Completeness results for metrized rings and lattices

The Boolean ring $B$ of measurable subsets of the unit interval, modulo sets of measure zero, has proper radical ideals (e.g., $\{0\})$ that are closed under the natural metric, but has no prime ideals closed under that metric; hence closed radical ideals are not, in general, intersections of closed prime ideals. Moreover, $B$ is known to be complete in its metric. Together, these facts answer a question posed by J.Gleason. From this example, rings of arbitrary characteristic with the corresponding properties are obtained. The result that $B$ is complete in its metric is generalized to show that if $L$ is a lattice given with a metric satisfying identically either the inequality $d(x\vee y,\,x\vee z)\leq d(y,z)$ or the inequality $d(x\wedge y,\,x\wedge z)\leq d(y,z),$ and if in $L$ every increasing Cauchy sequence converges and every decreasing Cauchy sequence converges, then every Cauchy sequence in $L$ converges; i.e., $L$ is complete as a metric space. We show by example that if the above inequalities are replaced by the weaker conditions $d(x,\,x\vee y)\leq d(x,y),$ respectively $d(x,\,x\wedge y)\leq d(x,y),$ the completeness conclusion can fail. We end with two open questions.


Overview: a ring-theoretic question, culminating in a lattice-theoretic result
A standard result of ring theory says that if I is an ideal of a commutative ring R, then the nil radical of I (the ideal of elements having some power in I) is the intersection of the prime ideals of R containing I [1, Proposition 10.2.9, p. 352].
Jonathan Gleason 1 (personal communication) asked the present author about a possible generalization of that result. Namely, suppose R is a topological commutative ring. For any ideal I of R, let √ I denote the least closed ideal J ⊇ I such that J contains every element x such that x n ∈ J for some n ≥ 1. Must √ I be the intersection of all closed prime ideals containing I ? If not in general, does this become true if R is complete with respect to the given topology ? We shall see that the answer is negative: If R is the Boolean ring of measurable subsets of the unit interval modulo sets of measure zero, topologized using the metric given by the measure of the symmetric difference of such sets, then R is complete in that metric, and {0} = {0} (defined as above); but R has no closed prime ideals, so {0} is not an intersection of such ideals. We give the details in §2, and note in §3 how to get, from this characteristic-2 example, examples of arbitrary characteristic.
The one not-so-obvious property of our example is the completeness of B as a metric space. In §4 (which is independent of § §2-3) we note that this can be deduced from a standard result of measure theory, and then prove a general result on when a metrized lattice is complete, which yields an alternate proof.
In §5 we give a curious counterexample to that general completeness result under a weakened hypothesis.

The Boolean example
Most of the desired properties of the example sketched above are straightforward to verify.
Recall that for any set X, the subsets of X form a Boolean ring under the operations (1) 0 = ∅, 1 = X, S + T = S ∪ T \ (S ∩ T ), S T = S ∩ T.
For S, T ∈ B 0 , let Then d 0 is a pseudometric, i.e., for all S, T, U ∈ B 0 , Here (4)-(6) are immediate. To get (7), note that the second equality of (2) gives the inequality µ(S)+µ(T ) ≥ µ(S + T ). Putting S + T and T + U in place of S and T in that relation gives (7). The definition (3), and the identities of Boolean rings, show that the Boolean operations behave nicely under d 0 : These relations show in particular that if S and T are close to one another under d 0 , then the results of adding U to S and T are also close to one another, as are the results of multiplying S and T by U, in each case in a uniform way, whence addition and multiplication are continuous with respect to d 0 . Now let B be the quotient ring Then (4)-(9) clearly carry over to d, with the " =⇒ " of (5) strengthened to " ⇐⇒ ". Thus, d is a metric, and So P indeed has elements arbitrarily close to 1. Multiplying arbitrary [V ] ∈ B by such elements, we see that P has elements arbitrarily close to [V ]; so the closure of P contains every [V ] ∈ B, as claimed.
Hence no prime ideal P is itself closed, giving the final assertion.
is not an intersection of closed prime ideals. Since B is complete in the metric µ (to be proved in two ways in §4), B answers the strongest form of Gleason's question.
(To be precise, Gleason's question concerned completeness of a topological ring R in the uniform structure arising from additive translates of neighborhoods of 0. Our metrized ring has additive-translation-invariant metric by (8), so completeness in the uniform structure so arising from the metric topology is equivalent to completeness in the metric.)

Non-Boolean algebras
Is the behavior of above example limited to Boolean rings, or perhaps to rings of finite characteristic? No. We note below how to generalize the construction of the preceding section to algebras in the sense of General Algebra (a.k.a. Universal Algebra), and observe that when the algebras in question are rings (of arbitrary characteristic), these give more varied examples of the properties proved for B.
We start with the analog of B 0 .
Definition 3. For X a set, let X [0,1] denote the set of all X-valued functions on the unit interval, and for each f ∈ X [0,1] and x ∈ X, let Let X ′ denote the subset of X [0, 1] consisting of those f such that the image of f, that is, {x ∈ X | f x = ∅}, is countable (i.e., finite or countably infinite).
We now want to deduce the corresponding results with the set of functions X ′ replaced by the set of equivalence classes of such functions under the relation of differing on a set of measure zero. We will need the following observation. Let S 0 , S 1 , . . . be a countable (i.e., finite or countably infinite) family of elements of B 0 such that and the T i partition [0, 1], i.e., satisfy the two conditions and These sets are clearly measurable and partition [0, 1].
Since the sets S j whose members are removed from S i in (23) have, by (18), only a set of measure zero in common with S i , we see that for i > 0, T i differs from S i in a set of measure zero, giving (20) for such i. Also by (23), no T i with i > 0 contains elements of S 0 , so by (24), T 0 ⊇ S 0 ; hence to prove the i = 0 case of (20), it suffices to show that µ(T 0 ) = µ(S 0 ). To do this, note that since the . If we subtract from that relation the equations µ(T i ) = µ(S i ) for all i > 0, which follow from the cases of (20) already obtained, we get the desired i = 0 case. Now -still assuming the completeness of B, to be obtained in the next section -we can get Proposition 6. For X a set, let X * denote the quotient of X ′ (defined in Definition 3) by the equivalence relation d ′ (f, g) = 0, and let d * be the metric on X * induced by d ′ .
Then X * under the metric d * is a complete metric space.

Proof. Consider any Cauchy sequence
is nonempty is countable by (14); hence there exists a countable (possibly finite) list of distinct such elements: Now (writing d 0 and d, as in the preceding section, for our pseudometric on B 0 and metric on B), we have for all i, m, n, xi ], . . . converges to an element which we shall write [S i ], choosing an arbitrary representative S i ∈ B 0 of the limit of that sequence in B.
I claim that these sets S i satisfy (18) and (19). To get the first of these equations for given i = j, note that for any ε > 0, one can choose n such that d 0 (f are disjoint, we see that S i and S j intersect in a set of measure at most ε. Since this holds for all ε > 0, they intersect in a set of measure 0. To get (19), note that for any ε > 0 we may choose m such that for all n ≥ m we have Since f x0 , f x1 , . . . partition [0, 1], we can also choose j such that and since a subset of [0, 1] cannot have measure larger than 1, we get (19).
Lemma 5 now gives us a partition of [0, 1] into sets T i which differ from the S i by sets of measure zero.
Remark: If in (14) we had allowed uncountable cardinalities, we would not have been able to use basic properties of measure, e.g., in concluding that the set in the middle term of (15) was measurable, and in proving in Lemma 4 that u [0,1] carries (X ′ ) n to X ′ . On the other hand, if we had required the set of x making f x nonempty to be finite, our X * would not have been complete, except in the case where X was finite. So countability is the only choice that gives our construction X * the desired properties.
We have not yet called on (16). It implies that our construction behaves nicely on algebras: Proposition 7. Suppose A is an algebra in the sense of General Algebra, that is, a set given with a (finite or infinite) family of operations, each of finite arity. Then for each operation u : A n → A of A, the operation u * of A * described by is well-defined, and uniformly continuous in the metric d * ; indeed, it satisfies Lipschitz condition The algebra A * satisfies all identities satisfied by A. In fact, every finite set of elements of A * is contained in a subalgebra of A * isomorphic to a countable direct product of copies of A. are constant. Dropping those intersections that have measure zero, and looking at the algebra of members of A ′ that are constant on the remaining countably many subsets, we see that this contains f (0) , . . . , f (N −1) , and has as its image in A * a subalgebra isomorphic to a countable direct product of copies of A.
Finally, some observations specific to rings: Proposition 8. Let A be an associative unital ring. Then in the complete metrized ring A * arising by the construction of Proposition 7, the closure of every prime ideal is all of A * ; hence A * has no closed prime ideals.
On Thus, for A a commutative ring without nilpotents, the rings A * generalize the properties of the example B of the preceding section. That B is, of course, the case of this construction with A = Z/2Z.
Remark: The development of the above results in terms of measurable X-valued functions on [0, 1], modulo disagreement on sets of measure zero, feels artificial. Surely one should be able to perform our constructions abstractly in terms of the set X, the Boolean ring B, and the real-valued function on B induced by the measure on [0, 1], and then generalize it to get such results with B replaced by any Boolean ring with an appropriate real-valued function.
If we were interested in maps [0, 1] → X assuming only finitely many values, then the analog of X * could be described as the set of continuous functions from the Stone space of B to the discrete space X. But for maps allowed to assume countably many values, the function corresponding to the metric seems to be needed in defining X * . I leave the proper formulation and generalization of that construction to experts in the subject. Cf. [2,Chapter 31].
In contrast, the results of the next section will be obtained in a satisfyingly general context.

Completeness
When I first suggested the Boolean ring B of measurable subsets of [0, 1], modulo sets of measure zero, as an answer to J. Gleason's question, the one property not clear to me was completeness in the natural metric, though it seemed intuitively likely.
One might naively hope to prove completeness by showing that every sequence of measurable subsets of [0, 1] whose images in B form a Cauchy sequence "converges almost everywhere" on [0, 1]; i.e., that almost every t ∈ [0, 1] belongs either to all but finitely many members of the sequence, or to only finitely many. But this is not so; a counterexample [3, Exercise 22(6), p.94] is the sequence whose first term is [0, 1], whose next two are [0, 1/2] and [1/2, 1], whose next three are [0, 1/3], [1/3, 2/3], [2/3, 1], and, generally, whose 1+2+ . . . +(n−1)+i-th term for 1 ≤ i ≤ n is [(i − 1)/n, i/n]. The measures of these sets approach zero, so the sequence approaches ∅ in our metric; but clearly every t ∈ [0, 1] occurs in infinitely many of these sets. Looking at this example, one might still hope that given a Cauchy sequence in B, almost every t ∈ [0, 1] has the property that the terms S i which contain t are either "eventually scarce", or have eventually scare complement. But this, too, fails; to see this, take the above example, and "stretch it out" by repeating the m-th term 2 m times successively, for each m.
However, an online search turned up a proof of the desired completeness statement in a set of exercises [5] (in particular point 6 on p. 2). I cited that in the first draft of this note as the only reference for the result that I could find. David Handelman then pointed out that the desired statement follows immediately from the standard fact that L 1 of the unit interval is complete in its natural metric ( In all these sources, the key to the proof of completeness is to pass from an arbitrary Cauchy sequence to a subsequence with the property that the distance between the i-th and i+1-st terms is ≤ 2 −i . Rather magically, a sequence with this property does indeed converge almost everywhere, giving a limit of the original Cauchy sequence. In fact, this trick can be abstracted from the context of measure theory to that of lattices (or even semilattices) as in the next theorem, from which we will recover, as a corollary, the result on measurable sets modulo null sets.
Since we no longer need the notation " f x " of the preceding section for the point-set at which a function takes on the value x, we will henceforth use subscripts in the conventional way to index terms of sequences.
We remark that the condition that a metrized lattice be complete as a metric space, obtained in the theorem, is independent of its completeness as a lattice, i.e., the existence of least upper bounds and greatest lower bounds of not necessarily finite subsets (though the condition that certain infinite least upper bounds and greatest lower bounds exist will be key to the argument). For instance, any lattice, given with the metric that makes d(x, y) = 1 whenever x = y, is complete as a metric space, and, indeed, satisfies the hypotheses of the next theorem, but need not be complete as a lattice. Inversely, the totally ordered subset of the real numbers {−2} ∪ (−1, 1) ∪ {2} is complete as a lattice, but not as a metric space.
Theorem 9. Let L be a lattice, whose underlying set is given with a metric d which satisfies identically at least one of the inequalities (or, more generally, let L be an upper semilattice satisfying (33), or a lower semilattice satisfying (34)). Suppose moreover that in L Then every Cauchy sequence in L converges; i.e., L is complete as a metric space.
Proof. It suffices to prove the case where L is an upper semilattice satisfying (33), since this includes the case where L is a lattice satisfying (33), while the cases where L is a lower semilattice or lattice satisfying (34) follow by duality. So assume L such an upper semilattice. In proving L complete, it suffices to show convergence of sequences x 0 , x 1 , . . . such that (37) i≥0 d(x i , x i+1 ) < ∞, since every Cauchy sequence has such a subsequence (e.g., one chosen so that d(x i , x i+1 ) ≤ 2 −i , as in [3], [4] and [5]), and if a subsequence of a Cauchy sequence converges, so does the whole sequence.
So let the sequence x 0 , x 1 , . . . satisfy (37), and let us define Note that if in (33) we put x = x h,j , y = x j , z = x j+1 , we get Also, for h ≤ j ≤ k, the triangle inequality (applied k − j − 1 times) gives d(x h,j , x h,k ) ≤ j≤ℓ<k d(x h,ℓ , x h,ℓ+1 ). Applying (39) to each term of this summation, we get (40) d(x h,j , x h,k ) ≤ j≤ℓ<k d(x ℓ , x ℓ+1 ). In particular, for each j, the distance from x h,j to any of the later terms x h,k is ≤ j≤ℓ<∞ d(x ℓ , x ℓ+1 ). As j → ∞, this sum approaches 0, so (still for fixed h) the elements x h,j (j = h, h+1, . . . ) form an increasing Cauchy sequence. By (35) this sequence will converge; let Note that, by (40), (41), and the continuity of d in the topology it defines, we have I claim that this implies that Indeed, (43) and (44) are respectively equivalent to the conditions d(x h,j , x h,j ∨ x i,j ) = 0 and d(x h,∞ , x h,∞ ∨ x i,∞ ) = 0, and the latter can be obtained from the former using (41). So the elements x h,∞ (h = 0, 1, . . . ) form a decreasing sequence. I claim that this sequence, too, is Cauchy; in fact, that Namely, by essentially the same argument used to prove (40), one sees that for every j ≥ i, d(x h,j , x i,j ) ≤ h≤ℓ<i d(x ℓ , x ℓ+1 ); and by continuity of d, this again carries over to the limit as j → ∞. Hence by (36), the terms of (44) converge, and we can define Finally, note that for every h ≥ 0, and that this sum approaches 0 as h → ∞. Hence the x h converge, completing the proof of the theorem.
The first assertion of the following corollary clearly includes the completeness result called on in § §2-3. The remaining two assertions are further generalizations. Alternatively, if, in the latter situation, we define B fin to be the nonunital Boolean ring of measurable sets of finite measure modulo sets of measure zero, then B fin is again complete with respect to the metric d of (49).
In all these cases, the operations of our Boolean ring are continuous in the metric named.
Proof. In each of these cases, the lattice operations on our structure are easily shown to satisfy (33)-(36) with respect to the indicated metric. (In the case of the metric d C of (50), note that in any Cauchy sequence, all but finitely many terms must have the property that their distances from later terms are all ≤ C, so that the definition (50), applied to those distances, reduces to (49).) Hence Theorem 9 gives completeness.
The standard result mentioned earlier, that L 1 of the unit interval (indeed, of any measure space) is complete in its natural metric, follows similarly, on regarding L 1 as a lattice under pointwise max and min.
We remark that in any partially ordered set with a metric, the conjunction of conditions (35) and (36) above is easily shown equivalent to the single condition that every Cauchy sequence whose members form a chain under the partial ordering converges. But the pair of conditions as stated seems easier to work with. In particular, it is easy to see that it holds for measurable sets modulo sets of measure zero in a measure space.
Concerning conditions (33) and (34), note that these are equivalent to Lipschitz continuity of ∨, respectively ∧, with Lipschitz constant 1. One could generalize the proof of Theorem 9 to allow any Lipschitz constant; in fact, I suspect that a version of the theorem could be proved -at the cost of more complicated arguments -with these conditions weakened to say that ∨, respectively, ∧, is uniformly continuous; equivalently, that there exists a function u from the positive reals to the positive reals satisfying (51) lim t→0 u(t) = 0 such that respectively, The idea would be to choose from a general Cauchy sequence a subsequence for which the distance between i-th and i+1-st terms decreases rapidly enough not only to make these distances have a convergent sum, but to have the corresponding property after taking into account the effect of the u in (52) or (53) under the iterated application of that inequality in the proof. But I don't know whether there are situations where metrics arise that would make it worth trying to prove such a result.

Counterexample: a natural lattice under a strange metric
Another pair of conditions weaker than (33) and (34), which I at one point thought might be able to replace those two hypotheses in Theorem 9, are One can in fact prove from (54) that ; by the triangle inequality, the right-hand side is ≤ d(x 0 , x 1 )+ d(x 1 , x 1 ∨· · · ∨x i ). The second of these terms is (if i > 1) similarly bounded by d(x 1 , x 2 ) + d(x 2 , x 2 ∨ · · · ∨ x i ), and this procedure, iterated, gives (56). But one cannot similarly get . , x j ∈ L) as would be needed to carry out the argument used in the proof of Theorem 9.
I give below examples showing that that theorem in fact does not hold with (54) and (55) in place of (33) and (34). We will first get an example for upper semilattices and (54), then note how to modify it to make the semilattice into a lattice. Applying duality, one gets examples for the remaining two cases of the theorem.
To start the construction, let M be any metric space, with metric d M , and for any finite nonempty subset S of M, define its diameter, It is straightforward that d L is a metric on L; the only step requiring thought is the triangle inequality d L (S, U ) ≤ d L (S, T ) + d L (T, U ) in the case where the three sets S, T and U are distinct and the maximum defining the left-hand side of the desired inequality is given by distance between some x ∈ S and some y ∈ U. In that case, taking any z ∈ T, one sees that d L (S, Under this metric, (54) is also immediate: writing that relation as d(S, S ∪ T ) ≤ d(S, T ), we see that unless T ⊆ S, the two sides both equal diam(S ∪ T ), while if T ⊆ S, the left-hand side is zero.
I claim next that L has no infinite strictly increasing Cauchy sequences. Indeed, given S 0 S 1 · · · ∈ L, the set S 1 must have more than one element, hence have nonzero diameter; and from (59) we see that for every i ≥ 1, d L (S i , S i+1 ) ≥ diam(S 1 ), so the distances between successive terms of the sequence do not approach 0. L also has no infinite strictly decreasing Cauchy sequences, since any strictly decreasing sequence of sets starting with a finite set is finite. So, trivially, (35) and (36) hold.
Note also that every non-singleton S ∈ L has distance at least diam(S) from every other element of L, so it is an isolated point. It follows that the set of non-singleton elements of L is open in L, so the set of singleton elements is a closed set, which is easily seen to be isometric to M : d L ({x}, {y}) = d M (x, y).
Hence if we take for M a non-complete metric space, then a non-convergent Cauchy sequence in M yields a non-convergent Cauchy sequence in L. So the semilattice L is non-complete, despite satisfying (54), (35) and (36).
To get an example which is a lattice, we pass from L as above to L ′ = L ∪ {∅}, which is clearly a lattice under union and intersection. The only problem is how to extend the metric d L to L ′ . We may in fact use any extension of this metric to that set that does not sabotage the non-completeness of L; for (54) holds automatically when x and y are comparable, and ∅ is comparable to every element of L ′ . So, for instance, we might choose a fixed P ∈ L, and define (60) d L ′ (∅, S) = 1 + d L (P, S) for all S = ∅.
Since this makes ∅ an isolated point, it leaves the image of M in L ′ closed, so L ′ remains non-complete. We remark that the join operation of L, and hence of L ′ , is in general discontinuous; for given any noneventually-constant sequence x 0 , x 1 , . . . in M that approaches a limit y ∈ M, we know that {x 0 }, {x 1 }, . . . approach {y} in L; but for any z = y, if we apply − ∨ {z} we get the sequence {x 0 , z}, {x 1 , z}, . . . , which cannot approach the isolated point {y, z}.
So is it plausible that continuity of the meet and join operations, combined with (54), (35) and (36), would imply completeness of a metric lattice? Still no: if we apply the above construction of L with M taken to be a discrete non-complete metric space (e.g., {n −1 | n ≥ 1} ⊆ [0, 1]), then L is also discrete. (The only elements we don't already know are isolated are the singletons {x} for x ∈ M ; but taking ε such that the ball of radius ε about x in M contains no other points, we find that the ball of radius ε about {x} in L also contains no other points.) Hence L ′ , metrized as in (60), is also discrete; and any operation on a discrete space is continuous, though L ′ is, we have shown, non-complete.

Open questions
I have not examined the question of whether the conjunction of (54) and (55) might somehow force a metrized lattice satisfying (35) and (36) to be complete as a metric space. (In the discrete L ′ constructed above satisfying (54), the inequality (55), i.e., d L ′ (S, S ∩ T ) ≤ d L ′ (S, T ), holds when S ∩ T = ∅, but not, in general, when S ∩ T = ∅.) The referee has asked whether Gleason's original question has the same answer if restricted to the case where R an integral domain. I do not know whether this is so, with or without the assumption that R is complete in the given topology. It seems an interesting question.

Acknowledgements
I am indebted to David Handelman and Hannes Thiel for pointing me to results in the literature on completeness of metric spaces arising in measure theory, and to the referee for several useful suggestions.