# Sub-(complete lattice)? - what's the correct terminology?

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.
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