# What is lattice with arbitrary suprema called?

Topology is a system of sets, which is closed under arbitrary unions and finite intersections.

Is there a name for lattice, which has analogous properties, i.e., every subset has supremum. (And also every finite subset has infimum, but this is already in the definition of lattice.

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A lattice with suprema for all subsets is called a complete lattice, or perhaps a cocomplete lattice if you are category-theoretically inclined. Here's a fun fact: any lattice that has suprema for all subsets also has infima for all subsets. The infimum of a subset is, of course, the supremum of all the lower bounds for that subset.

Unfortunately, a homomorphism of lattices need not preserve infinite suprema/infima, and here is where the subtlety is: for example, when one says "(co)complete join semilattice", one is thinking of homomorphisms that preserves all suprema, even though a (co)complete join semilattice is automatically a complete lattice. A continuous map of topological spaces induces a homomorphism on the lattice of open sets that preserves finite meets and infinite joins, which is precisely the definition of a homomorphism of frames. (A frame is a complete lattice that satisfies the infinite distributive law.)

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Normally one defines a complete lattice to be one that has suprema and infima for all subsets. –  Brian M. Scott Oct 13 '12 at 7:45
You fun fact is a relatively easy observation, I should have noticed that. But I'm glad I haven't - since otherwise I would not learn from your answer about relation of this question to frames. –  Martin Sleziak Oct 13 '12 at 7:51
In Gratzer's new book, this result is the very first result shown about complete lattices in the chapter on complete lattices, p.50. –  Martin Sleziak Oct 16 '12 at 11:31

$\newcommand{\int}{\operatorname{int}}$If $\langle X,\tau\rangle$ is a topological space, $\langle\tau,\subseteq\rangle$ is a complete distributive lattice: for any $\mathscr{U}\subseteq\tau$, $\bigvee\mathscr{U}=\bigcup\mathscr{U}$, and $\bigwedge\mathscr{U}=\int\bigcap\mathscr{U}$, and for any $U,V,W\in\tau$ we have $$U\land(V\lor W)=U\cap(V\cup W)=(U\cap V)\cup(U\cap W)=(U\land V)\lor(U\land W)$$ and $$U\lor(V\land W)=U\cup(V\cap W)=(U\cup V)\cap(U\cup W)=(U\lor V)\land(U\lor W)\;.$$ What distinguishes the two is infinite distributivity: for any $V\in\tau$

$$V\land\bigvee\mathscr{U}=V\cap\bigcup\mathscr{U}=\bigcup_{U\in\mathscr{U}}(V\cap U)=\bigvee_{U\in\mathscr{U}}(V\land U)\;,$$

but it’s not true in general that

$$V\lor\bigwedge\mathscr{U}=V\cup\int\bigcap\mathscr{U}\overset{?}=\int\bigcap_{U\in\mathscr{U}}(V\cup U)=\bigwedge_{U\in\mathscr{U}}(V\lor U)\;.$$

What you really want to talk about, I suspect, are frames; see also pointless topology (a name that I occasionally find regrettably apt!).

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