Let $L$ be a bounded distributive lattice and let $PF(L)$ denote its set of prime filters.
It is well known that $PF(L)$ is a Stone space if it is equipped with the topology that has sets of the form $\phi(a):=\{P\in PF(L)|a\in P\}$ and their complements as subbase.
A proof that I encountered looked quite complicated especially on the part where compactness of $PF(L)$ is shown. For instance there it was used that for disjoint filter $F$ and ideal $I$ there always exist a prime filter containing $F$ and disjoint with $I$.
I found an alternative proof of my own but do not trust myself. This also because distributivity is not used in it. Please be so kind to check me on errors.
At first I observe that compact Hausdorff space $2^{L}$ equipped with product topology is a Stone space, and consequently every closed (hence compact) subset of it is a Stone space. For completeness let me say that here $2=\{0,1\}$ is equipped with discrete topology.
Then I identify $PF(L)$ with the set of indicator sets of prime filters in lattice $L$. Actually you can think of them as the lattice-morphisms $L\to2$. Then $PF(L)$ is a subset of $2^L$ and it is enough now to prove that it is closed.
Then I observe that function $f$ is an element of $PF(L)$ iff it satisfies:
- $f(0)=0$ and $f(1)=1$
- $f(a\wedge b)=\min\{f(a),f(b)\}$ for every $a,b\in L$
- $f(a\vee b)=\max\{f(a),f(b)\}$ for every $a,b\in L$
But it seems to me that for every mentioned condition we can find a closed set in $2^L$ such that demanding that $f$ satisfies the condition is the same thing as demanding that $f$ is an element of that closed set.
This becomes more clear if we rewrite e.g. the second condition by means of (continuous) projections as:
$|\pi_{a\wedge b}(f)-\min\{\pi_a(f),\pi_b(f)\}|=0$
This shows that $f$ must be an element of the preimage of closed set $\{0\}$ with respect to a continuous function.
If that is correct then $PF(L)$ is the intersection of closed sets hence is closed itself (and we are ready).
Any mistake in this?
I am curious.
Yes, your proof is correct. To explain why anyone would ever use the more complicated proof that you encountered, let me point out two things. First, your proof relies on Tychonoff's theorem to conclude that $2^L$ is compact, and so is essentially hiding much of the complexity of the argument inside the proof of Tychonoff's theorem. Second, the more complicated proof actually shows something stronger: it can also be used to show that the canonical map $L\to 2^{PF(L)}$ is a lattice embedding. This stronger result really does require distributivity. Indeed, since $2^{PF(L)}$ is distributive, any lattice that can embed in it must be distributive as well.