# Can a path avoid a dense set?

My actual question is related to an unusual circumstance in a game which I am playing. We are trying to move one point-like object from one part of the moon to another without passing through another player's land. However, due to a loophole, nothing prevented one friend from buying a dense set of measure zero. (The game encourages rules-lawyering so this isn't completely out of line). We now don't know what's going on.

I imagine the simplest version of this question would be if there is a continuous nontrivial curve in $\mathbb{R}^2$ that avoid every point in $\mathbb{Q}^2$, but now I'm curious if there's a way we that this statement could be generalized to arbitrary topological spaces (without necessarily having a measure).

Technically, according to this game the object has to have a velocity, so to resolve the confusion for us we'd need a curve that is also non-differentiable at at most countably many points, but I don't know enough about differentiablity to imagine the appropriate generalization.

-
Would you mind posting what the game is? I'm super curious. – anonymous Dec 8 '12 at 21:11
$X := \mathbb R^2 \setminus \mathbb Q^2$ is path connected. If you start with two points $(x_0,y_0)$ and $(x_1,y_1)$ in $X$, then you can assume for example that $x_0 \not \in \mathbb Q$ and $y_1 \not \in \mathbb Q$. Then move along the vertical line $x = x_0$ until you reach the ordinate $y_1$, and then move horizontally until you reach $(x_1,y_1)$. The other cases are treated similarly. – Mauro Porta Dec 8 '12 at 21:14
@anonymous: It's a game of en.wikipedia.org/wiki/Nomic. I'm actually not the record keeper for our game, so if you'd like to see the ruleset I could host it somewhere but you'd have to be patient. Also Nomic's infobox on wiki is hilarious. "Players: variable. Setup time: variable… Random Chance: variable. Skill(s) required: …variable." – Eric Stucky Dec 9 '12 at 13:56

## 1 Answer

To answer your simple version, the line $$y=\pi x+\sqrt{2}$$ avoids every point in $\mathbb{Q}^2$ since $\sqrt{2}$ is algebraic and $\pi$ is not.

There are dense sets of measure zero which prevent this sort of approach such as $$(\mathbb{R}\times \mathbb{Q}) \cup (\mathbb{Q}\times \mathbb{R})$$ so it rather depends on precisely what set your friend bought.

-
You don't need $\pi$; the line $y=x+\sqrt2$ avoids every point in $\mathbb Q^2$. – Andreas Blass Dec 8 '12 at 21:19
@Andreas: True enough. I had started with $y=\pi x$ then realised it hit one point in $\mathbb Q^2$ so added $\sqrt{2}$ – Henry Dec 8 '12 at 21:23
I believe the wording was a particular projection of $\mathbb{Q^2}$ onto the surface of the moon, which we are essentially assuming is spherical. – Eric Stucky Dec 9 '12 at 13:48