Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider the poset category $\mathbb{Q}^{op}$, i.e. where $p \rightarrow q$ iff $p \geq q$. Take any $q \in \mathbb{Q}$. Then how many sieves are there for $q$ in $\mathbb{Q}^{op}$?

Supposedly, the answer should be $\lbrace r \vert r \geq q, r \in \mathbb{R} \cup \lbrace \infty \rbrace \rbrace$. But I can only see there corresponding exactly one sieve for each rational $p \geq q$, namely the sieve $S_p = \lbrace r \in \mathbb{Q} \vert r \geq p \rbrace$. In which case there are only countably many sieves on $q$.

So I really cannot see how there could be uncountably many sieves on $q$ (in $\mathbb{Q}^{op}$) as the suggested answer suggests. I tried thinking in terms of Dedekind cuts on the rationals, but that doesn't change the fact that there is exactly one sieve for each rational greater than $q$. Namely, even if I label the sieves with reals, there's still going to be countably many reals.

I'm sure I've misunderstood something, but really cannot see what it is. Any help will be greatly appreciated.

share|improve this question

1 Answer 1

up vote 2 down vote accepted

If I correctly understand the definitions here and here, a sieve $S$ on $q$ is essentially just a subset of the up-set of $q$ such that whenever $p\in S$ and $r\ge p$, then $r\in S$. In other words, $S$ is closed under taking up-sets. You were right to think of Dedekind cuts, because they explain why there is a sieve on $q$ for each real number $x\ge q$, not just for each rational $p\ge q$. Suppose, for instance, that $q=1$. $\{p\in\mathbb{Q}:p\ge\sqrt 2\}$ is a subset of the up-set of $q$ that is closed under taking up-sets, and it’s not equal to $S_p$ for any $p\in\mathbb{Q}$.

More generally, let $x\in\mathbb{R}$ with $x\ge q$, and let $S_x=\{p\in\mathbb{Q}:p\ge x\}$; then $S_x$ is a sieve on $q$. Moreover, if $x\ne y$, then $S_x\ne S_y$. To see this, suppose that $x<y$; then there is a rational $p\in(x,y)$, and clearly $p\in S_x\setminus S_y$.

share|improve this answer

Your Answer

 
discard

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.