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The definition of a spectral space requires four conditions:

  1. The space is compact,
  2. The space is Kolmogorov (or $T_0$),
  3. The compact open subsets form a basis of the topology and are closed under finite intersection,
  4. The space is sober, i.e. every nonempty irreducible closed subset has a generic point.

Is there an example of a space satisfying (1), (2) and (3) but not (4)?

A simple example would be appreciated.

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  • $\begingroup$ This is not quite the definition of a spectral space: in a spectral space, the intersection of two quasi-compact opens must be quasi-compact. Note that my example has this additional property. $\endgroup$
    – Slade
    Jan 18, 2016 at 23:13

1 Answer 1

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The simplest example is an infinite set $X$ equipped with the cofinite topology.

This satisfies the first three conditions, but the whole space $X$ is an irreducible closed set without a generic point.

This also provides a good example of soberification: we can make $X$ into a sober space by adding a single dense point.


Note: I am taking "compact" to mean "quasi-compact". Compact is often taken to mean "Hausdorff and quasi-compact", but every Hausdorff space is sober, so there are no Hausdorff examples.

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