On spectral structure of bounded linear operators on reflexive Banach spaces

First Advanced Course in Operator Theory and Complex Analysis, University of Seville, June 2004 ON SPECTRAL STRUCTURE OF BOUNDED LINEAR OPERATORS ON REFLEXIVE BANACH SPACES STANISLAV A. SHKARIN AND OLEG G. SMOLYANOV Abstract. A descriptive characterization of point, continuous, and residual spectra of operators acting on a separable Hilbert space is obtained. The possible point spectra of bounded linear operators acting on `p, 1<p<∞are characterized. 1. Introduction As usual Cis the field of complex numbers, Ris the field of real numbers, Zis the set of integers and Nis the set of positive integers. We also denote N=N∪ {∞}. All vector spaces in this paper are assumed to be over the field Cand all topological vector spaces are assumed to be Hausdorff. For a vector space Xand a linear operator T:DT−→ Xdefined on a linear subspace DT of X, the set σp(T) = {z∈C: dim {x∈DT:Tx =zx}=ν(z)>0} is called the point spectrum [2] of T. The number ν(z)∈Nfor z∈σp(T) is called the multiplicity of z. For any n∈Nwe denote σp,n(T) = {z∈σp(T) : ν(z) = n}and σn p(T) = {z∈σp(T) : ν(z)⩾n}. Note that σp(T) = σ1 p(T) and σ∞ p(T) = σp,∞(T). For a topological vector space X, an operator T:DT−→ Xis said to be closed [2] if its graph ΓT={(x, Tx) : x∈DT}is closed in X×Xand Tis called 2000 Mathematics Subject Classification. 47A10. Revised September 14, 2004. Partially supported by “El Ministerio de Ciencia y Tecnolog´ıa, Spain” (BFM2003-00034) and by “Junta de Andaluc´ıa” (FQM-260). 133 134 S. A. SHKARIN and O. G. SMOLYANOV densely defined [2], if DTis dense in X. The spectrum of T:DT⊆X−→ X is [2] the set σ(T) = C\½z∈C:the operator T−zI has continuous densely defined inverse ¾, where I:X−→ Xis the identity operator. The sets σc(T) = {z∈σ(T)\σp(T) : the set (T−zI)(DT) is dense in X}and σr(T) = {z∈σ(T)\σp(T) : the set (T−zI)(DT) is not dense in X} are called continuous spectrum and residual spectrum [2] of T, respectively. Obviously, σ(T) is the disjoint union of three sets σp(T), σc(T) and σr(T). For any non-empty compact set K⊂R, Kalisch [3] constructed a bounded linear operator Tacting on a separable Hilbert space such that σ(T) = σp(T) = K. Using a similar construction, Nikolskaia [8] proved that a set A⊂Cis the point spectrum of a linear continuous operators on a separable Hilbert space if and only if Ais a bounded Fσ-set. R. Kaufmann [4, 5, 6, 7] proved that a set A⊂Cis a point spectrum of a bounded linear operator on a separable Banach space if and only if Ais bounded and is a Souslin set, that is, the continuous image of a complete separable metric space. The result of Kaufmann was strengthened by the authors in the following way [11]. Theorem S. Let A, B, C be three disjoint subsets of C. Then the following conditions are equivalent. (S1) There exists a bounded linear operator Tacting on a separable Banach space Xsuch that A=σp(T),B=σc(T)and C=σr(T). (S2) The set A∪B∪Cis non-empty and compact, the sets A,C\Band C\Care Souslin and there exists an Fσ-set Dand an operator Tsuch that σp(T)∪D=σp(T)∪σr(T). In the present work we provide the following characterization of the sets σp,n(T), σn p(T), σc(T) and σr(T) for bounded and for closed densely defined linear operators acting on a separable Hilbert space. Theorem 1. I. Let Tbe a closed densely defined linear operator acting on a separable Hilbert space H. Then σn p(T)is an Fσ-set for any n∈Nand σc(T) is a Gδ-set. II. Let K⊂Cbe non-empty and compact and K=A∪B∪C, where A∩B=A∩C=B∩C=∅,Ais an Fσ-set and Bis a Gδ-set. Let also Anbe a decreasing sequence of Fσ-sets such that A=A1. Then there exists a bounded linear operator Tacting on a separable infinite dimensional Hilbert space such that σn p(T) = Anfor any n∈N,σc(T) = Band σr(T) = C. III. Let K⊂Cbe a closed set and K=A∪B∪C, where A∩B=A∩C= B∩C=∅,Ais an Fσ-set and Bis a Gδ-set. Let also Anbe a decreasing sequence of Fσ-sets such that A=A1. Then there exists a closed densely defined linear operator Tacting on a separable infinite dimensional Hilbert space such that σn p(T) = Anfor any n∈N,σc(T) = Band σr(T) = C. BOUNDED LINEAR OPERATORS ON REFLEXIVE BANACH SPACES 135 Since the spectrum of a bounded linear operator on a Hilbert space is nonempty and compact and the spectrum of a closed densely defined linear operator on a Hilbert space is closed, Theorem 1 provides a complete description of all possible σn p(T), σc(T) and σr(T) for bounded and for closed densely defined operators acting on `2. Note also that Theorem 1, even in its point spectrum part, can not be obtained from the constructions used by Nikolskaia or Kalisch, because the latter do not provide the full variety of the parts σn p(T) of the point spectrum. Unfortunately the proof of Theorem 1 does not admit any straightforward modification applicable to any single Banach space different from `2. The question whether Theorem 1 remains true if one replaces the separable Hilbert space by, for instance, `pfor 1 <p<∞,p6= 2 remains open. However, using a completely different approach, we prove an analogue of Nikolskaia’s theorem for these spaces. Theorem 2. Let 1< p < ∞and A⊂C. Then there exists a bounded linear operator T:`p−→ `pfor which σp(T) = Aif and only if Ais a bounded Fσ-set. Note also that there are separable reflexive Banach spaces X, for which the family of the sets σp(T), σc(T) and σr(T) for bounded linear operators T acting on Xis much poorer than for X=`2. For instance, if one takes X being hereditarily indecomposable [1], then σr(T)∪σc(T)⊆ {0}and σp(T) is countable for any bounded linear operator Tacting on X. It is also worth noting that the sets appearing as spectra of continuous linear operators acting on a separable Fr´echet space were characterized by Shkarin [10] (necessary conditions on a set to be such a spectrum were earlier obtained by Slodowski [12]); namely, A⊂Cis the spectrum of some linear continuous operator acting on a separable Fr´echet space if and only if Ais a Gδσ-set. 2. Properties of spectral parts for general separable reflexive Banach spaces Proposition 1. Let Xbe a topological vector space and T:DT−→ Xbe a linear operator whose graph is a union of countably many metrizable compact sets. Then for any n∈N,σn p(T)is an Fσ-set. Proof. Let n∈Nand A,Bbe the sets defined by the formulas A={((x1, y1),...,(xn, yn), z)∈Γn T×C:yj=zxjfor 1 ⩽j⩽n}, B=½((x1, y1),...,(xn, yn), z)∈A:the vectors x1, . . . , xn are linearly independent¾. Since Ais closed in the space Γn T×C, which is a union of countably many metrizable compact sets, the set Ais itself a countable union of metrizable compact sets. One can easily verify that Bis open in A. Since an open subset of a metrizable compact set is a countable union of metrizable compact sets, 136 S. A. SHKARIN and O. G. SMOLYANOV we have that Bis a countable union of metrizable compact sets. Let now ϕ:B−→ C, ϕ((x1, y1), . . . , (xn, yn), z) = z. Since ϕis continuous and a continuous image of a compact set is again a compact set, we have that the set ϕ(B) = σn p(T) is σ-compact and, therefore, is an Fσ-set. ¤ Corollary 1. Let a topological vector space Xbe a countable union of metrizable compact sets and T:DT−→ Xbe a closed linear operator. Then for any n∈N,σn p(T)is an Fσ-set. For a locally convex topological vector space X, the symbol X0stands for the space of linear continuous functionals on X. As usual for y∈X0and x∈X we write (x, y) instead of y(x). If T:DT−→ Xis a densely defined linear operator, then the symbol T0stands for the dual operator T0:DT0−→ X0, that is, DT0is the set of ϕ∈X0for which the functional x7→ (Tx, ϕ) is continuous on DTwith respect to the topology of Xand T0ϕ∈X0is the (unique) continuous extension of this functional. Note that the operator T0is always closed when X0is endowed with the ∗-weak topology σ(X0, X) (see, for instance, [9]). Corollary 2. Let Xbe a separable metrizable locally convex topological vector space and T:DT⊆X−→ Xbe a densely defined linear operator. Then for any n∈N,σn p(T0)is an Fσ-set. Proof. Let {Un:n∈N}be a base of neighborhoods of zero in X. Then X0is the union of the sets U◦ n={y∈X0:|(x, y)|⩽1 for any x∈Un}. Alaoglu’s theorem [9] implies that U◦ nare compact in the ∗-weak topology σ(X0, X). Since Xis separable, the compact spaces (U◦ n, σ(X0, X)) are metrizable [9]. Since T0 is a closed operator on (X0, σ(X0, X)), it remains to apply Corollary 1. ¤ Proposition 2. Let Xbe a locally convex topological vector space and T:DT−→ Xbe a closed densely defined linear operator. Then σp(T)∪ σr(T) = σp(T)∪σp(T0). Proof. Let z∈σr(T). Then the linear space (T−zI)(X) is not dense in X. By Hahn–Banach theorem [9] there exists y∈X0\{0}such that ((T−zI)x, y) = 0 and therefore (Tx, y) = (x, zy) for any x∈X. Hence y∈DT0and T0y=zy. Thus z∈σp(T0). Let now z∈σp(T0). Then there exists y∈DT0\ {0}such that T0y=zy. Hence (Tx, y)=(x, T0y)=(x, zy) = (zx, y) and therefore ((T−zI)x, y) = 0 for any x∈DT. So we have (T−zI)(DT)⊆ker y. Thus the set (T−zI)(DT) is not dense in X. It follows that z∈σr(T)∪σp(T). ¤ Proposition 3. Let Xbe a separable reflexive Banach space and T:DT−→ Xbe a closed densely defined linear operator. Then for any n∈N,σn p(T)is an Fσ-set and σc(T)is a Gδ-set. Proof. Let Xσbe the space Xendowed with the weak topology. Since a linear subspace of a Banach space is closed if and only if it is weakly closed and is dense if and only if it is weakly dense, we have that Tis a closed densely BOUNDED LINEAR OPERATORS ON REFLEXIVE BANACH SPACES 137 defined linear operator on the space Xσ. Since closed balls of Xare metrizable and compact in the weak topology, we have that Xσis a countable union of metrizable compact sets. Applying Corollary 1, we obtain that σn p(T) is an Fσ-set for any n∈N. Corollary 2 implies that σp(T0) is an Fσ-set. According to Proposition 2, σp(T)∪σr(T) = σp(T)∪σp(T0). Hence σp(T)∪σr(T) is an Fσ-set. Since σ(T) is closed, we have that σc(T) = σ(T)\(σp(T)∪σr(T)) is a Gδ-set. ¤ Remark 1. Recall that a Banach space Xis called quasireflexive if dim X00/X < +∞. In particular, any reflexive Banach space is quasireflexive. Slightly modifying the first part of the proof, one can see that Proposition 3 remains true if reflexivity is replaced by quasireflexivity. 3. Proof of Theorem 1 We need some additional notation and auxiliary lemmas. Let S=S(R2) be the Schwarz space of rapidly decreasing infinitely differentiable functions on the plane; let S0be the dual space of the Fr´echet space S (it is usually called the space of Schwarz distributions [9]) and let Φ : S0−→ S0 be the Fourier transform. Let also α:R2−→ R+be given by the formula α(x, y) = (1 + x2)−1(1 + y2)−1. Consider the space E={f∈S0:α·Φf∈L2(R2)}endowed with the inner product (f, g)E= (α·Φf, α ·Φg)L2(R2)=ZZ R2 α2(x, y)Φf(x, y)Φg(x, y)dx dy. Since the map f7→ α·Φfis a linear homeomorphism of S0onto S0, we have that Eis a Hilbert space and the topology of Edefined by the inner product (·,·)Eis stronger than the topology of S0. Note also that L2(R2)⊂Eand the topology of Eis weaker than the natural Hilbert space topology of L2(R2). We shall also use the following notation. For two functions Aand Bdefined on the same set we write A¿Bif there exists c > 0 such that |A|⩽c|B|. For ϕ∈S denote p(ϕ) = ZZ R2¯¯¯¯µ1 + ∂4 ∂x4¶µ1 + ∂4 ∂y4¶ϕ(x, y)¯¯¯¯dx dy. Clearly pis a continuous norm on the locally convex topological vector space S. Lemma 1. kϕ·fkE¿p(ϕ)kfkEfor f∈Eand ϕ∈S. Proof. By definition kϕ·fkE=kα·Φ(ϕ·f)kL2(R2)¿ kα·(Φϕ∗Φf)kL2(R2), where ∗denotes the convolution of functions. Using the definition of pand the well-known properties of the Fourier transform, we obtain Φϕ¿p(ϕ)·β, 138 S. A. SHKARIN and O. G. SMOLYANOV where β(x, y) = (1 + x4)−1(1 + y4)−1. Hence kϕ·fk2 E¿p2(ϕ)kα·(β∗Φf)k2 L2(R2)=p2(ϕ)ZZ R2 α2(x, y)× ×ZZ R2 |Φf(u, v)|2β(x−u, y −v)du dvZZ R2 |Φf(s, t)|2β(x−s, y −t)ds dt dx dy = =p2(ϕ)ZZZZ R4 |Φf(u,v)Φf(s,t)|ZR β(x−u, x−s)dx (1 + x2)2ZR β(y−v, y −t)dy (1 + y2)2dudvdsdt. One can easily verify that ZR dx (1 + x2)2(1 + (x−u)4)(1 + (x−s)4)¿1 1 + (u−s)4·1 (1 + u2)2+1 (1 + s2)2¸. Thus kϕ·fk2 E¿p2(ϕ)ZZZZ R4 |Φf(u, v)Φf(s, t)| 1+(u−s)4·1 (1 + u2)2+1 (1 + s2)2¸× ×1 1+(v−t)4·1 (1 + v2)2+1 (1 + t2)2¸du dv ds dt. Performing the change of variables a=u−s,b=v−t(we pass from variables u, v, t, s to variables u, v, a, b) in the last integral and denoting g=α·Φf, we have kϕ·fk2 E¿p2(ϕ)ZZ R2  ZZ R2 |g(u, v)g(u−a, v −b)| 1 + a4·1+(u−a)2 1 + u2+1 + u2 1+(u−a)2¸× ×1 1 + b4"1+(v−b)2 1 + v2+1 + v2 1 + (v−b)2#du dv#da db. Since g∈L2(R2), we obtain ZZ R2 |g(u, v)g(u−a, v −b)|du dv ⩽kgk2 L2(R2)=kfk2 E. Hence, kϕ·fk2 E¿p2(ϕ)kfk2 EZZ R2 1 1 + a4sup u∈R·1+(u−a)2 1 + u2+1 + u2 1+(u−a)2¸1 1 + b4× ×sup v∈R·1 + (v−b)2 1 + v2+1 + v2 1+(v−b)2¸da db ¿p2(ϕ)kfk2 EZZ R2 1 1 + a2 1 1 + b2da db ¿p2(ϕ)kfk2 E. Thus kϕ·fkE¿p(ϕ)kfkE.¤ BOUNDED LINEAR OPERATORS ON REFLEXIVE BANACH SPACES 139 Let also γ:R2−→ C,γ(x, y) = x+iy and A⊂D={z∈C:|z|⩽1} be an infinite locally compact set. Then there exists a sequence of compact sets Kn⊂Csuch that Knis contained in the interior of Kn+1 in Afor any n∈Nand A=S∞ n=1 Kn. Since Ais locally compact, we also have that the set A\Ais compact. Hence for any n∈Nthere exists an infinitely differentiable function ϕn:R2−→ [0,1] with bounded support such that ϕn¯¯γ−1(Kn)≡0 and ϕn¯¯γ−1(A\A)≡1. Consider the space EA={f∈E: supp f⊆γ−1(A) and kfkA<+∞}, where kfk2 A=kfk2 E+P∞ n=1 kf·ϕnk2 Eand supp fis the support of the generalized function f. It is straightforward to verify that (EA,k · kA) is a Hilbert space. For any (x, y)∈R2, the symbol δx,y stands for the Dirac’s δ-function concentrated in the point (x, y). Since the function Φδx,y is bounded, we have that δx,y ∈E. If (x, y)∈R2is such that x+iy ∈A, then there exists n∈Nfor which x+iy ∈Kn. Hence δx,y ·ϕm= 0 for m⩾n. Therefore kδx,yk2 A=kδx,yk2 E+Ã1 + n−1 X k=1 ϕ2 n(x, y)!<+∞. It follows that δx,y ∈EAfor x+iy ∈A. Let us verify that the map x+iy 7→ δx,y is continuous from Ato the Hilbert space EA. Let xn+iyn∈A, (xn, yn∈R), xn→x,yn→yand x+iy ∈A. Then there exists n∈Nfor which x+iy ∈Kn. Since Knis contained in the interior of Kn+1 in A, we have xm+iym∈Kn+1 for sufficiently large m. For such m, kδxm,ym−δx,yk2 A=kδxm,ym−δx,yk2 E+ n X k=1 kϕk(xm, ym)δxm,ym−ϕk(x, y)δx,yk2 E. The definition of k·kEand the Lebesgue theorem imply that kδxm,ym−δx,ykA→ 0 as m→+∞. The continuity of the map x+iy 7→ δx,y is verified. Let now HAbe the closure in EAof the linear span of the set {δx,y :x+iy ∈A}. Since the map x+iy 7→ δx,y is continuous, HAis separable as a closed linear span of a separable set. Thus (HA,k · kA) is a separable Hilbert space. Consider now the operator T:S0−→ S0defined by the formula Tf =γ·f. Lemma 2. T(HA)⊆HAand the restriction TAof Tto HAconsidered as an operator on the Hilbert space HAis bounded. Moreover, σ(TA) = A, σp(TA) = σp,1(TA) = Aand the operator TA−zI has dense range in HAfor any z∈Cwhich is not an isolated point of A. Moreover, there exist a constant c⩾1and a decreasing continuous function a: (0,+∞)−→ (0,+∞), which do not depend on Asuch that kTAk⩽cand k(TA−zI)−1k⩽a( dist(z, A)) for z∈C\A. 140 S. A. SHKARIN and O. G. SMOLYANOV Proof. Let η∈S. According to Lemma 1 for any f∈HA, kη·fk2 A=kη·fk2 E+ ∞ X n=1 kη·f·ϕnk2 E ¿p2(η)Ãkfk2 E+ ∞ X n=1 kf·ϕnk2 E! =p2(η)kfk2 A. Choose an infinitely differentiable function γ0with compact support such that γ0(x, y) = γ(x, y) for x2+y2⩽1. Since the support of any f∈EAis contained in the set γ−1(A)⊆ {(x, y)∈R2:x2+y2⩽1}, we have Tf =γ0·fand supp Tf ⊆supp f⊆γ−1(A) for any f∈EA. From the fact that Tf =γ·f and Lemma 1, it follows that kTfkA¿p(γ0)kfkAfor any f∈EA, where pis the norm defined by at the beginning of Section 3. Hence there exists a constant c⩾1 such that kTfkA⩽ckfkAfor any f∈EA. Thus the operator T¯¯EAacts boundedly on the Hilbert space EAand kT¯¯EAk⩽c. Let x, y ∈R be such that x+iy ∈A. Then Tδx,y =γ(x, y)δx,y ∈HA. Hence Tmaps the dense in HAlinear span of the set {δx,y : (x, y)∈γ−1(A)}into HA. Since the operator T¯¯EAis bounded with respect to the norm k · kAand HAis complete, we obtain that T(HA)⊆HAand kTAk⩽c. Let ρ∈C∞[0,∞) be such that ρ¯¯[0,1/2)∪[9c,+∞)≡0, ρ¯¯[1,8c]≡1 and x0, y0∈ Rbe such that z=x0+iy0∈C\Aand |z|⩽2c. Denote η(x, y) = 1 x+iy −zρ³|x+iy −z| dist(z, A)´. Clearly ηis an infinitely differentiable function with bounded support and η¯¯γ−1(A)=1 γ−z¯¯γ−1(A). Since the support of any f∈HAis contained in the set γ−1(A), we have that η·(TA−zI)f= (T−zI)(η·f) = ffor any f∈HA. Since Tf =γ·f, the operator T−zI is invertible and k(TA−zI)−1k ¿ p(η). It is straightforward to verify that p(η)¿( dist(z, A))−6. Hence there exists a constant c2>0 for which k(TA−zI)−1k⩽c2dist(z, A)−6for z∈C,|z|⩽2c and z /∈A. If |z|>2cthen the estimate kTAk⩽cimplies that TA−zI is invertible and k(TA−zI)−1k⩽2|z|−1⩽4 dist(z, A)−1. Hence σ(TA)⊆A and k(TA−zI)−1k⩽a( dist(z, A)), where a(t) = max{c2t−6,4t−1}. Obviously, a: (0,+∞)−→ (0,+∞) is continuous and decreasing. Note that the spectrum of the operator Tis the entire complex plane C, is purely point spectrum of multiplicity 1 and for any z=x+iy ∈C(x, y ∈R) the one-dimensional space ker (T−zI) is spanned by δx,y. Hence σp(TA) = σp,1(TA) = {x+iy :x, y ∈Rand δx,y ∈HA}. For x+iy ∈A,δx,y ∈HAby definition of HA. If x+iy /∈A, then supp δx,y 6⊆ A and therefore δx,y /∈EA. Hence δx,y /∈HA. If x+iy ∈A\A, then ϕn(x, y) = 1 for any n∈N. Hence δx,y ·ϕn=δx,y for any n∈Nand the terms of the BOUNDED LINEAR OPERATORS ON REFLEXIVE BANACH SPACES 141 series from the definition of the norm kδx,ykAare all equal to the same positive number and therefore the series diverges. Hence δx,y /∈EAand therefore δx,y /∈HA. From the last display we have σp(TA) = σp,1(TA) = A. This equality together with the already proven inclusion σ(TA)⊆Aimply that σ(TA) = A. It remains to verify the density in HAof the range of the operator TA−zI when z=x0+iy0∈Cis not an isolated point of A. By definitions of TAand HA we have that δx,y ∈(TA−zI)(HA) for any (x, y)∈γ−1(A)\{(x0, y0)}. If z /∈A we have that {δx,y : (x, y)∈γ−1(A)} ⊂ (TA−zI)(HA). Let z∈A. Since zis not an isolated point of Aand the map x+iy 7→ δx,y from Ato HAis continuous, we have that δx0,y0is a limit point of the set ©δx,y : (x, y)∈γ−1(A)\{(x0, y0)}ª in HA. Thus, in any case the set {δx,y : (x, y)∈γ−1(A)}is contained in the closure of (TA−zI)(HA). Hence (TA−zI)(HA) is dense in HAsince the set {δx,y : (x, y)∈γ−1(A)}has dense linear span HA.¤ Lemma 3. Let K⊂Cbe a nonempty compact set. Then there exists a bounded linear operator CKacting on a separable infinite dimensional Hilbert space such that σ(CK) = σc(CK) = K. Proof. Let H0=L2[0,1] and T0:H0−→ H0be the classical Volterra operator: T0f(t) = t R0 f(s)ds. It is well-known that T0is a bounded linear operator and σ(T0) = σc(T0) = {0}. Let {zn}be a sequence dense in K,Hn=H0for any n∈Nand H= ∞ ⊕ n=1 Hnbe the Hilbert direct sum of the Hilbert spaces Hn. Then His a separable Hilbert space. Define the operator CK:H−→ Hby the formula (CKx)n= (T0−znI)xn. It is straightforward to verify that this operator satisfies the desired properties. ¤ Lemma 4. There exists a bounded linear operator T1acting on `2such that kT1k⩽1,σp(T1) = σ(T1) = σp,1(T1) = {0}and the range of T1is dense. Proof. One can easily verify that the weighted backward shift (T1x)n= xn+1/(n+ 1) satisfies the desired conditions. ¤ Lemma 5. There exist a constant c1>0and a decreasing continuous function a1: (0,+∞)−→ (0,+∞)such that for any non-empty σ-compact set A⊂ Dthere exists a bounded linear operator QAacting on a separable infinite dimensional Hilbert space such that σ(QA) = A,σp(QA) = σp,1(QA) = A, the range of the operator (QA−zI)is dense for any z∈C,kQAk⩽c1and k(QA−zI)−1k⩽a1( dist(z, A)) for z∈C\A. Proof. Pick an increasing sequence Kn(n∈N) of compact sets such that A=S∞ n=1 Kn. Let K0=∅and An=Kn\Kn−1for n∈N. Then the sets Anare locally compact as open subsets of compact spaces. The set An(as for any subset of a separable metrizable set) can be decomposed as An=Ac n∪Au n, where the set Ac nis finite or countable, Au nis closed in Anand does not have