3
$\begingroup$

Awodey makes a claim in Chapter 6 of his Category Theory.

A poset is (co)complete if it is so as a category, thus if it has all set-indexed meets $\bigwedge_{i\in I}a_i$ (resp. joins $\bigvee_{i\in I}a_i$). A lattice, Heyting algebra, Boolean algebra, etc. is called complete if it is so as a poset. For lattices, completeness and cocompleteness are equivalent.

I have a problem with the bolded statement. For example, a topology $\tau$ on a topological space $X$ is a lattice (with meet as intersection and join as union) since $\tau$ is closed under finite unions and intersections (and it also contains $\varnothing$ and $X$, the empty meet and join). It seems to me that $\tau$ is cocomplete as well, since it is closed under arbitrary unions, but not complete since open sets are not closed under arbitrary intersection.

Is there a gap in my understanding here? What's going on?

$\endgroup$
  • 2
    $\begingroup$ It's not because the meet is not the intersection that the meet doesn't exist. What do you think about the interior of the intersection ? (What's true is that it's not a sub-complete-lattice of $P(X)$) $\endgroup$ – Max Jul 5 '18 at 17:55
  • $\begingroup$ I do not understand the wording of your comment. Are you suggesting that I should take meets to be the interior of the intersection? If that's the case, that still doesn't solve my problem because, as I see it, the structure I have presented is indeed a lattice. So while taking meets to be the interior of intersections will complete the lattice, I don't see why my lattice should be complete, or isn't cocomplete (or even isn't a lattice). That is what I would like explained. $\endgroup$ – D. Brogan Jul 5 '18 at 19:27
  • $\begingroup$ I don't understand your objection. I'll write an answer, that might be clearer to understand $\endgroup$ – Max Jul 5 '18 at 19:29
5
$\begingroup$

Your topology $\tau$ is indeed a cocomplete lattice. But notice that the notion of "lattice" only includes the order $\leq$ and the finite meets and joins $\land, \lor$.

In your case, the finite meets are indeed intersections: a finite intersection of open sets is an open set, by definition.

The joins (arbitrary ones) are unions: an arbitrary union of open sets is open, by definition.

But, as you notice, an arbitrary intersection of open sets need not be open: it seems as though $\tau$ is not complete. But it is, because of the theorem you mention. So where did we go wrong ?

Well we went wrong when we went from "$\tau$ is not stable under arbitrary intersections" to "$\tau$ doesn't have arbitrary meets".

Let's see why that goes wrong on an easier example. Consider a lattice with $4$ elements :$a,b,c,d$ where $a\leq b$ and $b\leq c, b\leq d$. This is indeed a lattice (check it if you're not convinced !).

Now let's call it $L$ and let $\land_L$ denote the meet in $L$. In particular we have $c\land_L d= b$. But consider the subset $\{a,c,d\}$: it is a partially ordered set, but it's not stable under $\land_L$: indeed $b$ is not in it. Can we conclude that it's not a lattice ? No, indeed in this subset the meet of $c,d$ exists, and it's $a$, but it doesn't coincide with $\land_L$. T

hat can happen on this level, but it can also happen for complete lattices: that is we may have a complete lattice $L$ with a sublattice $L'$ such that both $L,L'$ are complete and $\land_L = \land_{L'}$ (finite meets !) but arbitrary meets in $L'$ don't need to be the same as in $L$.That's exactly what happens here: you're seeing $\tau$ as a sublattice of $\mathcal{P}(X)$ (power set of $X$, with $\subset$). You notice that the two have the same finite meets; and you notice that (denoting by $\bigwedge_X$ arbitrary meets in $\mathcal{P}(X)$) $\tau$ is not stable under $\bigwedge_X$. Does that mean that $\tau$ is not complete ?

Certainly not; just as above. In fact; if $(O_i)_{i\in I}$ is a family of open sets, then $\bigwedge_{i\in I}O_i$ in $\tau$ (not in $\mathcal{P}(X)$ !) is precisely $\mathrm{Int}(\displaystyle\bigcap_{i\in I}O_i)$. Indeed, this is clearly open, it's clearly included in each of the $O_i$'s; furthermore if $O$ is an open set such that $O\subset O_i$ for each $i$, then $O\subset \displaystyle\bigcap_{i\in I}O_i$ by definition of the intersection, and thus, by definition of the interior, $O\subset \mathrm{Int}(\displaystyle\bigcap_{i\in I}O_i)$: thus this is precisely the definition of a meet: it's a lower bound such that every other lower bound is smaller than it.

In fact, there's nothing special about open sets here. Consider the following : let $L$ be a cocomplete and complete (though this second condition is not necessary) lattice, $L'$ a sublattice of $L$ (that is, $L'$ is a subset that is closed under finite meets and finite joins in $L$) such that arbitrary joins in $L'$ exist, and are the same as those in $L$. Then for any subset $S\subset L'$, the meet of $S$ in $L'$ is the join in $L'$ (and thus in $L$) of $\{x\in L'\mid x\leq \displaystyle\bigwedge_L S\}$; and the proof is exactly the same as above ! Remember that the interior of a set is nothing but the union (join) of all open sets included in it.

$\endgroup$
  • $\begingroup$ This makes perfect sense. Thank you. $\endgroup$ – D. Brogan Jul 5 '18 at 19:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.