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Let $\mathbb{K}$ be either $\mathbb{R}$ or $\mathbb{C}$ and let $N\geq 2$. Similarly to the construction of the Zariski topology, take the collection of zero sets of $\mathbb{K}$-analytic functions to be the system of closed sets for a topology on $\mathbb{K}^N$.

Is this topology always strictly coarser than the Euclidean topology?

I am mainly interested in the complex case, but I have also included the real-analytic case for completeness. Note that in dimension $N=1$ the answer is clearly yes, since the non-trivial closed sets are then precisely the countable discrete subsets of $\mathbb{K}$. Also, trivially, a positive answer in the real case in dimension $2N$ would imply a positive answer in the complex case in dimension $N$ ($N\geq 2$).

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    $\begingroup$ 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. $\endgroup$
    – Ben McKay
    Commented Jan 21, 2018 at 17:31
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    $\begingroup$ 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. $\endgroup$
    – nfdc23
    Commented Jan 21, 2018 at 17:35
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    $\begingroup$ Out of curiosity, what about the topology on $\mathbb{R}^n$ whose closed sets are the zero sets of $C^{\infty}$ functions? $\endgroup$
    – Zach Teitler
    Commented Jan 21, 2018 at 20:31
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    $\begingroup$ @ZachTeitler: that is just the usual Euclidean topology, by a theorem of Hassler Whitney. $\endgroup$
    – Ben McKay
    Commented Jan 21, 2018 at 21:33
  • $\begingroup$ Thanks for all the comments! That was very enlightening. $\endgroup$
    – M.G.
    Commented Jan 22, 2018 at 20:35

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