Abstract Swiss Cheese Space and Classicalisation of Swiss Cheeses

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Introduction
Throughout, we use the term compact plane set to mean a non-empty, compact subset of the complex plane.Let X be a compact plane set.Then C(X) denotes the set of all continuous, complex-valued functions on X, and R(X) denotes the set of those functions f ∈ C(X) which can be uniformly approximated on X by rational functions with no poles on X.Both R(X) and C(X) are uniform algebras on X.We refer the reader to [1,2,7] and [13] for further definitions and background concerning uniform algebras and Banach algebras.
A Swiss cheese set is a compact subset of C obtained by deleting a sequence of open disks from a closed disk.Such sets have been used as examples in the theory of uniform algebras and rational approximation.Swiss cheese sets were introduced by Roth [12], where she gave the first known example of a compact plane set X such that R(X) = C(X) but X has empty interior.Since then there have been numerous applications of Swiss cheese sets in the literature.
One notable example of a Swiss cheese construction is due to McKissick [10].He gave an example of a Swiss cheese set X such that R(X) is regular but R(X) = C(X).(We will define regularity in Section 7.) The sequence of open disks used to construct this Swiss cheese set may touch or overlap, which means that the set X might have undesirable topological properties.To improve the topological properties of the resulting Swiss cheese set, while preserving the properties of the uniform algebra, a process that we call classicalisation was developed ( [5]).
We may consider a pair consisting of a closed disk and a collection of open disks in the plane, from which we obtain the desired Swiss cheese set (see Definition 2.1 below).We call such a pair a Swiss cheese and say it is classical if the collection of open disks and the complement of the closed disk have pairwise disjoint closures and the sum of the radii of all open disks is finite.Note that, in the literature, the term 'Swiss cheese' traditionally refers to what we call a Swiss cheese set.Feinstein and Heath [5] considered Swiss cheeses in which the sum of the radii of the open disks is strictly less than the radius of the larger, closed disk.They proved, using Zorn's lemma, that for such a Swiss cheese, the associated Swiss cheese set contains a Swiss cheese set associated to a classical Swiss cheese.Later, Mason [9] gave a proof of this theorem using transfinite induction.
Classical Swiss cheese sets have many desirable topological properties.For example, Dales and Feinstein [3] proved that given two points x, y in a classical Swiss cheese set there is a rectifiable path connecting x, y and such that the length of this path is no more than π|x − y|; in fact, the constant π can be replaced by π/2 here.After this observation it is easy to see that a classical Swiss cheese set is path connected (and hence connected), locally path connected (and hence locally connected), and uniformly regular, as defined in [3].Also as a consequence of connectedness, we see that a classical Swiss cheese set cannot have any isolated points.In [5] it was noted that every classical Swiss cheese set with empty interior is homeomorphic to the Sierpiński carpet as a consequence of a theorem of Whyburn [14].
Browder [1] notes that if X is a classical Swiss cheese set then R(X) is essential (see also [5]).In particular, R(X) = C(X), as originally proved by Roth [12].It follows from the Hartogs-Rosenthal theorem that X must have positive area.A direct proof that every classical Swiss cheese set has positive area is due to Allard, as outlined in [1, pp. 163-164].
Where existing examples of Swiss cheese sets in the literature are not classical, it is of interest to construct classical Swiss cheese sets which solve the same problems.As part of a general classicalisation scheme, we discuss some new techniques for constructing such classical Swiss cheese sets.
In this paper we consider what we call abstract Swiss cheeses, which are sequences of pairs consisting of a complex number and a non-negative real number.Each pair in this sequence corresponds to a centre and radius of a disk in the plane.We give the set of all abstract Swiss cheeses a natural topology and use this topology to give a new proof of the Feinstein-Heath theorem.We show that, under some conditions, we can classicalise Swiss cheese sets while only changing open disks which lie in certain regions.We prove an analogue of the Feinstein-Heath theorem for annuli.We give some results regarding regularity of R(X) for unions of compact plane sets, which will be used in the final section.Finally, we give an example of the application of a combination of these results to construct an example of a classical Swiss cheese set X such that R(X) is regular and admits a non-degenerate bounded point derivation of infinite order (as defined in Section 8), which improves an example of O'Farrell [11].This fits into our general classicalisation scheme.

Swiss cheeses and abstract Swiss cheese space
We denote the set of all non-negative real numbers by R + , the set of positive integers by N and the set of all non-negative integers by N 0 .Let a ∈ C and let r > 0. We denote the open disk of radius r and centre a by B(a, r) and the corresponding closed disk by B(a, r).We also set B(a, 0) = {a} and B(a, 0) = ∅.We say a disk with radius zero is degenerate.(a) The Swiss cheese set X E associated with the Swiss cheese E is defined by (e) The Swiss cheese E is finite if the collection D is finite and infinite otherwise.
The condition δ(E) > −∞ is equivalent to the sum of the radii of the open disks being finite.
We note that without some condition on the disks in D we can obtain every compact plane set as a Swiss cheese set with this definition.
Throughout this paper, we will work in what we call abstract Swiss cheese space F , where F = (C × R + ) N0 with the product topology.
We call A an abstract Swiss cheese, and we define the following.
(a) The significant index set of A is S A := {n ∈ N : r n > 0}.We say that A is finite if S A is a finite set, otherwise A is infinite.
(b) The associated Swiss cheese set X A is defined by (c) We say that A is semiclassical if ∞ n=1 r n < ∞, r 0 > 0 and for all k ∈ S A the following hold: r 0 > 0 and for all k ∈ S A the following hold: For α ≥ 1 we define the discrepancy function of order α , Note that in (2) we could instead write If A is semiclassical or classical, then πδ 2 (A) is the area of the Swiss cheese set X A .We will usually write A = ((a n , r n )) for an abstract Swiss cheese.All sequences, unless otherwise specified, will be indexed by N 0 .We also define the following functions on F .
Definition 2.3.The radius sum function is the map ρ : F → [0, ∞] defined by The centre bound function is the map µ : F → [0, ∞] defined by Let E ⊆ C. For an abstract Swiss Cheese A = ((a n , r n )) we define H A (E) to be the set of those n ∈ S A such that B(a n , r n ) ∩ E = ∅.The local radius sum function on E is the function It is easy to see that ρ and µ are both lower semicontinuous from F to [0, ∞].(For ρ, this is an easy consequence of Fatou's lemma for series.) We now explain the connection between Swiss cheeses, as in Definition 2.1, and abstract Swiss cheeses.We construct a many-to-one surjection of a subset of F onto the collection of all Swiss cheeses as in Definition 2.1.Let A = ((a n , r n )) be an abstract Swiss cheese with r 0 > 0. Then we can obtain an associated Swiss cheese E A by setting The associated Swiss cheese sets of A and E A are equal, and E is a finite Swiss cheese then there is a finite abstract Swiss cheese A such that E A = E.
Let E = (∆, D) be a Swiss cheese.If E is (semi)classical then there is an abstract Swiss cheese A with E A = E such that A is (semi)classical.Moreover, when the sum of the radii of open disks in D is finite, we can find an abstract Swiss cheese A = ((a n , r n )) with ρ(A) < ∞ and E = E A such that the sequence (r n ) ∞ n=1 is non-increasing.We denote the collection of all abstract Swiss cheeses A = ((a n , r n )) with ρ(A) < ∞ and (r n ) ∞ n=1 non-increasing by N .In addition, for each M > 0 and R > 0, we denote the set of all those abstract Swiss cheeses Proof.As for the lower semicontinuity of ρ, it is an easy consequence of Fatou's lemma for series that δ by the dominated convergence theorem, we have We remark that there are examples showing that δ 1 is only upper semicontinuous, but not continuous.(c) Suppose A ∈ N .Let a ∈ C and r > 0 and k, ℓ ∈ N with k = ℓ.We say an abstract Swiss cheese B = ((b n , s n )) is obtained from A by replacing the disks B(a k , r k ), B(a ℓ , r ℓ ) by B(a, r) if B is obtained by deleting the disks at indices k, ℓ and inserting the disk B(a, r) at the first index in N such that the sequence (s n ) ∞ n=1 is non-increasing.Note that, if A ∈ N , then the abstract Swiss cheese B obtained by deleting or replacing disks, as defined in Definition 2.5, is also in N .

Some geometric results
Throughout, we shall require the following elementary geometric lemmas.The first is probably well-known, and the proof is elementary.
The following two elementary lemmas are essentially those used in [5,9], but including some additional information distilled from the original proofs.These lemmas are summarised in Figure 1.In the first lemma, we allow for the line segment to be degenerate.Lemma 3.2.Let a 1 , a 2 ∈ C and r 1 , r 2 > 0. Then there exists a unique pair Moreover, the point a lies on the line segment joining a 1 and a 2 .Suppose further that B(a 1 , r 1 ) ∩ B(a 2 , r 2 ) = ∅.Then r ≤ r 1 + r 2 , and equality holds if and only if B(a 1 , r 1 ) ∩ B(a 2 , r 2 ) = ∅.
The cases in which equality holds in Lemmas 3.2 and 3.3 are illustrated in Figure 2.

Classicalisation of Swiss cheeses
We aim to give a topological proof of the Feinstein-Heath classicalisation theorem (Theorem 4.1), as described in the introduction, stated below in the language of abstract Swiss cheeses.Theorem 4.1.Let A = ((a n , r n )) be an abstract Swiss cheese with δ 1 (A) > 0.
Then there exists a classical, abstract Swiss cheese B ∈ F such that X B ⊆ X A and δ 1 (B) ≥ δ 1 (A).
We will see below that it is enough to prove this theorem for abstract Swiss cheeses where some redundancy has been eliminated, as the general case then follows.We first introduce the following terminology.
An elementary argument, which we leave to the reader, shows that it is easy to eliminate redundancy from abstract Swiss cheeses with finite radius sum, as in the following lemma.
Note that, since B(b 0 , s 0 ) = B(a 0 , r 0 ) and ρ(B) ≤ ρ(A) in the above lemma we actually have δ 1 (B) ≥ δ 1 (A), as we claimed before.It is clear, by Lemma 4.3, that to prove Theorem 4.1 it is enough to consider A such that δ 1 (A) > 0 and A is redundancy-free.
We now define a relation on F which will help us to construct a compact subset of F .Then we prove the existence of classical abstract Swiss cheeses with desired properties in this compact subset.

Fix a redundancy-free abstract Swiss cheese
By our conditions on A it is clear that A ∈ S(A).We now prove that S(A) is compact.
Lemma 4.5.The set S(A) is a compact subset of F .
Proof.As noted earlier, it is enough to prove that S(A) is closed in N (M, R) and that the 0-th coordinate projection is bounded on S(A).The latter is clear from the definition of S(A), so we prove that n=0 be an abstract Swiss cheese in S(A), and suppose the sequence ( It is easy to see (by Lemma 3.1, for example) that B(b and the result is trivial, so we may assume that k ∈ S A .First assume that there exists for all m ≥ n 0 by Lemma 3.1.Letting m → ∞, we obtain |a k − a 0 | ≥ r k + r 0 , and so, by Lemma 3.1 again, B(a k , r k ) ⊆ C\ B(b 0 , s 0 ).
Otherwise for each n 0 ∈ N 0 , there exist m ≥ n 0 and ℓ m ∈ N such that ℓm ). ( By passing to a subsequence of A (m) if necessary, we can assume (5) holds for all m ∈ N 0 .For each m, since r (m) ℓm ≥ r k , by (4) we have ℓ m ≤ R/r k .Thus there must be a p ∈ N that appears infinitely many times in the sequence (ℓ m ) m .Passing to a subsequence again if necessary, we may assume ℓ m = p for all m.
p ), it is again easy to show, using Lemma 3.1, that B(a k , r k ) ⊆ B(b p , s p ). Thus B is partially above A and we have proved that S(A) is closed.
Since δ 1 is upper semicontinuous and S(A) is compact and non-empty, δ 1 attains a maximum value on S(A) and this value is at least δ 1 (A) > 0. Let which is also compact and non-empty.
In this case, we have We now show that B ′ ∈ S(A).Clearly B ′ ∈ N by our definition of replacing disks in an abstract Swiss cheese.Since b lies on the line segment connecting b k and b ℓ , it follows that µ(B ′ ) ≤ µ(B) and since s ≤ s k +s ℓ we have ρ(B ′ ) ≤ ρ(B).
then, with q as the index where B(b, s) was inserted, we have B(a p , r p ) ⊆ B(b ′ q , s ′ q ).If m = k, ℓ, then there exists q ∈ S B ′ such that B(b ′ q , s ′ q ) = B(b m , s m ).Thus B(a p , r p ) ⊆ B(b ′ q , s ′ q ).Hence B ′ is partially above A, and so B ′ ∈ S(A) as required.
(b) Let B(b, s) be the closed disk obtained by applying Lemma 3.3 to the disks B(b 0 , s 0 ) and B(b ) be the abstract Swiss cheese obtained by deleting the disks at indices 0 and k and inserting the disk B(b, s) The proof that B ′ ∈ S(A) is similar to the proof in part (a).
We are now ready to prove the main results of this section.The remaining case is where there is a k ∈ S B with B(b k , s k ) B(b 0 , s 0 ).We have δ 1 (B) ≥ δ 1 (A) > 0 so that s k < s 0 .By Lemma 4.6(b) there exists B ′ ∈ S(A) with δ 1 (B ′ ) > δ 1 (B), which is a contradiction.
Since S 1 is compact and non-empty, δ 2 attains both maximum and minimum values on S 1 .Let which is again non-empty and compact.Since all the abstract Swiss cheeses in S 1 are semiclassical, πδ 2 (B) is the area of X B for all B ∈ S 1 , and hence for all B ∈ S 2 .So the abstract Swiss cheeses in S 2 are obtained by finding those B ∈ S 1 for which the area of X B is minimal on S 1 .
Theorem 4.8.All abstract Swiss cheeses in S 2 are classical.
Otherwise there exists k ∈ S B with B(b k , s k ) B(b 0 , s 0 ).Note that s k < s 0 since δ 1 (B) > 0. By Lemma 4.6(b) there exists B ′ ∈ S(A) such that either In either case we obtain a contradiction since B ∈ S 2 .
In the next theorem, we show that if X A has empty interior then we do not have to minimise δ 2 on S 1 to find classical abstract Swiss cheeses.n , s ′ n )) such that B ′ ∈ S 1 (following the proof of Lemma 4.6).Let p be the index at which the disk B(a, r) was inserted.Since X B has empty interior, there exists m ∈ S B with m = p such that B(a, r) ∩ B(b m , s m ) = ∅.Let q ∈ S B ′ be such that B(b ′ q , s ′ q ) = B(b m , s m ).Note that p = q.Applying Lemma 4.6(a) to p, q ∈ S B ′ and B ′ , we obtain an abstract Swiss cheese B ′′ ∈ S(A) which has δ 1 (B ′′ ) > δ 1 (B ′ ).But this is a contradiction.Now suppose there exists k ∈ S B with B(b k , s k ) B(b 0 , s 0 ).Let B(b, s) be the closed disk obtained by applying Lemma 3.3 to the disks B(b 0 , s 0 ) and B(b k , s k ).Since B is semiclassical, we have s = s 0 − s k (as in Figure 2b).By deleting the disks at indices 0 and k and inserting B(b, s) at index 0, we obtain a new abstract Swiss cheese (again following the proof of Lemma 4.6).Since X B has empty interior, there exists q ∈ S B ′ such that B(b q , s q ) B(b, s).Applying Lemma 4.6(b) to q and B ′ , we obtain an abstract Swiss cheese B ′′ ∈ S(A) which has δ 1 (B ′′ ) > δ 1 (B ′ ).But this is a contradiction.
Let B be an abstract Swiss cheese satisfying δ 1 (B) > 0, so that B satisfies the conditions of Theorem 4.1.Then we can apply Lemma 4.3 to obtain a redundancy-free abstract Swiss cheese A ∈ N with X A = X B and such that δ 1 (A) ≥ δ 1 (B).We can then apply the above constructions to A. Each abstract Swiss cheese A ′ from the corresponding non-empty set S 2 is classical by Theorem 4.8 and has X A ′ ⊆ X A = X B and δ 1 (A ′ ) ≥ δ 1 (A) ≥ δ 1 (B).So we obtain the Feinstein-Heath classicalisation theorem as a corollary of Theorem 4.8.
Figure 3: The two cases in the proof of Theorem 4.9

Controlled classicalisation
In this section we discuss some situations in which it is possible to make a Swiss cheese classical without changing certain disks.This process we call "controlled classicalisation".Recall that, for E ⊆ C and an abstract Swiss cheese A = ((a n , r n )), the set Proof.Since U is open and A (m) → A as m → ∞, for each k ∈ H A (U ) there exists m 0 ∈ N 0 such that, for all m ≥ m 0 , we have k ∈ S A (m) and B(a For the rest of this section A = ((a n , r n )) ∈ N will be a fixed redundancy-free abstract Swiss cheese.Note that both ρ(A) and µ(A) are finite and r n ≤ ρ(A)/n for all n ∈ N. We define the (classical) error set of A to be Note that if E(A) ⊆ B(a 0 , r 0 ) then B(a n , r n ) ⊆ B(a 0 , r 0 ) for all n ∈ S A .We aim to prove that, under suitable conditions, we can classicalise A while leaving many of the open disks unchanged.
As in Section 4, we seek to construct a compact subset of F on which the function δ 1 can be maximised and then the function δ 2 minimised to give a suitable classical abstract Swiss cheese.
In the rest of this paper, we will frequently need to consider indexed collections of pairs of sets of the following form.Let I ⊆ N be non-empty.Let C = ((K n , U n )) n∈I , where each K n is a compact plane set and each U n is an open set with K n ⊆ U n .We call such an indexed collection a controlling collection of pairs.In the special case where I has only one member, we say C is a controlling pair and write C = (K, U ).We first require some preliminary lemmas.The following lemma is probably well-known and can be proved using a Hausdorff metric argument, but we include an elementary proof for the convenience of the reader.
Lemma 5.3.Let K be a compact plane set.Let (z n ) be a sequence in C, and let (t n ) be a sequence in R + .Suppose that B(z n , t n ) ∩ K = ∅ for all n, and that z n → z and t n → t as n → ∞.Then B(z, t) ∩ K = ∅.
Proof.For each n ∈ N 0 there exists a point Hence, taking the limit as k → ∞, we have |w − z| ≤ t so that w ∈ B(z, t) ∩ K as required.
We now prove that the space L A (C) is a compact subspace of F for an arbitrary countable collection C of pairs (K, U ) where K is a compact plane set and U an open neighbourhood of K. Lemma 5.4.Let C := ((K n , U n )) n∈I be a controlling collection of pairs.Then the set L A (C) ⊆ F is compact.
Proof.It is sufficient to show that L A (C) is closed in N (µ(A), ρ(A)), since the 0-th coordinate projection is clearly bounded on L A (C).For each m ∈ N 0 , let A (m) = ((a By Lemma 5.1 we see that B also satisfies (a), and it is immediate that (b) is also satisfied.
It remains to prove (c) and (d) hold for B. Fix k ∈ S A .Suppose that B(a k , r k ) ∩ V (C) = ∅.Since, for each m ∈ N 0 , we have A (m) ∈ L A (C) it follows that for each m there exists an integer ℓ m such that B(a k , r k ) = B(a for all m.But then there must exist an integer 1 ≤ p ≤ ρ(A)/r k such that ℓ k = p infinitely often so we can find a subsequence (A (mj ) ) j such that ℓ mj = p for all j.Since B(a k , r k ) = B(a ) for all j and A (mj ) → B as j → ∞, it follows that B(a k , r k ) = B(b p , s p ).This proves that (c) holds for B. Now suppose that B(a k , r k ) ∩ U = ∅ for some (K, U ) ∈ C. As above, for each m ∈ N 0 there exists an integer ℓ m such that B(a k , r k ) ⊆ B(a ℓm ≥ r k .We choose ℓ m as follows: if in A (m) there is an open disk B(a, r) = B(a k , r k ) then we pick ℓ m to be the index of that open disk, otherwise we choose ℓ m to be the index of an open disk B(a, r) that properly contains B(a k , r k ) and B(a, r) ∩ F (C) = ∅.Hence we have 1 ≤ ℓ m ≤ ρ(A)/r k for all m and so there exists an integer 1 ≤ p ≤ ρ(A)/r k such that ℓ m = p infinitely often.By considering a subsequence we can assume that ℓ m = p for all m.If B(a p ) = B(a k , r k ) holds for infinitely many m then there is a subsequence (A (mj ) ) j such that B(a k , r k ) = B(a p ) for only finitely many m then we must have for infinitely many m.Then there exists a subsequence (A (mj ) ) j such that B(a k , r k ) ⊆ B(a ) and B(a ) ∩ K = ∅ for all j.But then B(a k , r k ) ⊆ B(b p , s p ) and, by Lemma 5.3, we have B(b p , s p ) ∩ K = ∅.This proves that (d) holds for B.
Thus we have proved that B ∈ L A (C) and hence L A (C) is compact.
We are interested in those abstract Swiss cheeses B in a space L A (C) on which the discrepancy function δ 1 is maximised.These abstract Swiss cheeses have some desirable properties.Let L * A (C) denote the subset of L A (C) of all abstract Swiss cheeses where δ 1 achieves its maximum.Since L A (C) is nonempty and compact, L * A (C) is non-empty and compact.Recall that A ∈ N is assumed to be redundancy-free.Moreover, if B(b k , s k ) ∩ F (C) = ∅ then this ℓ ∈ S A is unique, and we have There exists a bijection σ : H 1 → H 2 satisfying the following condition: for each k ∈ H 1 and ℓ ∈ H 2 , we have then we can obtain an abstract Swiss cheese B ′ by deleting the disk at index ℓ which has Assume, for contradiction, there does not exist ℓ ∈ S A such that B(a ℓ , r ℓ ) ⊆ B(b k , s k ).Then we can delete the disk at index k from B to obtain an abstract Swiss cheese B ′ which has δ 1 (B ′ ) > δ 1 (B).It is clear that B ′ ∈ L A (C), which contradicts the maximality of δ 1 (B).Thus there exists ℓ ∈ S A such that B(a ℓ , r ℓ ) ⊆ B(b k , s k ).Now suppose, in addition, that B(b k , s k ) ∩ F (C) = ∅.We show that the ℓ ∈ S A found above with B(a ℓ , r ℓ ) ⊆ B(b k , s k ) is unique and that we have B(a ℓ , r ℓ ) = B(b k , s k ).Assume, for contradiction, that B(a ℓ , r ℓ ) = B(b k , s k ).Then, since A is redundancy-free, we must have B(a m , r m ) = B(b k , s k ) for all m ∈ S A .We claim that the abstract Swiss cheese B ′ obtained by deleting the disk at index k from B has B ′ ∈ L A (C); this will lead to a contradiction.
Clearly B ′ ∈ N (µ(A), ρ(A)) and it is also clear that B ′ satisfies conditions But now δ 1 (B ′ ) > δ 1 (B), which contradicts the maximality of δ 1 (B).Thus we must have B(a ℓ , r ℓ ) = B(b k , s k ).The uniqueness of ℓ follows from the fact that A is redundancy-free.
(c) Note that if, for some k ∈ S B and ℓ Combining this with (b), for each k ∈ H 1 there exists a unique ℓ ∈ H 2 such that B(b k , s k ) = B(a ℓ , r ℓ ).Thus we may define σ(k) = ℓ for such k, ℓ.We must show that σ is a bijection.By (a), σ is injective.Let ℓ ∈ H 2 .By 5.2(c), there exists k ∈ S B with B(b k , s k ) = B(a ℓ , r ℓ ).By the remark above, k ∈ H 1 , and so σ(k) = ℓ.This proves that σ is surjective.It is now immediate that n∈H1 s n = n∈H2 r n .This completes the proof.
In order to obtain a controlled classicalisation theorem, we need to impose some technical conditions on C. Recall that if E ⊆ C is non-empty and z ∈ C then we define the distance of z to E by dist(z, E) := inf{|z − x| : x ∈ E}.For a non-empty compact set K ⊆ C and positive real number M we define U (K, M ) := {z ∈ C : dist(z, K) < M }.Lemma 5.6.Let I ⊆ N be non-empty.Let (K n ) n∈I be a collection of compact plane sets and let (M n ) n∈I be a collection of positive real numbers.For each It remains to show that Note here that the classical, abstract Swiss cheese B obtained from this theorem is an element of L * A (C) and therefore satisfies properties (a)-(d) of Definition 5.2, and the conclusion of Lemma 5.5 holds for B. Note also that, in contrast to the Feinstein-Heath classicalisation theorem, δ 1 (B) may be negative here.We can obtain similar results using transfinite induction.
Taking I to have just one element in Theorem 5.7, we obtain the following corollary, which we use in Section 8.
Corollary 5.8.Let K be a compact plane set and let M be a positive real number.Let U = U (K, M ) and let C be the controlling pair (K, U ). Suppose that ρ U (A) < M and E(A) ⊆ K. Then there exists B = ((b n , s n )) ∈ L * A (C) such that X B \ U = X A \ U and B is classical.
In Section 8 we give an application of controlled classicalisation to construct an example of a classical Swiss cheese set X such that R(X) is regular and admits a non-degenerate bounded point derivation of infinite order, which improves the example constructed by O'Farrell [11].First we need to discuss annular classicalisation and discuss regularity of R(X).

Annular classicalisation
In this section we give some results about Swiss cheese like sets obtained by deleting open disks from a closed annulus, rather than a closed disk.If K is a closed annulus in the plane, we can write K = B(a 0 , r 0 ) \ B(a 1 , r 1 ) for some a 0 = a 1 ∈ C and r 0 > r 1 > 0 real.We say an abstract Swiss cheese A = ((a n , r n )) is annular if a 0 = a 1 and 0 < r 1 < r 0 and let K A denote the annulus B(a 0 , r 0 )\B(a 1 , r 1 ).We shall usually omit 'abstract' from the statement A is an annular abstract Swiss cheese.
Note that, in the previous lemma, K B = K A and ρ ann (B) ≤ ρ ann (A) together imply that δ ann (B) ≥ δ ann (A).
For the rest of this section, let A = ((a n , r n )) be an annular Swiss cheese with δ ann (A) > 0, such that µ(A) < ∞ and (r n ) ∞ n=2 is non-increasing.Proof.It is easy to see that the family A is pointwise bounded by properties (b),(c) and (e) so it remains only to prove that A is closed.For each m ∈ N 0 , let Since each A (m) ∈ A we have r 0 ≥ r Hence A is closed and pointwise bounded and is therefore compact by Tychonoff's theorem.Let B = ((b n , s n )) ∈ A with δ ann (B) > 0. Then we have b 0 = b 1 and δ ann (B) > 0 and this implies that s 0 > s 1 and it follows that B is annular.
The proof that δ ann is upper semicontinuous is an immediate consequence of Fatou's lemma for series, similar to the upper semicontinuity of δ 1 .
To prove that the restriction of δ 2 to A is continuous note that, for n ∈ N with n ≥ 2, we have s The result then follows from the dominated convergence theorem as in the proof of Lemma 2.4.
It is clear that A ∈ A and so A is non-empty.For all B ∈ A we also have X B ⊆ X A .We require one additional lemma before we prove the main theorem.Lemma 6.5.Let A be as in Lemma 6.4.Let B = ((b n , s n )) ∈ A be an annular Swiss cheese such that δ ann (B) ≥ δ ann (A).Suppose there exists k ) be the abstract Swiss cheese obtained by deleting the disks at indices k, ℓ from B and inserting the disk B(b, s) at the first index in so that δ ann (B ′ ) ≥ δ ann (B).By the maximality of δ ann (B), equality must hold here and in (6).Thus s = s k + s ℓ and s 2 = (s k + s ℓ ) 2 > s 2 k + s 2 ℓ so that δ 2 (B ′ ) < δ 2 (B).This contradicts the minimality of δ 2 (B).It follows that no such k, ℓ exist.Now suppose there exists k ∈ S B \ {1} such that B(b k , s k ) ∩ C \ K B = ∅ and s k > 0. By Lemma 6.5 there exists an annular Swiss cheese B ′ ∈ A with δ ann (B ′ ) ≥ δ ann (B) such that, if δ ann (B ′ ) = δ ann (B) then δ 2 (B ′ ) < δ 2 (B).This is a contradiction, so no such k can exist.It follows that B is classical.

Regularity of R(X)
Let X be a compact plane set.We say that R(X) is regular if, for all closed sets E ⊆ X and points x ∈ X \ E, there exists a function f ∈ R(X) such that f (x) = 1 and f (y) = 0 for all y ∈ E. We say that R(X) is normal if, for each pair of disjoint closed sets E, F ⊆ X, there exists a function f ∈ R(X) such that f (x) = 0 for all x ∈ E and f (x) = 1 for all x ∈ F .It is standard that R(X) is regular if and only if it is normal (see [2,Proposition 4.1.18]).
In order to avoid ambiguity, we introduce the following notation to clarify in which topological space we are taking the interior.Let X be a compact plane set and E ⊆ X.Then int X E denotes the interior of E in the topological space X.
Definition 7.1.Let X be a compact plane set, and let x ∈ X.We denote by M x the ideal of all functions in R(X) which vanish at x.We denote by J x the ideal of all functions in R(X) which vanish on a neighbourhood of x.We say x is an R-point for R(X) if, for all y ∈ X with y = x, we have J x M y .
It is standard that R(X) is regular if and only if every point x ∈ X is an R-point of R(X).The following proposition is a special case of [6,Corollary 4.7].Proposition 7.2.Let X be a compact plane set such that R(X) is not regular.Let E denote the set of non-R-points for R(X).Then E contains a non-empty perfect subset.In particular, E is uncountable.
such that R(X m ) is regular, where X m := X A (m) .Note that, for each m ∈ N, since A (m) is annular we have a The set {(a n ) : m, n ∈ N, n ≥ 2} is countably infinite so we may enumerate it as a sequence of pairs (a n , r n ), so that each pair occurs exactly once.Let a 0 := 0 and r 0 := 1 and let A = ((a n , r n )) be the resulting abstract Swiss cheese.It is clear that 0 ∈ X A .It is easy to check that ∞ m=1 int XA X m = X A \ {0} so that R(X A ) is regular by Corollary 7.6.
Using the notation for closed annuli from Section 6, for each m ∈ N, let W m = K A (m) ∪ K A (m+1) .Then, by (7), we see that We also have By an application of Lemma 4.3, we may assume that A is redundancy-free and A ∈ N while preserving X A , the regularity of R(X A ) and the inequalities (8) and (9).Since each disk meets at most two of the E m , R(X B ) admits a non-degenerate bounded point derivation of infinite order at 0 by Hallstrom's theorem [8] (see also [11]).
We raise the following open question related to regularity and bounded point derivations.
Question 8.7.Let X be a compact plane set such that R(X) has no non-zero bounded point derivations.Is R(X) necessarily regular?
We would like to thank the referee for helpful comments.
For a non-degenerate open or closed disk D in the plane, let r(D) denote the radius of D; for a degenerate disk D we define r(D) = 0.The following is the definition of a Swiss cheese used in [5].Definition 2.1.Let ∆ ⊆ C be a non-degenerate open disk and let D be a countable collection of non-degenerate, open disks in the plane.Then the ordered pair E = (∆, D) is a Swiss cheese.We also define the following.
for each D ∈ D we have D ⊆ ∆, and for each D ′ ∈ D with D = D ′ we have D ∩ D ′ = ∅.In this case we say the Swiss cheese set associated to E is semiclassical.(d) The Swiss cheese E is classical if δ(E) > −∞, for each D ∈ D we have D ⊆ ∆, and for each D ′ ∈ D with D = D ′ we have D ∩ D ′ = ∅.In this case we say the Swiss cheese set associated to E is classical.

Definition 2 . 5 .
Let A = ((a n , r n )) be an abstract Swiss cheese.(a) Let a ∈ C and r > 0 and let m ∈ N 0 .We say an abstract Swiss cheese B = ((b n , s n )) is obtained from A by inserting a disk B(a, r) at index m if, for 0 ≤ n < m, we have b n = a n , s n = r n ; for n > m we have b n = a n−1 , s n = a n−1 , and b m = a, s m = r.(b) Let m ∈ N 0 .We say an abstract Swiss cheese B = ((b n , s n )) is obtained from A by deleting the disk at index m if, for 0 ≤ n < m, we have b n = a n , s n = r n and for all n ≥ m we have b n = a n+1 , s n = r n+1 .

Figure 1 :
Figure 1: Elementary lemmas for combining and pulling in disks.

Figure 2 :
Figure 2: Extreme cases in the combining and pulling in lemmas.

Lemma 4 . 3 .
Let A = ((a n , r n )) ∈ F with ρ(A) < ∞.Then there exists a redundancy-free abstract Swiss cheese B = ((b n , s Proof.(a) Let B(b, s) be the open disk obtained by applying Lemma 3.2 to the disks B(b k , s k ) and B(b

Theorem 4 . 7 .
All abstract Swiss cheeses in S 1 are semiclassical.Proof.Let B = ((b n , s n )) ∈ S 1 .Suppose for contradiction that B is not a semiclassical abstract Swiss cheese.Consider first the case where there are distinct k, ℓ ∈ S B with B(b k , s k ) ∩ B(b ℓ , s ℓ ) = ∅.By Lemma 4.6(a) there exists B ′ ∈ S(A) with δ 1 (B ′ ) > δ 1 (B), which is a contradiction.

Theorem 4 . 9 .
If int X A = ∅ then each abstract Swiss cheese in S 1 is classical.Proof.Let B = ((b n , s n )) ∈ S 1 .Then, by Theorem 4.7, B is semiclassical.Suppose for contradiction that B is not classical.Then there are two cases summarised in Figure 3. First suppose there exist distinct k, ℓ ∈ S B with B(b k , s k ) ∩ B(b ℓ , s ℓ ) = ∅.Then by Lemma 3.2, since B(b k , s k ) ∩ B(b ℓ , s ℓ ) = ∅, there exists an open disk B(a, r) ⊇ B(b k , s k ) ∪ B(b ℓ , s ℓ ) with r = s k + s ℓ .By replacing the disks B(b k , s k ) and B(b ℓ , s ℓ ) with B(a, r) we obtain a new abstract Swiss cheese B ′ = ((b ′ Then χ m converges pointwise to χ := χ HA(U) as m → ∞.Since r (m) k → r k as m → ∞ for each k, by Fatou's lemma for series, we have

Definition 5 . 2 .
Let C = ((K n , U n )) n∈I be a controlling collection of pairs.Define V (C) := n∈I U n , F (C) := n∈I K n .Let L A (C) denote the set of all B = ((b n , s n )) ∈ N (µ(A), ρ(A)) such that: (a) for each (K, U ) ∈ C we have ρ U (B) ≤ ρ U (A); (b) B(b 0 , s 0 ) = B(a 0 , r 0 ); (c) for all k ∈ S A with B(a k , r k ) ∩ V (C) = ∅ there exists ℓ ∈ S B such that B(b ℓ , s ℓ ) = B(a k , r k );(d) for each n ∈ I and for all k ∈ S A with B(a k , r k ) ∩ U n = ∅ : (i) there exists ℓ ∈ S B with B(b ℓ , s ℓ ) = B(a k , r k ); or (ii) there exists ℓ ∈ H B (K n ) with B(a k , r k ) ⊆ B(b ℓ , s ℓ ).Note that A ∈ L A (C), and if B ∈ L A (C) then B is partially above A. Thus if B ∈ L A (C) then X B ⊆ X A .The properties (a)-(d) reflect the properties we desire for the final abstract Swiss cheese.We will use the open sets U to bound the error set E(A).Under some technical assumptions, conditions (c) and (d) ensure that abstract Swiss cheeses maximising δ 1 in L A (C) have the property that any open disk which lies outside V (C) is the same as an open disk from A.

Lemma 5 . 5 .
Let C := ((K n , U n )) n∈I be a controlling collection of pairs.Let B = ((b n , s n )) ∈ L * A (C). Then B has the following properties.(a) For all k, ℓ ∈ S B with k = ℓ, we have B(b k , s k ) = B(b ℓ , s ℓ ).(b) For each k ∈ S B , there exists ℓ ∈ S A such that B(a ℓ , r ℓ ) ⊆ B(b k , s k ).
(a) and (b) of Definition 5.2(a).Since B(a m , r m ) = B(b k , s k ) for all m ∈ S A , it follows that 5.2(c) remains true for B ′ .Similarly, since B(b k , s k ) ∩ F (C) = ∅, 5.2(d) remains true for B ′ .This proves our claim.
Proof.Let B(b, s) be the disk obtained by the application of Lemma 3.2 to the disks B(b k , s k ) and B(b ℓ , s ℓ ).LetB ′ = ((b ′ n , s ′ n )) be an abstract Swiss cheese obtained from B by replacing the disks at indices k, ℓ with the disk B(b, s).SinceB ∈ L A (C) we have ρ Um (B) ≤ ρ Um (A) < M m /2, so that s ≤ s k + s ℓ < M m /2.Since B(b k , s k )∩K m = ∅, we must have B(b, s) ⊆ U m and hence B(b, s)∩U n = ∅ for all n ∈ I with n = m.It is clear now that either δ 1 (B ′ ) > δ 1 (B), when s < s k + s ℓ , or we have δ 1 (B ′ ) = δ 1 (B) and δ 2 (B ′ ) < δ 2 (B), when s = s k + s ℓ , so it remains to show that B ′ ∈ L A (C).By construction, and since B(b, s) ⊆ U m and B(b, s) ∩ U n = ∅ for n ∈ I with n = m, we have B ′ ∈ N (µ(A), ρ(A)) and satisfies (a) and (b) in Definition 5.2.We may assume, without loss of generality, that B(b k , s k ) ∩ K m = ∅ for some m ∈ I.It follows that s k , s ℓ < M m /2 and B(b k , s k ) ⊆ U m and B(b ℓ , s ℓ )∩U m = ∅.Let B(b, s) be the open disk obtained by an application of Lemma 3.2 to the disks B(b k , s k ) and B(b ℓ , s ℓ ).Then, by Lemma 5.6, the abstract Swiss cheese B ′ ∈ L A (C) obtained by replacing the disks B(b k , s k ) and B(b ℓ , s ℓ ) with B(b, s) has either δ 1 (B ′ ) > δ 1 (B) or δ 1 (B ′ ) = δ 1 (B) and δ 2 (B ′ ) < δ 2 (B).Both of these cases are impossible since we assumed that δ 1 was maximised on B and δ 2 was minimised on B. It follows that no such pair k, ℓ can exist and hence B is classical.

Lemma 6 . 4 .
Let A be the family of all B = ((b n , s n )) ∈ F such that (a) the sequence (s n ) n≥2 is non-increasing, (b) ρ ann (B) ≤ ρ ann (A), (c) µ(B) ≤ µ(A), (d) B is partially above A, and (e) b 0 = b 1 = a 0 , and r 0 ≥ s 0 ≥ s 1 ≥ r 1 .Then A is compact in F , each abstract Swiss cheese B ∈ A with δ ann (B) > 0 is annular.Moreover, the function δ ann | A : A → R is upper semicontinuous and the function δ 2 | A : A → R is continuous.
(a)-(d) (as in the proof of Lemma 4.5).Since convergence is pointwise, we have b 0 = a 0 and b 1 = a 1 .Since A was annular, it follows that b 0 = b 1 .