# I need a $T_D$ topological space that is not sober

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

• 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... – Henno Brandsma Apr 9 at 8:39
• 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 – Alessandro Nanto Apr 9 at 9:50

## 1 Answer

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 $$T_D$$?