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Is it possible to have two topologies, one strictly finer than the other, yet all of the accumulation points are the same?

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Since the closure of a set is the union of the set with the set of its accumulation points, if all accumulation points of all sets are the same, the closures of all sets are the same, and therefore the set of closed sets is the same. Because the open sets are the complements of the closed sets, then also the set of open sets is the same. But the set of open sets is the topology, therefore the topologies are the same.

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  • $\begingroup$ Pardon my ignorance but ... an accumulation point is something defined in terms of sequences so it might be possible that one topology restricts the direction of approach while the other allows it, right? $\endgroup$ – SolutionExists Jun 10 '20 at 5:35
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    $\begingroup$ @SolutionExists: No, an accumulation point $x$ of a set is a point where in every neighbourhood of $x$ there is a point of the set other than $x$. In certain topological spaces, this implies that you can construct a sequence of such points that converges to $x$, but that's not its defining property. $\endgroup$ – celtschk Jun 10 '20 at 5:57

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