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Dec 4 at 23:10 comment added Corentin B My bad, I got way too confident
Dec 4 at 19:40 comment added Andy Putman @HJRW: The sequence $(e_n)$ doesn't just have a convergent subsequence, it actually converges to $0$.
Dec 4 at 19:02 vote accept Dominic van der Zypen
Dec 4 at 18:45 answer added Jeremy Brazas timeline score: 8
Dec 4 at 17:33 comment added HJRW It may be that, in order to think about this group, one needs to use the techniques referenced in the answers to this MSE question: math.stackexchange.com/questions/667035/… .
Dec 4 at 17:26 comment added HJRW @CorentinB: It's natural to suspect that it's a graph, but I don't think it is. The product topology is different from the weak topology. For instance, let $e_n$ be the $n$th unit vector. The sequence $(e_n)$ has no convergent subsequence in the weak topology. But $[0,1]^{\mathbb{N}}$ is compact by Tychonoff's theorem, so the sequence does have a convergent subsequence in the product topology. Now, it may still be an increasing union of graphs, and that may be enough to get the conclusion, but it doesn't follow immediately, I think.
Dec 4 at 16:52 comment added Jeremy Brazas @CorentinB $X$ is path connected because $X$ has the product topology you can create infinite concatenations of paths that move in one coordinate at a time.
Dec 4 at 16:20 comment added Corentin B I mean, this is a simplicial graph, and $\pi_1$ of graphs are free.
Dec 4 at 16:19 comment added Corentin B $X$ is not path-connected, and any path-connected component is a increasing union of grids in finite dimension, $\pi_1(X)$ is just going to be a free group of countably infinite rank (as for the grid in $\mathbb R^2$).
Dec 4 at 15:02 comment added Moishe Kohan Did you think through the case of a grid in $\mathbb R^4$? What group did you get in this case?
Dec 4 at 14:42 history asked Dominic van der Zypen CC BY-SA 4.0