '14 at 05:43
  • @user99680 Obviously, since they are not graphs of maps from $\mathbb R$ to itself. Otherwise, the question would just be a matter of cardinality (there is a bijection between $\mathbb R$ and $\mathbb R^2$) rather than topology. – Andrés E. Caicedo Jun 17 '14 at 05:43
  • @AndresCaicedo: Ah yes, I missed that obvious part. – user99680 Jun 17 '14 at 05:52
  • Given that JDH is still active at mathoverflow, you should probably ask your question there. (See his answer http://math.stackexchange.com/questions/35606/lebesgue-measure-of-the-graph-of-a-function) – Moishe Kohan Jun 17 '14 at 17:35
  • Does the following hold? Work in polars $(r,\theta)$. Define an equivalence relation on $r$ in the manner of Vitali, i.e. $r_1\sim r_2$ if $r_1-r_2\in\mathbb{Q}$. Take a member from each equivalence class to construct a set $R$. Do the same with theta to construct a set $\Theta$. Now we have two sets with the same cardinality, so can construct a bijection between $R$ and $\Theta$. We can now translate the image of this bijection through all rationals in the direction of $r$ and $\theta$ to cover all points in $\mathbb{R}^2$. –  Jun 18 '14 at 06:24
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    This is a really cool question. – goblin GONE Jun 20 '14 at 22:07

2 Answers2

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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.

hot_queen
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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.

Andrés E. Caicedo
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    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. – Vladimir Reshetnikov Jun 17 '14 at 14:52
  • Ah, I made a mistake (somehow mixed up nowhere dense with empty interior). – Andrés E. Caicedo Jun 17 '14 at 15:22
  • 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. – Dave L. Renfro Jun 20 '14 at 21:02
  • 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. – Eran Jun 20 '14 at 21:42