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On $\Bbb R^2$, let $\tau$ be the collection of subsets which contain an open line segment in each direction about each if its points. We claim that $\tau$ form a topology on $\Bbb R^2$. Clearly, $\Bbb R^2$ and $\emptyset$ are in $\tau$. For any $A,B\in \tau$, suppose $x\in A\cap B$ and $m\in\Bbb R\cup \{\infty\}$. Since $A,B\in \tau$, there exist some open line segments $l_A$, $l_B$ of slope $m$ which contain $x$ in $A$ and $B$, respectively. Since $l_A\cap l_B$ is an open line segment of slope $m$ which contains $x$ in $A\cap B$, $\tau$ is closed under finite intersections. Also, $\tau$ is closed under arbitrary unions.

Clearly, The Euclidean topology is a subset of $\tau$.
Question. Is it true that $\tau$ is a subset of the Euclidean topology? Thanks.

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    $\begingroup$ Hint: Try showing $(\mathbb{R}^2 \setminus S^1) \cup \{ (1, 0) \}$ is in $\tau$ but not open in the Euclidean topology (where $S^1$ is the unit circle). $\endgroup$ – Daniel Schepler Sep 7 '18 at 0:52
  • $\begingroup$ I understood. Thanks! $\endgroup$ – Ergin Suer Sep 7 '18 at 2:29
  • $\begingroup$ $\tau$ is much stronger than the Euclidean (standard)(usual) topology. $\endgroup$ – DanielWainfleet Sep 7 '18 at 3:18
  • $\begingroup$ @ErginSuer Perhaps you should write an official answer to your own question to make it visible at first glance that the question is no longer open. $\endgroup$ – Paul Frost Sep 7 '18 at 14:04
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We show that set $U:=(\mathbb{R}^2 \setminus S^1) \cup \{ (1, 0) \}$ belongs to $\tau$. Let $x\in U$. Then either $x\in \Bbb R^2\setminus S^1$ or $x\in \{(1,0)\}$. Suppose $x\in \Bbb R^2\setminus S^1$. Since $R^2\setminus S^1 \in \tau_{Euclid}$, there is some $\epsilon>0$ such that the open ball $B(x,\epsilon)$ is a subset of $R^2\setminus S^1$. But $B(x,\epsilon)\in \tau$ and $B(x,\epsilon)\subseteq U$. Now, suppose $x\in\{(1,0)\}$ and $l$ be any line passes through the point $x$ in $\Bbb R^2$. If $l$ does not intersect any point in $S^1$ other than the point $(1,0)$, we are done. But if it is so, say it $p$, then we can pick $\epsilon>0$ less than the distance between $p$ and $x$. And the open line segment $B(x,\epsilon)\cap l$ is a subset of $U$.

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