in the book Frames and Locales by Jorge Picardo are defined two types of spaces:

  • Sober spaces where the only meet-irreducible open sets are those in the form $X\setminus x^-$, where $x^-$ is the closure of the point $x$. Here an open set $O$ is meet irreducible if, for any two open sets $A,B$ of $X$, $A\cap B\subseteq O\rightarrow A\subseteq O\vee B\subseteq O$
  • $T_D$ spaces where each point $x$ have an open neighborhood $U$ such that $U\setminus\{x\}$ is open

Can someone show me a sober, not $T_D$ space? Thank you

  • $\begingroup$ non-$T_1$ does not imply non-$T_D$, we just know $T_1$ implies $T_D$ but not the other way around. sober and $T_D$ both imply $T_0$ so any non-$T_0$ space is neither sober nor $T_D$.... Also, a cofinite space is not sober... $\endgroup$ – Henno Brandsma Apr 9 at 8:39
  • $\begingroup$ yes, sorry. I made a mistake, what I wanted was a sober, not $T_D$ space. About the counterexample, that was a mistake too, I took an example of a $T_1$, not sober space $\endgroup$ – Alessandro Nanto Apr 9 at 9:50

A $T_D$ space that is not sober you already mention: $\mathbb{N}$ in the cofinite topology. This is $T_1$ (singletons are closed because they are finite) so certainly $T_D$ (recall that $T_1 \implies T_D \implies T_0$) and not sober: $X$ is irreducible closed but not the closure of a singleton (which is the more usual equivalent condition for sober spaces).

An almost sober space that is not $T_D$ is $X=\mathbb{R}$ in the upper topology $\{\emptyset,\Bbb R\} \cup \{(a,+\infty): a \in \Bbb R\}$. All closed sets of the form $(-\infty,x]=\overline{\{x\}}$ are irreducible and the closures of its generic point. But the irreducible $\Bbb R$ has no generic point... But no $\{x\}$ is open in its closure (as another formulation of the $T_D$ axiom states) so it's not $T_D$. We can (I think) modify it adding $+\infty$ to the space and working with open sets $(a,+\infty]$. (Or use $X=(0,1]$ and open sets $\emptyset$ and all $(x,1]$ for $x \in (0,1)$ instead). The result will then be sober as $X$ gets a generic point.

A simple space that does work is $X=\mathbb N \cup \{\infty\}$ with the topology $\{\emptyset\} \cup \{X\setminus F: F \subset \Bbb N \text{ finite }\}$ Here $\infty$ is not open in its closure $X$, so the space $X$ is not $T_D$ while the closed irreducible sets are $\{n\}, n \in \Bbb N$ which are their own closures and $X$ which is the closure of $\{\infty\}$, so $X$ is sober.

A sober $T_1$ (so $T_D$) space that is not Hausdorff (any Hausdorff space is of course both $T_1$ and sober) can be found on the aforementioned wikipedia page: $\Bbb R \cup \{\infty\}$ where $\Bbb R$ keeps its usual topology and a neighbourhood of $\infty$ is that singleton together with a cofinite subset of $\Bbb R$.

Any two point (or more) indiscrete space is of course neither $T_D$ nor sober. So any combination of $T_D$ and sober is possible; they're independent of each other.

Finally, (wondering aloud), probably spaces of the form $\textrm{Spec}(R)$ (in the Zariski topology) where $R$ is a commutative ring (which are all sober) can be found that are not $T_D$? I'm not quite sure about those kind of spaces; too little algebra experience with prime ideals... Is there a characterisation of those rings where $\text{Spec}(R)$ is $T_D$?


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