Not understanding this definition in Lattice Theory: First Concepts and Distributive Lattices by Gratzer

lattice-ordersorder-theory

In the third section "Some Algebraic Concepts" he gives a definition for sublattices. Verbatim:

let $$ A_\lambda, \lambda \in \Lambda $$ be sublattices of L. Then
$$ \bigcap (A_\lambda | \lambda \in \Lambda ) $$ (the set theoretic intersection) is also closed under meet and join; thus for every H subset of L, H nonempty, there is a smallest [H] subset of L containing H and closed under meet and join. The sublattice [H] is called the sublattice of L generated by H, and H is a generating set of [H].

I do not understand what big-lambda is. What does it mean to say little-lambda is an element of big lambda? He never defined big lambda, I have no idea what this set is. I understand what a sublattice is, I understand meet and join homomorphism (and just lattice homomorphism), I get isotone mappings and isomorphisms but I don't understand what this lambda is. He never defined this set. Is it just the set of joins? That doesn't make sense.You're taking intersections of these sublattice but I don't understand what the subscript is and what lambda is, and I don't understand what this idea of "sublattices of L generated by H" and this idea of a generating set [H].

He goes on to talk about convex lattices, ideals and prime ideals of lattices which I understand but then I am quickly blocked out from continuing because I do not understand what this idea of "sublattices generated by H" or what this lambda means.

Best Answer

$\Lambda$ is just some arbitrary indexing set. We could phrase this without an indexing set at all, as:

Let $\mathcal{A}$ be a set of sublattices of $L$. Then $$\bigcap_{X\in\mathcal{A}}X$$ (the set-theoretic intersection [of the sublattices in that set]) is also closed under meet and join. etc.

"$\{A_\lambda: \lambda\in\Lambda\}$" is the same as the $\mathcal{A}$ here, except that we've explicitly "named" the elements of $\mathcal{A}$ for easy reference, with $\Lambda$ being the set of "names" we're using.

In this specific context I think the use of an indexing set isn't helpful, but in other contexts it can substantially simplify things.


As to the definition of $[H]$, the point is that we're taking $\mathcal{A}$ to be the set of all sublattices of $L$ containing $H$. The intersection $\bigcap_{X\in\mathcal{A}}X$ is then (by the remark above) a sublattice of $L$, and it clearly contains $H$ and is a subset of every sublattice of $L$ containing $H$. So we have:

Let $H\subseteq L$ and let $\mathcal{A}$ be the set of sublattices of $L$ containing $H$. Then $$[H]:=\bigcap_{X\in\mathcal{A}}X$$ (the set-theoretic intersection of those sublattices) is the smallest sublattice of $L$ containing $H$; that is, $[H]$ is a sublattice of $L$, $H\subseteq [H]$, and whenever $J$ is a sublattice of $L$ with $H\subseteq J$ we have $[H]\subseteq J$.


An aside:

It's worth noting that any set can be "converted" into an indexed set, albeit in a rather silly way: use the set itself as the indexing set! That is, suppose I have a set $\mathcal{S}$. Let $\Lambda=\mathcal{S}$ and for $\lambda\in \Lambda$ - that is, for $\lambda\in S$ - let $S_\lambda=\lambda$. Then $\mathcal{S}=\{S_\lambda: \lambda\in \Lambda\}$.

This is admittedly really really silly, but it does tell us that we can always switch from sets to indexed sets without actually doing anything; it's just an introduction of extra language and symbols, which hopefully(!) improves the presentation. Now like I said, in this specific case I think the use of an indexing set actually just makes things messier, but it is often quite convenient.

(That's not to say that there aren't subtleties around indexed sets. For example, the axiom of choice is equivalent to "For every set $\mathcal{S}$ there is some indexing set $\Lambda$ and some 'naming surjection' $f:\Lambda\rightarrow\mathcal{S}$ - basically, $f$ tells us how $\Lambda$ indexes $\mathcal{S}$ - such that $\Lambda$ is well-orderable." But as long as we just want some indexing set, that's not a problem.)