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.

Let $X$ denote a set, let $\mathcal{O}$ denote the open sets of a topological space with carrier $X$, and let $\mathcal{P}$ denote the powerset of $X$. Furthermore, let $\leq_\mathcal{O}$ denote the restriction of $\subseteq$ to $\mathcal{O}$, and let $\leq_\mathcal{P}$ denote the restriction of $\subseteq$ to $\mathcal{P}$.

Then $(\mathcal{O},\leq_\mathcal{O})$ is a sublattice of $(\mathcal{P},\leq_\mathcal{P})$

Thus, since $(\mathcal{O},\leq_\mathcal{O})$ is complete when viewed as a lattice, I feel it would be appropriate to say "$(\mathcal{O},\leq_\mathcal{O})$ is a complete sublattice of $(\mathcal{P},\leq_\mathcal{P})."$ Is this how the statement would usually be phrased?

Similarly, since the induced meet operations $\bigwedge_\mathcal{O}$ and $\bigwedge_\mathcal{P}$ don't agree, I think it would be appropriate to say "its not the case that $(\mathcal{O},\leq_\mathcal{O})$ is a "sub-(complete lattice)" of $(\mathcal{P},\leq_\mathcal{P})."$ Is this how the statement would usually be phrased? Is there perhaps a better way of saying it?

share|improve this question
add comment

1 Answer

The best way of saying it is this: $\mathcal{O}$ is a subframe of $\mathcal{P}$. Here is the general definition:

A frame is a partially ordered set $F$ that has joins for all subsets, meets for all finite subsets, and in which finite meets distribute over all joins. A subframe of $F$ is a subset $F' \subseteq F$ such that $F'$ is closed in $F$ under joins for all subsets of $F'$ and $F'$ is closed in $F$ under meets for all finite subsets of $F'$.

I think the usual understanding of the phrase ‘complete sublattice’ is that the subset is closed under joins and meets for all subsets, i.e. what you might call a ‘sub-(complete lattice)’. Following this pattern, you can say that $\mathcal{O}$ is a complete join sub-semilattice of $\mathcal{P}$. But this terminology is potentially confusing, as you have observed.

share|improve this answer
    
Zhen, the first part of your answer is mistaken. A frame (or "complete Heyting Algebra") is a complete lattice in which finite meets distribute over arbitrary joins. And $(\mathcal{O},\leq_\mathcal{O})$ is not necessarily a subframe of $(\mathcal{P},\leq_\mathcal{P}),$ because infinite meets needn't coincide. –  user18921 Apr 26 '13 at 11:20
    
Anyway, the issue is that, for example: a sublattice is more than just a subposet that happens to be a lattice. The join and meet operations also need to coincide. Similarly, a complete sublattice and a sub-(complete lattice) are different, although I'm unsure of whether this is standard terminology. –  user18921 Apr 26 '13 at 11:24
    
Oh, and note that if a poset is closed under arbitrary meets, then it is necessarily closed under arbitrary joins as well. And vice versa. –  user18921 Apr 26 '13 at 11:49
    
No: the data of a frame only includes finite meets and possibly-infinite joins, so to be a subframe, you only need to have the same finite meets and possibly-infinite joins. Saying that a poset is "closed" under [operation] is bad terminology: it suggests there is an absolute notion of [operation] in some universal poset, and this is not the case. So a subposet of a frame that is closed under joins need not be closed under meets: it will have meets, but they won't be the same in general. –  Zhen Lin Apr 26 '13 at 11:58
1  
Along similar lines, it is very bad form to conflate frames and complete Heyting algebras. The data of a complete Heyting algebra includes the Heyting operation, which automatically exists in any frame, but a frame homomorphism need not preserve the Heyting operation. –  Zhen Lin Apr 26 '13 at 11:59
show 3 more comments

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.