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The intersection with every line L in $\mathbb{R^2}$ is open in $L$ with the topology

$\mathcal{T}_L = \{G\cap L : G \text{ open in } \mathbb{R^2} \text{ with the Euclidean Topology } \} $, but the set is not open in $\mathbb{R^2}$ with the Euclidean Topology ?

I couldnt find and such set but am pretty sure that there exist such a set , if you have any ideas let me know ! thanks in advance !

orangeskid
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ChrisNick92
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    Consider the origin together with an open interval (0,t) in direction t as t ranges from 0 to $2 \pi$. – Ittay Weiss Mar 28 '18 at 18:56
  • @IttayWeiss then its intersection with the line in direction t would be the interval [0,t), which is not an open subset of the line? – anonymous67 Mar 28 '18 at 19:02
  • Yes i also think that this example doesnt work :( – ChrisNick92 Mar 28 '18 at 19:10
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    @IttayWeiss Ok i understand what you mean something like this $A = \cup_{\theta \in [0,2\pi]}A_\theta $ such that $A_\theta = {(r \cos\theta,r\sin\theta) : r \in [0,\theta)}$ then $A$ is not open in $\mathbb{R^2}$ but open in $L$ for every lane L. – ChrisNick92 Mar 28 '18 at 19:39
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    Another example: take a parabola, remove one point, then take the complement. – Dap Mar 29 '18 at 10:54
  • yea seems to be working ! nice ! – ChrisNick92 Mar 29 '18 at 11:11

2 Answers2

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(To kick it from the unanswered queue)

Summarizing the comments:

Take $A=\Bbb R\setminus P$ where $P$ is a parabola with one point removed.

YuiTo Cheng
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By a classical result of Mazurkiewicz cited in this paper by Mauldin, assuming the axiom of choice there is a so-called "$2$-point set", i.e., a set $S\subset\mathbb R^2$ which intersects every line in exactly two points. In particular a $2$-point set intersects every line in a closed set, so its complement has your property, it intersects every line in a relatively open set.

By a result of Larman quoted in this MathOverflow question, a $2$-point set can't be an $F_\sigma$ set,
so its complement can't be a $G_\delta$ set, much less an open set. It can easily be arranged that the
$2$-point set is not even a Borel set, and in fact it's a well known open problem whether a $2$-point set can be a Borel set.

bof
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