Timeline for Is the topology generated by the complements of analytic subsets strictly coarser than the Euclidean topology in dimensions $\geq 2$?

7 events

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Jan 22, 2018 at 20:35	comment	added	M.G.		Thanks for all the comments! That was very enlightening.
Jan 21, 2018 at 21:33	comment	added	Ben McKay		@ZachTeitler: that is just the usual Euclidean topology, by a theorem of Hassler Whitney.
Jan 21, 2018 at 20:31	comment	added	Zach Teitler		Out of curiosity, what about the topology on $\mathbb{R}^n$ whose closed sets are the zero sets of $C^{\infty}$ functions?
Jan 21, 2018 at 17:35	comment	added	nfdc23		Any analytic set in $\mathbf{K}^N$ with non-empty interior (for the usual topology) must be the entire space due to analytic continuation reasons, so a proper analytic set is "thin". Moreover, at least over $\mathbf{C}$, the "analytic Zariski topology" satisfies features of the usual Zariski topology such as a good theory of irreducibility and irreducible components (locally finite for the usual topology) and a "descending chain condition" (locally for the usual topology). These matters and much more (over $\mathbf{C}$) are discussed in the wonderful book Coherent Analytic Sheaves.
Jan 21, 2018 at 17:31	comment	added	Ben McKay		Yes: the closed sets are as you say, so they do not include the graphs of continuous but not analytic functions. Hence not the usual topology, and contained in it, so coarser.
Jan 21, 2018 at 16:47	history	edited	Martin Sleziak		edited tags
Jan 21, 2018 at 16:43	history	asked	M.G.	CC BY-SA 3.0