Call an open set essential to a generating set of a sieve if the sieve generated upon exclusion of that open set is smaller. Given a topological space $X$ and nonempty open set $U,$ consider the proposition:

There exists a set $S\ni U$ generating a sieve covering $X$ to which $U$ is essential.

If $X$ is T1 (one-point sets are closed), then this proposition is true: choosing a point $p\in U,$ consider $U$ together with $X-\{p\}.$ This generates a sieve covering $X,$ and upon exclusion of $U,$ generates no open sets containing $p.$

If $X$ is the two-point set $\{0,1\}$ with topology $\{\emptyset,\{0\},X\},$ then the proposition is not true for $U=\{0\}:$ any set $S$ generating a sieve covering $X$ would have $X$ itself as an element, and $X$ is enough to generate the entire topology as a sieve.

My question is: is the proposition

Every nonempty open set $U$ is essential to some set generating a sieve covering $X$

equivalent to the T1 axiom? If not, is it equivalent to some weaker separation axiom?

This came up when poring through the definitions in a friend's attempt to generalize the functor $\text{Spec}:\text{Rng}\to\text{Top}$ to a functor $\text{Spec}:\mathcal{C}\to\mathcal{D}$ for any category $\mathcal{C}$ with arbitrary products and Grothendieck topology $\mathcal{D}.$


1 Answer 1


What a coincidence: I have been thinking about something very similar for the same reason!

The proposition is not equivalent to the $T_1$ property. For instance, every open subset of $\operatorname{Spec} \mathbb{C}[x]$ is "essential". In some sense this is because the proposition is about the underlying locale of a topological space, and $\operatorname{Spec} \mathbb{C}[x]$ is the soberification of $\operatorname{MaxSpec} \mathbb{C}[x]$, which is a $T_1$ space.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .