Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\mathcal{D}.$

share|cite|improve this question

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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