Skip to main content
edited tags
Link
Martin Sleziak
  • 4.7k
  • 4
  • 35
  • 40
Source Link
M.G.
  • 7.1k
  • 3
  • 46
  • 60

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

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 $N$ ($N\geq 2$).