fact,E has a Hamel basisB. As a linear subspace of the product spaceRB, RBf is a topological linear space that is linearly isomorphic toE by the linear isomorphism ϕ : RBf → E defined by ϕ(x) =
v∈Bx(v)v. Then, ϕ induces a topology that makes E a topological linear space. In the next section, it will be seen that if E is finite-dimensional, then such a topology is unique.
However, an infinite-dimensional linear space has various topologies for which the algebraic operations are continuous.
In the following proposition, we present the basic properties of a neigh-borhood basis at0in a topological linear space.
2.4.1 Proposition LetEbe a topological linear space andU be a neighbor-hood basis at 0in E. Then,U has the following properties:
(1) For eachU, V ∈ U, there is someW ∈ U such thatW ⊂U∩V; (2) For eachU ∈ U, there is someV ∈ U such thatV +V ⊂U; (3) For eachU ∈ U, there is someV ∈ U such that[−1,1]V ⊂U;
(4) For eachx∈E andU ∈ U, there is somea >0 such thatx∈aU;
(5)
U ={0}.
Conversely, let E be a linear space with U a collection of subsets satisfy-ing these conditions. Then, E has a topology such that addition and scalar multiplication are continuous andU is a neighborhood basis at0.
Proof.Property (1) is trivial; (2) comes from the continuity of addition at (0,0)∈E×E; (3) is obtained by the continuity of scalar multiplication at each (t,0)∈[−1,1]×Eand the compactness of [−1,1]; (4) follows from the continuity of scalar multiplication at (0, x)∈Ê×E; the Hausdorffness of Eimplies (5).
GivenU with these properties, an open set in E is defined as a subset W⊂E satisfying the condition that, for eachx∈W, there is someU ∈ U such thatx+U ⊂W. (Verify the axioms of open sets, i.e., the intersection of finite open sets is open; Every union of open sets is open.)
For each x ∈ E and U ∈ U, x+U is a neighborhood of x in this topology.8 Indeed, let
W ={y∈E| ∃V ∈ U such thaty+V ⊂x+U}.
Then,x∈W ⊂x+U because of (5). For eachy∈W, we haveV ∈ Usuch thaty+V ⊂x+U. TakeV′ ∈ U so that V′+V′ ⊂V as in (2). Then, y+V′ ⊂W because (y+y′) +V′ ⊂ y+V ⊂x+U for every y′ ∈ V′. Therefore,W is open inE, so x+U is a neighborhood ofxinE. By the definition of the topology,{x+U |U∈ U}is a neighborhood basis atx. In particular,U is a neighborhood basis at0.
Since {x+U | U ∈ U} is a neighborhood basis at x, the continuity of addition follows from (2). Using (3), we can show that the operation x→ −xis continuous.
8If E is a topological linear space, x+U is a neighborhood of x ∈ E for any neighborhoodU of0.
2.4 Topological Linear Spaces 97 For scalar multiplication, letx∈E,α∈Ê, andU ∈ U. Because of the continuity ofx→ −x, it can be assumed thatα0. Then, we can write α=n+t, where n∈ωand 0 t <1. Using (2) inductively, we can find V1⊃ · · · ⊃Vn⊃Vn+1inU such that
V1+· · ·+Vn+ (Vn+1+Vn+1)⊂U.
By (3), we haveW∈ Usuch that [−1,1]W ⊂Vn+1. Then,x∈rWfor some r >0 by (4). Choose δ >0 so thatδ < min{1/r,1−t}. Let y ∈ x+W and|α−β|< δ. Then, we can writeβ=n+s, wheret−δ < s < t+δ. It follows that
βy−αx= (n+s)y−(n+t)x=n(y−x) +s(y−x) + (s−t)x
∈nW + [−1,1]W+δ[−1,1](rW)
⊂nVn+1+Vn+1+Vn+1
⊂Vn+1+· · ·+Vn+1
n+ 2 many
⊂U,9
henceβy∈αx+U.
To see the Hausdorffness, letx=y∈E. By (5), we have U ∈ U such thatx−y∈U. By (2) and (3), we can findV ∈ U such thatV −V ⊂U. Then,x+V andy+V are neighborhoods ofxandy, respectively. Observe that (x+V)∩(y+V) =∅.
It is said thatA⊂E iscircled iftA⊂Afor everyt∈[−1,1]. It should be noted that the closure of a circled setAis also circled.
Indeed, letx∈clAandt∈[−1,1]. Ift= 0, thentx= 0∈A⊂clA. When t= 0, for each neighborhoodU of txinE, since t−1U is a neighborhood oft−1x,t−1U∩A =∅, which implies thatU∩tA=∅. BecausetA ⊂A, U∩A=∅. Thus, it follows thattx∈clA.
In (3) above, W = [−1,1]V is a neighborhood of0∈E thta is circled, i.e., tW ⊂W for everyt∈[−1,1]. Consequently, (3) is equivalent to the following condition:
(3)’ 0∈E has a neighborhood basis consisting of circled (open) sets.
Atopological groupGis a group with a topology such that the algebraic operations of multiplication (x, y)→xyand taking inversesx→x−1are both continuous.10Then,Gishomogeneous, that is, for each distinctx0, x1∈G, there is a homeomorphismh:G→Gsuch thath(x0) =x1. Such anhcan be defined byh(x) =x0x−1x1, where not onlyh(x0) =x1 but alsoh(x1) =x0. Every topological linear space is a topological group with respect to addition, so it is homogeneous.
9It should be noted that, in general, 2V ⊂V +V butV +V ⊂2V.
10These two operations are continuous if and only if the operation (x, y) →x−1y is continuous.
2.4.2 Proposition Every topological group G has a closed neighborhood basis at each g ∈ G, i.e., it is regular.11 For a topological linear space E, 0∈E has a circled closed neighborhood basis.
Sketch of Proof.Each neighborhoodU of the unit1∈Gcontains a neigh-borhoodV of1such thatV−1V ⊂U. For eachx∈clV, we havey∈V x∩V. Consequently,x∈V−1y⊂V−1V ⊂U, so we have clV ⊂U.
For the additional statement, recall that ifV is circled then clV is also circled.
2.4.3 Proposition LetGbe a topological group andH be a subgroup ofG.
(1) IfH is open inGthenH is closed in G.
(2) The closureclH ofH is a subgroup ofG.
Sketch of Proof.(1): For eachx∈G\H,Hxis an open neighborhood ofx inGandHx⊂G\H.
(2): For eachx, y∈clH, show thatx−1y∈clH, i.e., each neighborhood W ofx−1ymeetsH. To this end, choose neighborhoodsU andV ofxand y, respectively, so thatU−1V ⊂W.
Due to Proposition 2.4.3(1), a connected topological groupGhas no open subgroups except forG itself. Observe that every topological linear spaceE is path-connected. Consequently, E has no open linear subspaces except for E itself, i.e., every proper linear subspace ofE is not open inE.
The continuity of linear functionals is characterized as follows:
2.4.4 Proposition Let E be a topological linear space. For a linear func-tionalf :E→Rwithf(E)={0}, the following are equivalent:
(a) f is continuous;
(b) f−1(0)is closed inE;
(c) f−1(0)is not dense inE;
(d) f(V)is bounded for some neighborhoodV of 0∈E.
Proof. The implication (a) ⇒ (b) is obvious, and (b) ⇒ (c) follows from f(E)={0} (i.e.,f−1(0)=E).
(c)⇒(d): We havex∈Eand a circled neighborhoodV of0∈Esuch that (x+V)∩f−1(0) =∅. Then,f(V) is bounded. Indeed, iff(V) is unbounded, then there is some z ∈ V such that |f(z)| > |f(x)|. In this case, f(tz) = tf(z) = −f(x) for some t ∈ [−1,1], which implies that −f(x) ∈ f(V). It follows that 0 ∈ f(x) +f(V) = f(x+V), which contradicts the fact that (x+V)∩f−1(0) =∅.
11A topological group G is assumed to be Hausdorff, but it suffices to assume axiom T0. In fact, axiomT0 impliesT1 for a topological groupGbecause of the homogeneity ofG.
2.4 Topological Linear Spaces 99 (d)⇒ (a): For each ε >0, we have n ∈N such that f(V)⊂(−nε, nε).
Then, n−1V is a neighborhood of 0in E and f(n−1V)⊂(−ε, ε). Therefore, f is continuous at0∈E. Since f is linear, it follows thatf is continuous at every point ofE. ⊓⊔
2.4.5 Proposition LetEbe a topological linear space and A, B⊂E.
(1) IfB is open inE thenA+B is open inE.
(2) IfA is compact andB is closed inE thenA+B is also closed inE.
Sketch of Proof.(1): Note thatA+B=Ëx∈A(x+B).
(2): To show thatE\(A+B) is open inE, let z ∈E\(A+B). For eachx∈A, becausez−x∈E\B, we have open neighborhoodsUx, Vx of x, zinEsuch thatVx−Ux⊂E\B. SinceAis compact,A⊂Ëni=1Uxi for somex1, . . . , xn∈A. Then,V =Ìni=1Vxi is an open neighborhood ofzin E. We can show thatV ∩(A+B) =∅, i.e.,V ⊂E\(A+B).
Remark 8.In (2) above, we cannot assert that A+B is closed in E even if both A and B are closed and convex in E. For example, A = R× {0} and B ={(x, y)∈ R2 | x > 0, y x−1} are closed convex sets inR2, but A+B=R×(0,∞) is not closed inR2.
2.4.6 Proposition LetF be a closed linear subspace of a topological linear spaceE. Then, the quotient linear space E/F with the quotient topology is also a topological linear space, and the quotient map q : E → E/F (i.e., p(x) =x+F ∈E/F) is open, hence if U is a neighborhood basis at0in E, thenq(U) ={q(U)|U ∈ U}is a neighborhood basis0in E/F.
Sketch of Proof.Apply Proposition 2.4.5(1) to show that the quotient map q:E→E/F is open. Then, in the diagrams below,q×qand q×idÊare open, so they are quotient maps:
E×E
q×q
//E
q
E/F×E/F //E/F,
E×Ê
q×idR
//E
q
E/F×Ê //E/F.
Accordingly, the continuity of addition and scalar multiplication are clear.
Note thatE/F is Hausdorff if and only ifF is closed inE.
For convex sets in a topological linear space, we have the following:
2.4.7 Proposition For each convex set C in a topological linear space E, the following hold:
(1) clC is convex andrclC⊂clC, hencerclC=C ifC is closed inE;
(2) intFC=∅ for any flatF withflCF;
(3) intflCC=∅ impliesintflCC= coreflCC= rintC.
Proof. By the definition and the continuity of algebraic operations, we can easily obtain (1). For (2), observe intFC ⊂ coreFC. If intFC = ∅ then flC=F by the Fact stated in the previous section.
(3): Due to Proposition 2.3.2, coreflCC = rintC. Note that intflCC ⊂ coreflCC. Without loss of generality, we may assume that0∈intflCC. Then, for eachx∈rintC, we can find 0 < s <1 such that x∈sC. Since (1−s)C is a neighborhood of0=x−xin flC, we have a neighborhoodU ofxin flC such thatU −x⊂(1−s)C. Then, it follows that U ⊂(1−s)C+sC =C.
Therefore,x∈intflCC. ⊓⊔
Remark 9.In the above, we cannot assert any one of clC= rclC, intflCC= coreflCC, or intflCC =∅. For example, [−1,1]Nf is a convex set in RN such that rcl[−1,1]Nf = [−1,1]Nf but cl[−1,1]Nf = [−1,1]N. Note that fl[−1,1]Nf =RNf. Regard [−1,1]Nf as a convex set inRNf. Then,
intRNf[−1,1]Nf =∅ but coreRNf[−1,1]Nf = rint[−1,1]Nf = (−1,1)Nf. By Proposition 2.4.7(1), ifA is a subset of a topological linear space E, then clAis the smallest closed convex set containingA, which is called the closed convex hullofA.
Remark 10.In general, Ais not closed in E even if A is compact. For ex-ample, letA={an |n∈ω} ⊂ℓ1, wherea0(i) = 2−i for everyi∈Nand, for eachn∈N, an(i) = 2−i ifinandan(i) = 0 if i > n. Then,Ais compact andA=
n∈Na0, a1, . . . , an. For eachn∈N, let
xn = 2−na0+ 2−1a1+· · ·+ 2−nan ∈ a0, a1, . . . , an.
Then, xn(i) = 2−2i+1 ifi n and xn(i) = 2−n−i ifi > n. Hence, (xn)n∈N
converges tox0∈ℓ1, wherex0(i) = 2−2i+1for eachi∈N. However,x0∈ A. Otherwise,x0∈ a0, a1, . . . , anfor some n∈N, where we can write
x0=
n
i=0
z(i+ 1)ai, z∈∆n. Then, we have the following:
z(1)a0(n+ 1) =x0(n+ 1) = 2−2n−1= 2−na0(n+ 1) and z(1)a0(n+ 2) =x0(n+ 2) = 2−2n−3= 2−n−1a0(n+ 2),
hencez(1) = 2−n andz(1) = 2−n−1. This is a contradiction. Therefore,A is not closed inℓ1.
The following is the topological version of the Separation Theorem 2.3.5:
2.4 Topological Linear Spaces 101 2.4.8 Theorem (Separation Theorem) LetAandBbe convex sets in a topological linear space E such that intA = ∅ and (intA)∩B = ∅. Then, there is a continuous linear functional f :E →Rsuch that f(x)< f(y)for eachx∈intAandy∈B, and supf(A)inff(B).
Proof. First, intA = ∅ implies coreA = intA = ∅ by Proposition 2.4.7(3).
Then, by the Separation Theorem 2.3.5, we have a linear functionalf :E→R such thatf(x)< f(y) for everyx∈intAandy∈B, and supf(A)inff(B).
Note thatB−intAis open inEandf(z)>0 for every z∈B−intA. Thus, f−1(0) is not dense inE. Therefore,f is continuous by Proposition 2.4.4. ⊓⊔ A topological linear spaceE is locally convexif0∈E has a neighbor-hood basis consisting of (open) convex sets; equivalently, open convex sets make up an open basis for E. It follows from Proposition 2.4.6 that for each locally convex topological spaceEand each closed linear subspaceF ⊂E, the quotent linear spaceE/F is also locally convex. For locally convex topological linear spaces, we have the following separation theorem:
2.4.9 Theorem (Strong Separation Theorem) LetAandBbe disjoint closed convex sets in a locally convex topological linear spaceE. If at least one ofAandBis compact, then there is a continuous linear functionalf :E→R such that supf(A)<inff(B).
Proof. By Proposition 2.4.5(2), B −A is closed in E. Since A∩B = ∅, it follows that0∈B−A. Choose an open convex neighborhoodU of0so that U ∩(B−A) = ∅. By the Separation Theorem 2.4.8, we have a nontrivial continuous linear functional f : E → R such that supf(U) inff(B − A). Then, supf(A) + supf(U)inff(B), where supf(U)>0 by the non-triviality off. Thus, we have the result. ⊓⊔
As a particular case, we have the following:
2.4.10 Corollary Let E be a locally convex topological linear space. For each pair of distinct pointsx, y∈E, there exists a continuous linear functional f :E→Rsuch thatf(x)=f(y). ⊓⊔
Concerning the continuity of sublinear functionals, we have the following:
2.4.11 Proposition Letp: E → Rbe a non-negative sublinear functional of a topological linear spaceE. Then,pis continuous if and only ifp−1([0,1)) is a neighborhood of 0∈E.
Proof. The “only if” part follows fromp−1([0,1)) =p−1((−1,1)). To see the
“if” part, let ε > 0. Since p−1([0, ε)) = εp−1([0,1)) is a neighborhood of 0∈E, eachx∈Ehas the following neighborhood:
U =
x+p−1([0, ε))
∩
x−p−1([0, ε)) .
For eachy∈U, sincep(y−x)< εandp(x−y)< ε, it follows that p(y)p(y−x) +p(x)< p(x) +ε and
p(y)p(x)−p(x−y)> p(x)−ε, which means thatpis continuous atx. ⊓⊔
For each convex set C ⊂ E with 0 ∈ intC, we have intC = coreC = p−C1([0,1)) by Propositions 2.3.4 and 2.4.7(3). Then, the following is obtained from Proposition 2.4.11.
2.4.12 Corollary Let E be a topological linear space. For each convex set C ⊂E with0∈intC, the Minkowski functional pC : E →Ris continuous.
Moreover, p−C1([0,1)) = intC = rintC and p−C1(I) = clC = rclC, hence p−C1(1) = bdC=∂C. ⊓⊔
The boundedness is a metric concept, but it can be extended to subsets of a topological linear spaceE. A subset A ⊂E istopologically bounded12 provided that, for each neighborhoodU of0∈E, there exists somer >0 such that A⊂rU. IfA⊂E is topologically bounded andB ⊂A, thenB is also topologically bounded. Recall that every neighborhood U of0∈E contains a circled neighborhood V of0∈E (cf. Proposition 2.4.1(3)). SincesV ⊂tV for 0< s < t, it is easy to see that every compact subset ofE is topologically bounded. WhenEis a normed linear space,A⊂Eis topologically bounded if and only ifAis bounded in the metric sense. Applying Minkowski functionals, we can show the following:
2.4.13 Theorem LetE be a topological linear space. Each pair of topolog-ically bounded closed convex sets C, D ⊂E with intC = ∅ and intD = ∅ are homeomorphic to each other by a homeomorphism ofEonto itself, hence (C,bdC)≈(D,bdD)and intC≈intD.
Proof. Without loss of generality, we may assume that0∈intC∩intD. Let pC and pD be the Minkowski functionals forC and D, respectively. By the topological boundedness ofC andD, it is easy to see thatpC(x), pD(x)>0 for every x∈ E\ {0}. Then, we can define maps ϕ, ψ : E → E as follows:
ϕ(0) =ψ(0) =0, ϕ(x) = pC(x)
pD(x)x and ψ(x) =pD(x)
pC(x)x for eachx∈E\ {0}.
It follows from the continuity ofpC andpD (Corollary 2.4.12) thatϕ andψ are continuous at eachx∈E\ {0}.
12Usually, we say simply “bounded” but here add “topologically” in order to dis-tinguish the metric sense. It should be noted that every metrizable space has an admissible bounded metric.
2.4 Topological Linear Spaces 103 To verify the continuity ofϕat0∈E, letU be a neighborhood of0∈E.
Since D is topologically bounded and C is a neighborhood of0, there is an r >0 such that D ⊂rC. Then, pC(x)rpD(x) for every x∈ E. Choose a circled neighborhoodV of 0∈E so that rV ⊂U. Then,ϕ(V)⊂U. Indeed, for eachx∈V \ {0},
ϕ(x) = pC(x)
pD(x)x∈ pC(x)
pD(x)V ⊂rV ⊂U.
Similarly,ψis continuous at0∈E.
For eachx∈E\ {0}, sinceϕ(x)=0,
ψϕ(x) =pD(ϕ(x)) pC(ϕ(x))ϕ(x) =
pC(x) pD(x)pD(x) pC(x) pD(x)pC(x)
· pC(x) pD(x)x=x.
Hence,ψϕ= id. Similarly,ϕψ = id. Therefore, ϕis a homeomorphism with ϕ−1=ψ. Moreover, observe thatϕ(C)⊂Dandψ(D)⊂C, henceϕ(C) =D.
Thus, we have the result. ⊓⊔
The norm of a normed linear spaceE is the Minkowski functional for the unit closed ball BE ofE. Since bdBE is the unit sphereSE of E, we have the following:
2.4.14 Corollary Let E = (E, · ) be a normed linear space. For every bounded closed convex set C ⊂ E with intC = ∅, the pair (C,bdC) is homeomorphic to the pair(BE,SE)of the unit closed ball and the unit sphere ofE. ⊓⊔
It is easy to see that every normed linear spaceE = (E, · ) is homeo-morphic to the unit open ball B(0,1) =BE\SE ofE.
In fact, the following are homeomorphisms (each of them is the inverse of the other):
E∋x→ 1
1 +xx∈B(0,1); B(0,1)∋y→ 1
1− yy∈E.
By applying the Minkowski functional, this can be extended as follows:
2.4.15 Theorem Every open convex setV in a topological linear spaceEis homeomorphic toE itself.
Proof. Without loss of generality, it can be assumed that0∈intV =V. Then, we have V = intV = p−V1([0,1)) by Corollary 2.4.12. Using the Minkowski functionalpV, we can define mapsϕ:V →Eandψ:E→V as follows:
ϕ(x) = 1
1−pV(x)x forx∈V; ψ(y) = 1
1 +pV(y)y fory∈E.
Observe thatψϕ= idV andϕψ= idE. This means thatϕis a homeomorphism withψ=ϕ−1. ⊓⊔