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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?

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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.

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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. – goblin 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. – goblin 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. – goblin 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
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

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