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?

  • 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

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.

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

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