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Does there exist a locally compact Hausdorff topological group that is not sequential?

Edit: What is a sequential space? Let $X$ be a topological space. A subset $F$ of $X$ is sequentially closed if, whenever $(x_n)$ is a sequence in $F$ converging to $x$, then $x$ must also be in $F$. Every closed set in $X$ is sequentially closed. If every sequentially closed set in $X$ is closed, then $X$ is called a sequential space. Roughly, sequentially spaces are those whose topology is determined by the convergent sequences in the space. All metric spaces and all first-countable spaces are sequential.

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  • $\begingroup$ What does “sequential” mean? $\endgroup$ – José Carlos Santos Jan 20 '18 at 21:13
  • $\begingroup$ @JoséCarlosSantos: A topological space is sequential if every subset which is closed under taking limits of sequences is closed. $\endgroup$ – Eric Wofsey Jan 20 '18 at 21:19
  • $\begingroup$ @EricWofsey Thank you. $\endgroup$ – José Carlos Santos Jan 20 '18 at 21:20
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Sure. For instance, if $I$ is an uncountable set, the product $\{0,1\}^I$ is not sequential, but is a compact Hausdorff topological group by taking addition mod $2$ on each coordinate. (To see it is not sequential, note that the set of elements which are $0$ on all but countably many coordinates is sequentially closed but not closed.)

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  • $\begingroup$ This was the first example that came to mind for me as well. $\endgroup$ – Henno Brandsma Jan 20 '18 at 21:31

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