# If a subset of the plane has open intersection with every line, is it open?

Suppose that $U \subset \mathbb{R}^2$ is such that $U \cap L$ is open in $L$ for any line $L \subset \mathbb{R}^2$ where $L$ inherits the subspace topology from $\mathbb{R}^2$ (ie. $L \cong \mathbb{R}$). Does it follow that $U$ is open? I keep thinking I have a counterexample and then changing my mind...

Not sure if the topological vector space tag is appropriate. I thought if the question had a positive answer then it might have something to do with the fact that every finite dimensional vector space has a unique Hausdorff topology compatible with the operations.

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I removed the TVS tag as (in my humble opinion) it is irrelevant. I won't argue if someone better-informed than I am puts it back. –  Gerry Myerson Oct 19 '11 at 3:36

Let $$p_n = \left\langle \cos\frac{\pi}{2^{n+1}},\sin\frac{\pi}{2^{n+1}} \right\rangle,$$ let $P=\{p_n:n\in\omega\}$, and let $S=\mathbb{R}^2\setminus P$. The intersection of any line with $P$ is either a singleton or a doubleton, so every line intersects $S$ in an open set, but $\langle 1,0 \rangle \in S \cap \operatorname{cl}P$, so $S$ isn't open.
I take it $\omega$ is your preferred symbol for the natural numbers? –  Gerry Myerson Oct 19 '11 at 3:34
Excellent! I guess you could also just take the complement of $\{ ( \cos \theta, \sin \theta) : \theta \in [0, \pi/2] \}$? –  Mike F Oct 19 '11 at 3:36
@Gerry: Yes, assuming that by natural numbers you mean the non-negative integers; it comes naturally, since I’ve a strong set-theoretic background. I’d use $\mathbb{N}$ in other contexts except that too many people seem to use that for what I call $\mathbb{Z}^+$. –  Brian M. Scott Oct 19 '11 at 3:37
@Mike: Almost: you do have to omit one point! Make it $\{\langle\cos\theta,\sin\theta\rangle:\theta\in(0,2\pi)\}$. –  Brian M. Scott Oct 19 '11 at 3:39