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DEFINITION- A function $f:X \to Y$ is called D-supercontinuous if inverse image of every open set is open $F_{\sigma}$ set.

I am looking for an example of a non metrizable topological group (Hausdorff) in which the group operations are also D-supercontinuous.

Obviously every metric group is such a group but I am having a hard time finding a nontrivial or non metric topological group. I know it must not be second countable.

Any help to point me in the right direction is appreciated. Thanks.

Eric Wofsey
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blabla
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  • The point being: in a non-metric space, there could be open sets that are not $F_\sigma$ sets. An example of a non-metric topological group: $\mathbb T^A$ where $\mathbb T = \mathbb R/\mathbb Z$ and $A$ is an uncountable set. I do not know if this is D-supercontinuous, but maybe we should attempt to find out. – GEdgar Dec 25 '19 at 12:12
  • More simply, you just want a non-metrizable topological group in which every open set is $F_\sigma$. – Eric Wofsey Dec 25 '19 at 16:21

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Consider $\mathbb{R}^\infty$ with the weak topology, i.e. the colimit of the sequence of inclusions $\mathbb{R}^0\to\mathbb{R}^1\to\mathbb{R}^2\to\dots$ (concretely, $\mathbb{R}^\infty$ is the set of finite-support sequences of real numbers, with the topology that a set is open iff its intersection with $\mathbb{R}^n$ is open for each $n$). This is a topological group with respect to coordinatewise addition (this is not obvious--to prove addition is continuous, you have to show $\mathbb{R}^\infty\times\mathbb{R}^\infty$ also has the weak topology; see for instance Theorem A.6 in Hatcher's Algebraic Topology). It is not first countable and thus not metrizable (again, this is not obvious--you can use an argument similar to the argument in this answer, picking sequences converging to $0$ along each coordinate axis). However, every open subset of $\mathbb{R}^\infty$ is $F_{\sigma}$: an open set is the union of its intersections with $\mathbb{R}^n$ for each $n$, and the intersection with $\mathbb{R}^n$ is an open subset of $\mathbb{R}^n$ and thus a countable union of closed subsets of $\mathbb{R}^n$, which are then also closed in $\mathbb{R}^\infty$.

Eric Wofsey
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  • To find such a topological ring, I can just add trivial multiplication operation to $R^{\infty}$? Correct? – blabla Dec 26 '19 at 11:18
  • Can you give me some reference to some book or link where I can read more about this space and the topology you mentioned on it? – blabla Dec 26 '19 at 11:20
  • Sure, if you don't require your rings to have unit. If you want a unital ring then you could give $\mathbb{R}^{\infty}$ the ring structure of a polynomial ring $\mathbb{R}[x]$, for instance. I don't know a reference about this space particularly but colimit constructions like this are used all the time in algebraic topology (usually in the form of CW-complexes), for instance as discussed in the appendix to Hatcher's Algebraic Topology. – Eric Wofsey Dec 26 '19 at 15:26
  • So $R^{\infty}$ is a non-metrizable topological group in which every open set is $F_{\sigma}$. Are there anymore such examples, which are easier to understand. I am not versed in algebraic topology. – blabla Dec 27 '19 at 08:48
  • Also can you tell whether $R^{\infty}$ is connected or disconnected? – blabla Dec 27 '19 at 08:49
  • It should be connected, being countable product of connected spaces, right? – blabla Dec 27 '19 at 12:47
  • It is not a countable product of connected spaces (it has only the finite-support sequences), but it is connected. – Eric Wofsey Dec 27 '19 at 14:38
  • In similar fashion, $C^{\infty}$ should work too, right? – blabla Dec 28 '19 at 17:11
  • You mean a colimit of $\mathbb{C}^0\to\mathbb{C}^1\to\mathbb{C}^2\to\dots$? That's the same thing as $\mathbb{R}^\infty$, since $\mathbb{C}^n$ is just $\mathbb{R}^{2n}$. – Eric Wofsey Dec 28 '19 at 17:17
  • Sure, that's fine. – Eric Wofsey Jan 09 '20 at 14:59