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Everybody knows that with respect to Zariski's topology, every open set is dense. I want to know if a topological space $X$ has such property, can we deduce that $X$ is homeomorphic to an affine set $A^n$ over a field $K$?

Any, hint is appreciated.

Ramand
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    There are plenty of irreducible topological spaces which aren't homeomorphic to $\Bbb A^n_k$ for any $n,k$. Even staying in the realm of algebraic geometry, take $\Bbb P^2_k$ for instance: proving this is a nice exercise which you should try yourself. – KReiser Nov 18 '21 at 22:30
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    There are a lot of topologies like this. If $(X,\mu)$ is a measure space, we can define a topology with closed sets those having measure $0,$ or when $\mu(X)=+\infty,$ closed sets having finite outer measure. Basically, this will be true for most topologies where the closed sets are “small.” A simple example is $C\subset \mathbb N$ is closed iff $$\sum_{n\in C}\frac1n<+\infty.$$ Not sure how to show this is or isn’t a Zariski topology. – Thomas Andrews Nov 18 '21 at 22:31
  • @Thomas Andrews: if it’s homeomorphic to Zariski topology on some variety, then the same holds for the the set of squares (a closed subset), but this one is discrete infinite – isn’t this impossible? – Aphelli Nov 18 '21 at 22:40
  • My algebraic geometry is rusty, so I couldn’t say. My comment was purely about a class of topologies. @Mindlack – Thomas Andrews Nov 18 '21 at 22:44
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    Such a space is called hyperconnected in general topology. I think the final question is probably false; consider ultrafilter spaces e.g. – Henno Brandsma Nov 18 '21 at 22:44
  • @HennoBrandsma as I have stated upthread, there is no need to go to all of that trouble - a counterexample exists in the same field and it ought to be quite accessible to any student of the subject. – KReiser Nov 18 '21 at 22:50
  • @KReiser I'm a noob in algebraic geometry (but with a PhD in general topology), so I look for examples where I am familiar, sorry. Probably the OP will find your idea nicer and more appealing.. – Henno Brandsma Nov 18 '21 at 22:54
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    Thomas Andrews’ final example cannot be the Zariski topology of a “naive” algebraic variety – because such topological spaces are Noetherian and a closed discrete subspace of a Noetherian space must be finite. I think, more generally, that it can’t be the subspace of closed points of a Noetherian scheme, but I’m not entirely sure. – Aphelli Nov 18 '21 at 22:58
  • @Mindlack Your guess is correct: any subset of any noetherian topological space is quasi-compact in the induced topology, but as you have identified the subset of squares is infinite discrete. – KReiser Nov 19 '21 at 02:46
  • Btw, one should exclude $\emptyset$. Unless I’m overlooking something, already taking small examples with the indiscrete topology gives a counterexample. – Qi Zhu Nov 19 '21 at 07:19

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