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Let $A \subset \mathbb{R}^2$ be countable. Then it is not too hard to show that $\mathbb{R}^2 \setminus A$ is path-connected. However it is not always Manhattan connected since if $A = \mathbb{Q}^2 \setminus \{(0,0)\}$, the origin cannot be connected to any other point by a path moving only horizontally or vertically.

Say a subset of $\mathbb{R}^2$ is skew-Manhattan connected if there exist two directions so that any two points in the set can be connected by a path which moves in only those two directions.

What I want to decide is whether $\mathbb{R}^2 \setminus A$ as above is always skew-Manhattan connected?

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The set of directions of the lines that go through two of your points is countable, so there is a direction $d$ such that all lines parallel to it contain at most one point from $A$. That direction and its orthogonal $d^\perp$ work.

Indeed: there is a line with direction $d^\perp$ which does not intersect $A$. Since we can move along it freely, it is enough ---in order to show skew-Manhattan connectedness--- to show that: if $L$ is a line with direction $d$ which contains a point $p$ from $A$, then we can connect points on one side of $p$ to points in the other side. This is easy, using cardinality considerations.

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    @J.M. Thank you very much. A nice feature.2011-09-22
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The set $A$ is countable. Therefore so is the set of lines connecting two points of $A$. Theses lines have a countable set of slopes (counting infinity, if necessary), so it is easy to select two directions such that on any line parallel to either there is at most a single point in $A$. You can even choose the two directions to be orthogonal.