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I came up with a certain topology and played around with it a little. I was wondering if it has been looked at. I wouldn't know where to begin looking.

Given a real vector space $V$, define a topology on $V$ as such: A set $U\subset V$ is open if and only if for all $x\in U$ and all $y\in V$, there exists $\epsilon>0$ such that for all $t\in (-\epsilon,\epsilon)$, we have $x+ty\in U$. This topology is typically strictly finer than that induced by a norm on $V$. I've deduced several properties of this topology myself, but I wanted to read more on it. This can also be generalized to vector spaces over any field which is a metric space.

Anonymous
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    This looks related: https://math.stackexchange.com/questions/128278/generic-topology-on-a-vector-space. – Martin R Dec 30 '18 at 10:43
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    $\epsilon$ here depends on $y$ right? – Alessandro Codenotti Dec 30 '18 at 10:45
  • @AlessandroCodenotti Definitely. – Anonymous Dec 30 '18 at 10:48
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    Is it really finer though ? At least in finite dimension this seems unreasonable for real vector spaces because there is a unique $T_2$ topology on these (making them topological vector spaces) if I remember correctly – Maxime Ramzi Dec 30 '18 at 11:00
  • @MartinR Yes! He is talking about the same topology. – Anonymous Dec 30 '18 at 11:10
  • @Max I don't think there is a unique $T_2$ on topology on real topological vector spaces of any nonzero dimension. To see why this topology is typically finer, consider $\mathbb{R}^2$. Then circles are discrete. – Anonymous Dec 30 '18 at 11:17
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    @Max On second thought, this topology might not satisfy the requirements of a topological vector space. – Anonymous Dec 30 '18 at 11:25
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    I don't see why circles would be discrete... but in any case I don't think it will be a topological vector space indeed – Maxime Ramzi Dec 30 '18 at 11:45
  • @Max You're right about Hausdorff topologies on finite dimensional topological vector spaces being unique. This definitely is not a topological vector space. Either addition or scalar multiplication must be discontinuous. I suspect both are. – Anonymous Dec 30 '18 at 11:47
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    @Max To see why circles are discrete let $C$ be a circle and let $x_0\in C$. I claim $U=(\mathbb{R}^2\setminus C)\cup{x_0}$ is open. Let $x\in U$. If $x\neq x_0$, then clearly for all $y\in \mathbb{R}^2$ there exists $\epsilon>0$ such that for all $t\in(-\epsilon,\epsilon)$ we have $x+ty\in U$. In the case $x=x_0$, this is still true because of the fact that the circle is curved, so any line segment through $x$ can be shrunk to not touch any other part of the circle. – Anonymous Dec 30 '18 at 11:55
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    It’s called the “radial topology” IIRC. It’s Hausdorff but not regular in general IIRC. Cf. https://math.stackexchange.com/q/2967185 e.g. – Henno Brandsma Dec 30 '18 at 12:48
  • @HennoBrandsma Thank you! The topology is trivially Hausdorff since it is finer than a topology induced by a norm. I've been having significant trouble, however, in determining whether it is regular, even in just $\mathbb{R}^2$. – Anonymous Jan 02 '19 at 11:18
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    I think Kunen has written a paper on the regularity of the radial topology. – Henno Brandsma Jan 02 '19 at 11:56

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