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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.2011-10-19

1 Answers 1

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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.

Any convergent sequence with no three points collinear will do.

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    @Mike: I figured that it was probably either a typo or a momentary mental glitch.2011-10-19