0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $T_D$?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.