# How come nowhere dense subsets implies discrete?

Hi, I am reading an article and have encountered a remark in a proof which is not clear to me. Maybe someone can help?

The proposition is: Let X be a topological space without isolated points having countable $\pi$-weight and such that every nowhere dense subset in it is closed. Then it is a Pytkeev space.

Here is the begining of the proof: Let $x \in Cl(A) \setminus A$. Then $x \in Cl(Int(Cl(A)))$, because every nowhere dense set is closed (and hence discrete)...

The thing which is not clear to me: Why can one conclude that every nowhere dense closed set is discrete? Suppose I take the set $\mathbb N$ with the cofinite topology. Then the finite sets are closed and nowhere dense. But as far as I undesrtand they are not discrete since every open set in the topology that contains a finit set also has to contain other points since it is infinite. Can somone see what am I missing?

The definition of a Ptkeev space: Let X be a topological space. A point x is called a Pytkeev point if whenever $x \in \overline {A\setminus{x}}$, there exists a countable $\pi$-net of infinite subsets of A. If every point of a space is a Pytkeev point then the space is called a Pytkeev space.

Thanks!

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Yes, I used a wrong definition for a discrete set.. Thanks! – tali11 Aug 22 '10 at 16:25
@François: Done. (I turned 3 comments from 11 hours ago into an answer, then deleted them.) – Jonas Meyer Aug 22 '10 at 17:06

Because a nowhere dense set minus a point is nowhere dense, the hypothesis implies that points are relatively open in nowhere dense subsets. Perhaps more to the point, every subset of a nowhere dense set is nowhere dense, hence closed by hypothesis.

As for your question about cofinite topologies: Finite subspaces of cofinite topological spaces are discrete, because finite subsets are closed. (The same is true in any $T_1$ space.) This doesn't mean that subsets of finite sets are open, but rather that they are relatively open.

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