BOREL CLASSES OF SETS OF EXTREME AND EXPOSED POINTS IN ℝ^{n}

PETR HOLICKÝ AND TAMÁS KELETI

Abstract. It is known that the sets of extreme and exposed points of a convex Borel subset of ℝ^{n} are Borel. We show that for n ≥ 4 there exist convex G_{δ} subsets of ℝ^{n} such that the sets of their extreme and exposed points coincide and are of arbitrarily high Borel class. On the other hand, we show that the sets of extreme and of exposed points of a convex set C ⊂ ℝ^{3} of additive Borel class α are of ambiguous Borel class α + 1. For proving the latter mentioned results we show that the union of the open and the union of the closed segments of C ∩∂C are of the additive Borel class α if C ⊂ ℝ^{3} is a convex set of additive Borel class α.

1. Introduction

Rogers showed in [?] that the set ext C of all extreme points of every convex Borel subset of ℝ^{n} is Borel. Jayne and Rogers proved in the introduction of [?] that also the set exp C of all exposed points of any convex Borel subset of ℝ^{n} is Borel. Thus they improved the result of [?] that exp C is Borel if C is a closed convex subset of a Euclidean space. The mentioned results and the methods of their proofs in [?] and [?] do not give any information about the dependence of the Borel classes of the sets ext C and exp C on the Borel class of C. The only published results in this direction we know about concern convex F_{σ} subsets of ℝ^{n}. Jayne and Rogers explain in [?, p. 252] that even for every convex K_{σ} subset C of any Banach space the set ext C is G_{δ} in C. Choquet, Corson and Klee give a more detailed description of exp C for closed convex subsets of ℝ^{n} in [?, Theorem 1.1] depending on the dimension n. For arbitrary n their result gives that exp C is of ambiguous class Δ_{3}^{0}. Theorem 1 and the remark “Added in proof” at the end of Section 1 in [?] give in particular that exp C is in Π_{3}^{0} for every convex K_{σ} subset of a Banach space.

In Example ?? we show that for convex G_{δ} subsets C of ℝ^{n}, n ≥ 4, one cannot get any uniform bound on the Borel classes of the sets of extreme and exposed points of C. On the other hand, we prove that in ℝ^{3} there is a bound β(α) of the Borel classes of ext C and exp C depending on the Borel class α of C only. We also give examples showing that our estimates are sharp.

1991 Mathematics Subject Classification. 04A15, 28A05, 52A20, 52A15.

Key words and phrases. Convex set, extreme point, exposed point, Borel class.

The first author was supported by the “Mathematics in Information Society” project carried out by Alfréd Rényi Institute of Mathematics - Hungarian Academy of Sciences, in the framework of the European Community’s “Confirming the International Role of Community Research” program, the research was partly supported also by grants GAČR 201/03/0931 and MSM 113200007.

The second author was supported by OTKA grants F 029768 and F 043620.

2. Preliminaries

For any two distinct points a, b ∈ ℝ^{n} we use the notation ]a, b[ for the open segment {(1−t)a+tb; 0 < t < 1} and [a, b] for the closed segment {(1−t)a+tb; 0 ≤ t ≤ 1}.

Let us recall that a point x of a convex set C is called an extreme point of C if x ∉]a, b[ for any distinct elements a and b of C.

The point x is called an exposed point of a convex set C if there is a continuous linear functional which attains its maximum on C at x and only at x. (Note that every exposed point of a convex set C is also an extreme point of C but the converse is not always true.)

As we already mentioned in the introduction, we use ext C and exp C to denote the sets of all extreme and exposed points of C, respectively.

Given a nontrivial linear form L : ℝ^{n} → ℝ and a real number r, we call the hyperplane S = {x ∈ ℝ^{n}; L(x) = r} a supporting hyperplane of a convex set C (at point x ∈ C) if L(x) = r and either L(y) ≤ r for all y ∈ C or L(y) ≥ r for all y ∈ C. So x is an exposed point of a convex set C if and only if {x} = C ∩ S for some supporting hyperplane of C.

If S is a supporting hyperplane of C and the affine hull of S ∩ C has dimension n, we say that S ∩ C is an n-dimensional face of C. (Let us notice that our ndimensional faces do not coincide with n-faces of [?].)

The boundary, the closure and the convex hull of a set A will be denoted by ∂A, \overline{A} and conv A, respectively.

We use the notation Π_{α}^{0}, Σ_{α}^{0} and Δ_{α}^{0} in M for the multiplicative, additive and ambiguous classes Π_{α}^{0}(M), Σ_{α}^{0}(M) and Δ_{α}^{0}(M) of Borel subsets of the metric space M, respectively:

Π_{1}^{0}(M) stands for the closed, Σ_{1}^{0}(M) for the open and Δ_{1}^{0}(M) for the clopen subsets of M,

Σ_{α}^{0}(M) = {⋃_{n∈ℕ} A_{n}; A_{n} ∈ Π_{α_{n}}^{0}(M), α_{n} < α},

Π_{α}^{0}(M) = {M \ A; A ∈ Σ_{α}^{0}(M)} and

Δ_{α}^{0}(M) = Π_{α}^{0}(M) ∩ Σ_{α}^{0}(M) for 1 < α < ω_{1} (cf. [?, p. 68]). We omit M if M = ℝ^{n} and the value of n is clear from the context. We preferably use the more descriptive notation G_{δ} for Π_{2}^{0}, F_{σ} for Σ_{2}^{0} etc.

3. Example in ℝ^{4}

The following example shows that, if n ≥ 4, then ext C and exp C can belong to arbitrary exact Borel class (except for Σ_{1}^{0}) even if C is a convex G_{δ} subset of ℝ^{n}. One has to realize only that in any interval there are Borel subsets of all exact Borel classes [?, Theorem 22.4].

Example 3.1. For any Borel subset E of a circle in ℝ^{4} (e.g. E ⊂ {x ∈ ℝ^{4}; x_{1}^{2} + x_{2}^{2} = 1, x_{3} = x_{4} = 0}), there is a bounded convex G_{δ} set C ⊂ ℝ^{4} such that ext C = exp C = E.

Proof. There are obvious examples if E = ∅. Thus we may assume without loss of generality that E ⊂ {x ∈ ℝ^{4}; x_{1}^{2} + x_{2}^{2} = 1, x_{3} = x_{4} = 0} and (1, 0, 0, 0) ∈ E. Denote B = {a ∈ [0, 2π); (cos a, sin a, 0, 0) ∉ E}. We have 0 ∉ B. By a classical theorem (see e.g. [?, p. 83]), there is a one-to-one continuous map f : F → B of a G_{δ} subset F of [0, 1] onto the Borel set B. We denote the graph of f by G and note that it is a G_{δ} subset of [0, 1] × [0, 2π) and the projection of G onto B is injective.

Now we define our set C by the equality C = C_{0} ∪ C_{1} ∪ C_{2}, where

C_{0} = {(x, y, q, h) ∈ ℝ^{4}; x^{2} + y^{2} < 1, q ∈ (−1, 1), h ∈ (−1, 1)} ,
C_{1} = {(x, y, 0, 0) ∈ ℝ^{4}; x^{2} + y^{2} = 1} and
C_{2} = {(cos a, sin a, hp, h) ∈ ℝ^{4}; (p, a) ∈ G, h ∈ (−1, 1)} .

It is clear that the set C is bounded and the sets C_{0} and C_{1} are G_{δ} sets. We shall use the well known theorem that homeomorphic images of G_{δ} subsets of complete metric spaces are always G_{δ} sets (see e.g. [?, §35, III]). The set S = {(a, hp, h) ∈ ℝ^{3}; (p, a) ∈ G, h ∈ (−1, 0) ∪ (0, 1)} being homeomorphic to G × ((−1, 0) ∪ (0, 1)) is a G_{δ} set, too. As the map taking each triple (a, q, h) ∈ S to the quadruple (cos a, sin a, q, h) ∈ C is a homeomorphism of S onto C_{2} \ C_{1} (since a cannot be 0 as we assumed that 0 ∉ B), we conclude that C = C_{0} ∪ C_{1} ∪ C_{2} is a G_{δ} subset of ℝ^{4}.

The set C_{0} is obviously convex. The set C contains C_{0} as a subset and it is contained in its closure. Thus, for proving that C is convex, it is enough to prove that any supporting hyperplane of \overline{C_{0}} intersects C in a convex set. By definition, C consists of C_{0}, of open segments {(cos a, sin a, hp, h) ∈ ℝ^{4}; h ∈ (−1, 1)} for a ∈ B (one for each a ∈ B since the projection of G onto B is injective) and of points (cos a, sin a, 0, 0) for a ∈ [0, 2π) \ B. For fixed a ∈ ℝ, at the points of ∂C_{0} with fixed first coordinates (cos a, sin a), the set \overline{C_{0}} has the same supporting hyperplane and this hyperplane intersects \overline{C_{0}} in the (2-dimensional) plane \{(cos a, sin a, q, h) ∈ ℝ^{4}; (q, h) ∈ ℝ^{2}}. Since the intersection of each such plane with C is either a segment or a point, C is a convex set.

The facts mentioned in the previous paragraph give also that the only exposed points of C are the points (cos a, sin a, 0, 0) for a ∈ [0, 2π) \ B and thus exp C = ext C = E.

4. Extreme and exposed points in ℝ^{3}

In the sequel we shall prove the following two theorems.

Theorem 4.1. Let C ⊂ ℝ^{3} be a convex set of Borel class Σ_{α}^{0}. Then the set ext C is of Borel class Π_{α}^{0} in C.

Theorem 4.2. Let C ⊂ ℝ^{3} be a convex set of Borel class Σ_{α}^{0}.

If α ≠ 2, then the set exp C is of Borel class Π_{α}^{0} in C.

If α = 2, then exp C = (F ∩ G) ∪ H, where F is an F_{σ} set in ℝ^{3}, G is a G_{δ} set in ℝ^{3} and H is a G_{δ} set in C.

As an immediate corollary of these theorems we get the following result concerning Borel classes in ℝ^{3}.

Corollary 4.3. Let C ⊂ ℝ^{3} be convex.

If C belongs to the additive Borel class Σ_{α}^{0}, then the sets ext C and exp C are of ambiguous class Δ_{α+1}^{0} in ℝ^{3}.

If C belongs to the ambiguous Borel class Δ_{α}^{0}, then the sets ext C (for any α) and exp C (for α ≠ 2) are of multiplicative class Π_{α}^{0} in ℝ^{3}.

In particular, if C belongs to the multiplicative Borel class Π_{α}^{0}, then the sets ext C (for any α) and exp C (for α ≠ 1) are of multiplicative class Π_{α+1}^{0} in ℝ^{3}.

Remark 4.4. In the particular case when C is closed (and more generally also if C ∈ Δ_{2}^{0}), the set H in Theorem ?? is G_{δ} also in ℝ^{3} (since it is G_{δ} in a G_{δ} set) and we get the same description of exp C as Choquet, Corson and Klee in [?, Theorem 1.1(b)].

The most complicated description of exp C for an F_{σ} set C is necessary. An example of Corson [?] shows that there is a compact convex set C ⊂ ℝ^{3} such that exp C is not the union of an F_{σ} set and a G_{δ} set. Let us also notice that Holický and Laczkovich [?] observed recently that there is a compact convex set C ⊂ ℝ^{3} such that exp C is not the intersection of an F_{σ} set and a G_{δ} set.

Before coming to the proofs of both theorems, we illustrate on the next examples that there is not much space for improving the estimates of the classes of ext C and exp C in Theorem ??, Theorem ?? and Corollary ??.

Example 4.5. There is a convex set D ⊂ ℝ^{3} of class Σ_{α}^{0} for α ≥ 2 such that ext D = exp D is not Σ_{α}^{0} in D.

Proof. Let B ⊂ S = {(x, y) ∈ ℝ^{2}; x^{2} + y^{2} = 1} be in Π_{α}^{0} \ Σ_{α}^{0}, α ≥ 2. Put

D = ({(x, y) ∈ ℝ^{2}; x^{2} + y^{2} < 1} × (−1, 1)) ∪ ((S \ B) × (−1, 1)) ∪ (S × {0}).

Then D is clearly convex and Σ_{α}^{0} but ext D = exp D = B × {0} is not Σ_{α}^{0} in D (otherwise it would be in Σ_{α}^{0} in ℝ^{3} because S × {0} ⊂ D is closed in ℝ^{3}).

Example 4.6. For any α ≥ 1, there is a convex Π_{α}^{0} set E ⊂ ℝ^{3} with ext E = exp E not Σ_{α+1}^{0} in E and so not Σ_{α+1}^{0} in ℝ^{3}.

Proof. Let S = {(x, y) ∈ ℝ^{2}; x^{2} + y^{2} = 1} and F_{n}, n ∈ ℕ, in Π_{α}^{0}(S) be such that their union ⋃_{n∈ℕ} F_{n} is not in Π_{α+1}^{0}(S). We may and shall suppose that F_{n} ⊂ F_{n+1} for n ∈ ℕ.

If α = 1, we put

E = conv ( ⋃_{n∈ℕ} (F_{n} × [−\frac{1}{n}, \frac{1}{n}]) ∪ (S × \{0\})) .

Using the well known fact that the convex hull of a compact set is compact (see e.g. in [?, Theorem 17.2]) we get that E is compact. It is easy to check that ext E ∩ S = exp E ∩ S = S \ ⋃_{n∈ℕ} F_{n}. Since S × {0} ⊂ E is closed and ⋃_{n∈ℕ} F_{n} is not G_{δ} in S, this implies that ext E and exp E are not F_{σ} in E. (Let us also note that the stronger examples of [?] or [?] also give the case α = 1 for exposed points.)

If α > 1, put

E = (\{(x, y) ∈ ℝ^{2}; x^{2} + y^{2} < 1\} × (−1, 1)) ∪ (S × \{0\}) ∪ ⋃_{n∈ℕ} (F_{n} × (−\frac{1}{n}, \frac{1}{n})) .

Let α_{k} = α − 1 if α is a nonlimit ordinal and let (α_{k}) be an increasing sequence of ordinals that converges to α if α is a limit ordinal. In both cases there exist sets G_{k}^{n} in Σ_{α_{k}}^{0}(S) such that F_{n} = ⋂_{k∈ℕ} G_{k}^{n} and G_{k+1}^{n} ⊂ G_{k}^{n} for n, k ∈ ℕ. Then one can easily check that

⋃_{n∈ℕ} F_{n} × (0, \frac{1}{n}) = ⋂_{k∈ℕ} ⋃_{n∈ℕ} G_{k}^{n} × (\frac{1}{n + 2}, \frac{1}{n}).

Hence E is the union of ({(x, y) ∈ ℝ^{2}; x^{2} + y^{2} < 1} × (−1, 1)) ∪ (S × {0}) and

⋂_{k∈ℕ} ⋃_{n∈ℕ} G_{k}^{n} × ((\frac{1}{n + 2}, \frac{1}{n}) ∪ ( − \frac{1}{n}, −\frac{1}{n + 2})) ,

which implies that E is a Π_{α}^{0} set. The set E is clearly convex. However, ext E = exp E = S \ ⋃_{n∈ℕ} F_{n} is not Σ_{α+1}^{0} in E.

5. Proof of Theorem ??

Lemma 5.1. Let C ⊂ ℝ^{3} be a convex set of Borel class Σ_{α}^{0}, and P_{0}, P_{1} be two distinct parallel planes in ℝ^{3}. Then the set

S = ⋃ { ]a, b[ ; [a, b] ⊂ ∂C, a ∈ C ∩ P_{0}, b ∈ C ∩ P_{1}}

is of class Σ_{α}^{0}, too.

Proof. Since the case α = 1 is trivial, we may suppose that α ≥ 2.

There are at most countably many maximal “quadrangles” Q_{k} = conv(J_{0} ∪ J_{1}), where J_{i} ⊂ ∂C ∩ P_{i}, i = 0, 1, are bounded or unbounded line segments. Let us denote by G_{k} the relative interior of Q_{k} in the affine span of Q_{k} and put G = ⋃_{k} G_{k}. Thus each G_{k} is an F_{σ} set in ℝ^{3} and open in ∂C.

The intersection S ∩ G is the countable union of S ∩ G_{k}, each of the form conv(I_{0} ∪ I_{1}) ∩ G_{k}, where I_{0} ⊂ J_{0} and I_{1} ⊂ J_{1} are convex subsets of a line, and so S ∩ G is an F_{σ} set.

The map taking each triple (x_{0}, x_{1}, t) ∈ (∂C ∩ P_{0}) × (∂C ∩ P_{1}) × (0, 1) to the point (1 − t)x_{0} + tx_{1} ∈ C is continuous. Let us denote by f its restriction to the set

K = \{(x_{0}, x_{1}, t) ∈ (∂C ∩ P_{0}) × (∂C ∩ P_{1}) × (0, 1) ; \frac{x_{0} + x_{1}}{2} ∈ ∂C \ G\}.

Note that if ]x_{0}, x_{1}[∩∂C ≠ ∅ for some (x_{0}, x_{1}) ∈ (\overline{C} ∩ P_{0}) × (\overline{C} ∩ P_{1}), then [x_{0}, x_{1}] ⊂ ∂C.

This implies that

f(K ∩ (C × C × (0, 1))) = S \ G.

The previous observation also implies that, whenever two distinct open segments with endpoints in ∂C ∩ P_{0} and ∂C ∩ P_{1} have nonempty intersection in ∂C, then their midpoints belong to G. This proves that f is injective.

The set K can be written as the countable union of compact sets K_{n}. Each restriction f ↾ K_{n} being continuous and injective on a compact set is a homeomorphism taking the Σ_{α}^{0} subset K_{n} ∩(C ×C ×(0, 1)) of K_{n} to a Σ_{α}^{0} set (in the compact set f(K_{n}) and so in ℝ^{3} as well) and thus S \ G = ⋃_{n} f(K_{n} ∩ (C × C × (0, 1))) is also a Σ_{α}^{0} set.

Proof of Theorem ??. For i ∈ {1, 2, 3} and a rational number r ∈ ℚ, we put P_{i}(r) = {(x_{1}, x_{2}, x_{3}) ∈ ℝ^{3} ; x_{i} = r}. Clearly,

C ∩ ∂C \ ext C = ⋃ { ]a, b[ ; ]a, b[ ⊂ C ∩ ∂C} =

= ⋃_{i,p,q} ⋃ { ]a, b[ ; [a, b] ⊂ ∂C, a ∈ C ∩ P_{i}(p), b ∈ C ∩ P_{i}(q)} ,

where the union ⋃_{i,p,q} is over all triples (i, p, q) ∈ {1, 2, 3} × ℚ × ℚ such that p < q. By Lemma ?? with P_{0} = P_{i}(p) and P_{1} = P_{i}(q), and by the countability of the index set {1, 2, 3} × ℚ × ℚ, the set C ∩ ∂C \ ext C is Σ_{α}^{0} in ℝ^{3} and thus ext C is Π_{α}^{0} in C.

6. Proof of Theorem ??

The next lemma enables us to reduce the investigation of the set of exposed points of convex sets C ⊂ ℝ^{3} to the study of the union U^{C} of the closed segments of C ∩ ∂C. We get it by modifying a description indicated in the proof of [?, Theorem 1.1].

Lemma 6.1. Let C be a convex subset of ℝ^{3}. Then

exp C = (C ∩ ∂C \ U^{C}) ∪ (E ∩ C \ ⋃{I ∩ C \ exp C; I ∈ I}) ,

where

U^{C} = ⋃{ [a, b] ; [a, b] ⊂ C ∩ ∂C},

E = \{x; \{x\} = S ∩ [x, c] for an [x, c] ⊂ ∂C and a supporting plane S of \overline{C}\}

and I = \{I; I = T ∩ S ∩ C, S, T are distinct supporting planes of \overline{C},

dim I = 1, dim(T ∩ \overline{C}) = 2\}.

Moreover, the sets E and ⋃{I ∩ C \ exp C; I ∈ I} are F_{σ} sets in ℝ^{3}.

Proof. Let x be an exposed point of C which belongs to some segment [a, b] ⊂ C ∩ ∂C. Then x ∈ E and thus the inclusion “⊂” is proved.

For proving the other inclusion first note that C ∩ ∂C \ U^{C} ⊂ exp C is clear. Thus it is enough to prove that x ∈ ⋃ I if x ∉ exp C and x ∈ E ∩ C.

So suppose now that x ∉ exp C and x ∈ E ∩C. Then there is a supporting plane S of \overline{C} with \{x\} = S ∩ [x, t] for some [x, t] ⊂ ∂C. As x ∉ exp C there is an s such that [x, s] ⊂ S ∩ C. There is a supporting plane T ⊃ [x, t] of \overline{C} and since t ∉ S we have that S ≠ T. The point x belongs to the line T ∩S and T ∩S ∩C ≠ {x} because otherwise x would be an exposed point of C (a suitable rotation P of the plane S around S ∩ T would be a supporting plane with P ∩ C = {x}). So I = S ∩ T ∩ C is a segment on the boundary of C. Moreover, T ∩ \overline{C} contains the two-dimensional set conv(I ∪ {t}) and so x ∈ ⋃ I and the equality is proved.

The family I is countable as the set of two-dimensional faces of \overline{C} is countable and the set of one-dimensional faces of each two-dimensional face is also countable. On the other hand, if I is a segment, and C is convex, the set I ∩ C \ exp C is also a segment, so it is F_{σ}. Therefore ⋃{I ∩ C \ exp C; I ∈ I} is F_{σ}.

The set E can be written as the countable union of sets E_{n} of all elements of E such that the plane S and the segment [x, c] from the definition of E can be found so that ∥c − x∥ ≤ 1 and the scalar product of c − x and of the unit normal vector to S does not belong to the open interval (−\frac{1}{n}, \frac{1}{n}). Then each E_{n} is a closed set which concludes the proof.

Due to Lemma ?? it is enough to prove the following proposition to conclude the proof of Theorem ??.

Proposition 6.2. Let C be a convex Σ_{α}^{0} subset of ℝ^{3}. Then the set

U^{C} = ⋃{ [a, b] ; [a, b] ⊂ C ∩ ∂C}

is a Σ_{α}^{0} set in ℝ^{3}.

Let us realize that we know that ⋃{]a, b[ ; ]a, b[ ⊂ C ∩ ∂C} is in Σ_{α}^{0} from (the proof of) Theorem ??. The point distinguishing the case of exposed points from the case of extreme points is that we need to include the endpoints in our description, which makes it more delicate (recall that in the proof of Example ?? the set S is the union of open segments in ℝ^{3}, it is a G_{δ} set but, if we take the union of the closed segments, we get a set which can be in arbitrarily high Borel class). We are going to describe U^{C} as the union of countably many sets which will be studied separately.

It will be useful to note that U^{C} can be also written as

U^{C} = { x ∈ C; ∃ c ∈ C, [x, c] ⊂ ∂C}.

We call a subset D of ℝ^{3} a rational disc if it is a closed disc with positive rational radius in a plane P ⊂ ℝ^{3}, P is perpendicular to one of the axes and the center of D has rational coordinates. Let D denote the countable set of all rational discs.

For a fixed rational disc D in a plane P we define the multivalued map Φ_{D} taking points of ℝ^{3} \ P to compact subsets of D by

Φ_{D}(x) = {y ∈ D; [x, y] ⊂ ∂C},

and we put

S_{D} = {x ∈ ℝ^{3} \ P ; Φ_{D}(x) is a singleton}.

By restricting Φ_{D} to S_{D}, and identifying the singletons Φ_{D}(x) with their elements, we define the function φ_{D} : S_{D} → D such that

{φ_{D}(x)} = Φ_{D}(x) for x ∈ S_{D}.

Claim 6.3. Let D ∈ D be arbitrary and P ⊃ D be a plane. Then the multivalued map Φ_{D} is upper semi-continuous (i.e. Φ_{D}^{−1}(F) = {x; Φ_{D}(x) ∩ F ≠ ∅} is closed in ℝ^{3} \ P for every closed F ⊂ D) and the set S_{D} is G_{δ}.

In particular, we get that Φ_{D}^{−1}(F) ∈ F_{σ} if F ∈ F_{σ} and the restriction φ_{D} : S_{D} → D of Φ_{D} to S_{D} is continuous.

Proof. The first claim follows since ∂C is closed and the disc D is compact. For the other one it is enough to realize that the sets F_{n} = \{x ∈ ℝ^{3} \ P ; diam Φ_{D}(x) ≥ \frac{1}{n}\} are closed in ℝ^{3} \ P for n ∈ ℕ, the set F = {x ∈ ℝ^{3} \ P ; Φ_{D}(x) ≠ ∅} is closed in ℝ^{3} \ P and S_{D} = F \ ⋃_{n∈ℕ} F_{n}.

Let

U_{f} = ⋃\{F ; F is a two-dimensional face of \overline{C}\} and

U_{s} = {x ∈ ∂C; there are at least three disjoint segments ]a_{i}, x[⊂ ∂C \ U_{f}}.

Claim 6.4. We have the equalities

U^{C} = C ∩ ⋃_{D∈D} Φ_{D}^{−1}(C ∩ D) = (C ∩ ⋃_{D∈D} φ_{D}^{−1}(C ∩ D)) ∪ (U_{f} ∩ U^{C}) ∪ (U_{s} ∩ U^{C}).

Proof. The first equality and the inclusion

U^{C} ⊃ (C ∩ ⋃_{D∈D} φ_{D}^{−1}(C ∩ D)) ∪ (U_{f} ∩ U^{C}) ∪ (U_{s} ∩ U^{C})

are clear.

Let x ∈ U^{C} \ (U_{f} ∪ U_{s}). Then [c, x] ⊂ C ∩ ∂C for some c. There is a rational disc D with D∩]c, x[ being a singleton and D does not meet any segment ]c′, x[ ⊂ ∂C\U_{f} such that ]c′, x[ ∩]c, x[ = ∅. As x ∉ U_{f}, there is no segment [c′′, x] ⊂ C ∩ ∂C ∩ U_{f}. By noting that for any segment ]a, b[⊂ ∂C either ]a, b[⊂ ∂C \ U_{f} or [a, b] ⊂ U_{f}, this implies that x ∈ S_{D} and φ_{D}(x) is the only element of ]c, x[ ∩D. Since ]x, c[⊂ C we get that φ_{D}(x) ∈ C ∩ D, so x ∈ φ_{D}^{−1}(C ∩ D).

Claim 6.5. The set U_{f} ∩ U^{C} is relatively F_{σ} in C.

Proof. Note first that there are countably many two-dimensional faces F_{k}, k ∈ ℕ, of \overline{C}. Let N_{i} = \{k ∈ ℕ; F_{k} ∩ C is an i-dimensional face of C\} for i = 0, 1, 2. Then

⋃_{k∈N_{1}∪N_{2}} F_{k} ∩ C ⊂ U_{f} ∩ U^{C} ⊂ ⋃_{k∈N_{0}∪N_{1}∪N_{2}} F_{k} ∩ C.

Since ⋃_{k∈N_{0}} F_{k} ∩ C is a countable set and each F_{k} ∩ C is relatively closed in C, this implies that U_{f} ∩ U^{C} is F_{σ} in C.

Lemma 6.6. Let K be a closed convex set and let U_{f} denote the union of the twodimensional faces of K. Then the set U_{s} of points x for which there exist three pairwise disjoint segments ]x, b_{1}[, ]x, b_{2}[, ]x, b_{3}[⊂ ∂K \ U_{f} is countable.

Proof. First we claim that an open segment in ∂K \ U_{f} cannot intersect any nonparallel closed segment of ∂K. Indeed, otherwise they would belong to a common supporting plane S of K and S ∩ K would be a two-dimensional face of K; so both segments would be contained in U_{f}.

For each point x ∈ U_{s} take three pairwise disjoint segments ]x, b_{1}[, ]x, b_{2}[, ]x, b_{3}[ ⊂ ∂K \ U_{f}, put a_{i} = (x + b_{i})/2 and assign the “spider” S_{x} = ⋃_{i=1}^{3}[x, a_{i}[ to x.

First we show that the spiders S_{x}, x ∈ U_{s}, are pairwise disjoint. Suppose that x ≠ x′ and y ∈ S_{x} ∩ S_{x′}. We can suppose that y ≠ x. Then y ∈]x, a_{i}[ ∩ [x′, a′_{j}] for some i, j. According to our first observation this may happen only if ]x, a_{i}[ and ]x′, a′_{j}[ lie in the same line. Then we must have x ∈]x′, b′_{j}[ or x′ ∈]x, b_{i}[. The assumption y ≠ x was used only to get the last statement that is symmetrical with respect to x and x′. So we may and shall suppose now that x ∈]x′, b′_{j}[. Then for at least one k ∈ {1, 2, 3} the intervals ]x′, b′_{j}[ and [x, b_{k}] are non-parallel non-disjoint segments in ∂K \ U_{f} contradicting our first observation.

One can easily write ∂K as a countable union of relatively open (in ∂K) sets T_{n} such that each T_{n} can be projected to a plane by an injective projection. By taking the spiders smaller if necessary, we can suppose that each spider is contained in one of the T_{n}’s. Since the projections are injective and the spiders are pairwise disjoint, we get that the projections of the spiders contained in a fixed T_{n} are also pairwise disjoint.

By the injectivity, the projection of a spider is of the form ⋃_{i=1}^{3}[y, c_{i}[, where ]y, c_{1}[, ]y, c_{2}[ and ]y, c_{3}[ are pairwise disjoint open segments. We call such plane sets “plane-spiders” for a while.

Therefore what remains to prove is the (probably well known) fact that there can be only countably many pairwise disjoint plane-spiders in a fixed plane. For proving this, assign to each plane-spider a triangle such that the “head” of the spider is inside the triangle, each “leg” intersects one of the sides of the triangle and all coordinates of all vertices of the triangle are rational.

The observation that to two disjoint plane-spiders distinct “rational” triangles are assigned concludes the proof.

Proof of Proposition ??. By the first equality of Claim ?? and the upper semicontinuity of the multivalued map Φ_{D} (Claim ??), we have that for α = 2 the set U^{C} is indeed F_{σ}.

The α = 1 case is obvious, so let α > 2. Then by Claim ??, U_{f} ∩ U^{C} is in Σ_{α}^{0}. Applying Lemma ?? for K = \overline{C}, we get that U_{s} is countable, so U^{C} ∩U_{s} ∈ F_{σ} ⊂ Σ_{α}^{0}. By the continuity of φ_{D} (Claim ??), the set φ_{D}^{−1}(C ∩ D) is Σ_{α}^{0} in S_{D}. Since (by Claim ??) the set S_{D} is G_{δ} and α > 2, this means that φ_{D}^{−1}(C ∩ D) is in Σ_{α}^{0} in ℝ^{3}, too. Therefore, by Claim ??, we get that U^{C} is indeed in Σ_{α}^{0}.

Let us recall that Lemma ?? and Proposition ?? imply Theorem ??.

References

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[3] P. Holický and V. Komínek, Two remarks on the structure of exposed and extreme points, Extracta Math. 15 (2000), 547–561.

[4] P. Holický and M. Laczkovich, On the descriptive properties of the set of exposed points of a compact convex subset of ℝ^{3}, Proc. Amer. Math. Soc., to appear.

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[8] R. T. Rockafellar, Convex Analysis, Princeton, New Jersey, Princeton University Press 1970.

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Department of Math. Anal., Charles University, Sokolovská 83, 186 00 Prague 8, Czech Republic

E-mail address: holicky@karlin.mff.cuni.cz

Department of Analysis, Eötvös Loránd University, Pázmány Péter sétány 1/C, Budapest, 1117, Hungary

E-mail address: elek@cs.elte.hu