2.5 Finite-Dimensionality 105
ϕk+1ψ(z, t) = (1−t)
k
i=1
z(i)vi+tvk+1 = (1−t)ϕk(z) +tvk+1. From the property of the topology ofFand the continuity ofϕk, it follows that ϕk+1ψ is continuous. For eachi= 1, . . . , k+ 1, let pi = pri|fl∆k: fl∆k →R be the restriction of the projection onto thei-th factor. Note that
ψ|fl∆k−1×(R\ {1}) : fl∆k−1×(R\ {1})→fl∆k\p−k+11 (1)
is a homeomorphism. Hence,ϕk+1|fl∆k\p−k+11 (1) is continuous. Replacing the (k+ 1)-th coordinates with thei-th coordinates, we can see the continuity of ϕk+1|fl∆k\p−i1(1). Since fl∆k =k+1
i=1(fl∆k\p−i1(1)), it follows thatϕk+1
is continuous. Thus, the claim can be obtained by induction.
It remains to show the openness off. On the contrary, assume that f is not open. Then, we have x ∈ Rn and ε > 0 such that f(B(x, ε)) is not a neighborhood off(x) inF. Since bd B(x, ε) is a bounded closed set ofRn, it is compact, hence f(bd B(x, ε)) is closed inF. Then, F \f(bd B(x, ε)) is a neighborhood off(x) inF. Using the compactness ofI, we can find an open neighborhoodU off(x) inF such that
(1−t)f(x) +tU⊂F\f(bd B(x, ε)) for every t∈I.
Then, U∩f(bd B(x, ε)) =∅. Sincef(B(x, ε)) is not a neighborhood off(x), it follows thatU ⊂f(B(x, ε)), and so we can take a pointy∈U\f(B(x, ε)).
Now, we define a linear path g:I→Rn byg(t) = (1−t)x+tf−1(y). Since f is affine andy∈U, it follows that
f g(t) = (1−t)f(x) +ty∈F\f(bd B(x, ε)) for every t∈I.
Sincef is a bijection, we have
g(I)⊂Rn\bd B(x, ε) = B(x, ε)∪(Rn\B(x, ε)).
Then,g(0) =x∈B(x, ε) andg(1) =f−1(y)∈Rn\B(x, ε), which contradicts the connectedness ofI. Thus,f is open.
In the case whenE is a topological linear space, to prove thatF is closed inE, take a pointx∈E\F and consider the flatFx= fl(F∪ {x}). It is easy to construct an affine bijectionf :Rn+1→Fxsuch thatf(Rn× {0}) =F. As we saw in the above, f is a homeomorphism, hence F is closed inFx. Since Fx\F is open inFx, we have an open setU in Esuch thatU∩Fx=Fx\F.
Then,U is a neighborhood ofxinEandU ⊂E\F. Therefore,E\F is open in E, that is,F is closed inE. ⊓⊔
If a linear spaceE has a topology such that the operation E×E×R∋(x, y, t)→(1−t)x+ty∈E
is continuous, then scalar multiplication and addition are also continuous with this topology because they can be written as follows:
E×R∋(x, t)→tx= (1−t)0+tx∈E;
E×E∋(x, y)→x+y= 21
2x+12y
∈E.
Then, the following is obtained by Proposition 2.5.1:
2.5.2 Corollary Every finite-dimensional linear space E has the unique (Hausdorff) topology compatible with the algebraic operations (addition and scalar multiplication), and then it is linearly homeomorphic to Rn, where n= dimE. ⊓⊔
Moreover, we have the following:
2.5.3 Corollary Let E be a topological linear space and F a finite-dimen-sional flat in another topological linear space. Then, every affine function f :F→E is continuous, and iff is injective thenf is a closed embedding.
Proof. By Proposition 2.5.1, F can be replaced with Rn, where n = dimF. Then, we can write
f(x) =
1−
n
i=1
x(i)
f(0) +
n
i=1
x(i)f(ei) for eachx∈Rn,
where e1, . . . ,en is the canonical orthonormal basis for Rn. Thus, the conti-nuity of f is obvious. Since f(Rn) is a finite-dimensional flat inE, f(Rn) is closed inE by Proposition 2.5.1. Iff is injective then f :Rn →f(Rn) is an affine bijection, which is a homeomorphism by Proposition 2.5.1. Hence,f is a closed embedding. ⊓⊔
Combining Proposition 2.2.2 and Corollary 2.5.3, we have
2.5.4 Corollary Let E be a topological linear space and C a finite-dimen-sional convex set in another topological linear space. Then, every affine func-tionf :C→E is continuous. Moreover, iff is injective thenf is an embed-ding. ⊓⊔
For finite-dimensional convex sets in a linear space, we have the following:
2.5.5 Proposition LetCbe a finite-dimensional convex set in an arbitrary linear space E. Then,rintC= intflCC with respect to the unique topology forflC as in Proposition 2.5.1.
2.5 Finite-Dimensionality 107 Proof. We may assume that E is a topological linear space. By Proposition 2.4.7(3), it suffices to show that intflCC =∅. We have affinely independent v0, v1, . . . , vn ∈ C with flC = fl{v0, v1, . . . , vn}, where n = dimC. We have an affine bijection f : Rn → flC such that f(0) = v0, f(e1) = v1, . . ., f(en) =vn. Then,f is a homeomorphism by Proposition 2.5.1, hence
intflCC⊃intflCv0, v1, . . . , vn=f(intRn0,e1, . . . ,en)=∅. ⊓⊔ Note that every compact set in a topological linear space is topologically bounded and closed. For ann-dimensional convex setC in a linear space, the flat hull flC is affinely isomorphic toRn. Combining Propositions 2.5.1 and 2.5.5 with Corollary 2.4.14, we have the following:
2.5.6 Corollary For everyn-dimensional compact convex set C in an arbi-trary topological linear spaceE, the pair(C, ∂C)is homeomorphic to the pair (Bn,Sn−1)of the unit closedn-ball and the unit(n−1)-sphere. ⊓⊔
Remark 11.It should be noted that every bounded closed set in Euclidean spaceRn is compact. More generally, we can prove the following:
2.5.7 PropositionLetEbe an arbitrary topological linear space andA⊂E with dim flA < ∞. Then, A is compact if and only if A is topologically bounded and closed inE.
Sketch of Proof.Using Proposition 2.5.1, this can be reduced to the case of
Ê
n.
The following convex version of Proposition 2.5.1 is not trivial.
2.5.8 Proposition Let C be an n-dimensional convex set in an arbitrary linear spaceE. If (1)C is the convex hull of a finite set13 or (2)C= rintC, thenChas the unique (Hausdorff) topology such that the following operation is continuous:
C×C×I∋(x, y, t)→(1−t)x+ty∈C.
In case (1),rclC=Cand(C, ∂C)≈(Bn,Sn−1); in case (2),C≈Rn. Proof. Like Proposition 2.5.1, it suffices to see the uniqueness and the addi-tional statement. To this end, suppose that C has such a topology, but it is unknown whether this is induced from a topology of flC or not.
Case (1): Let C = v1, . . . , vk and define f : ∆k−1 → C by f(z) = k
i=1z(i)vi. In the same way as for the Claim in the proof of Proposition 2.5.1, we can see that the continuity of the operation above induces the continuity of f. Since ∆k−1 is compact, f is a closed map, hence it is quotient. Thus,
13In this case,Cis called acellor a (convex)linear cell(cf.§3.1).
the topology ofC is unique andC is compact with respect to this topology.
Giving any topology on E so that E is a topological linear space, we have rclC = C by Proposition 2.4.7(i) and (C, ∂C) ≈ (Bn,Sn−1) by Corollary 2.5.6.
Case (2): Letf :Rn →flC be an affine bijection, where n= dim flC = dimC. SinceD=f−1(C) is ann-dimensional convex set inRn,D= rintD= intD is open in Rn by Proposition 2.5.5, hence D ≈ Rn by Proposition 2.4.15. Then, it suffices to show that f|D : D → C is a homeomorphism.
For each x ∈ D, choose δ > 0 so that x+δBn = B(x, δ) ⊂ D. Let v0 = x−δ∆ˆn−1, where ˆ∆n−1 is the barycenter of the standard (n− 1)-simplex ∆n−1=e1,e2, . . . ,en ⊂Rn. For eachi= 1, . . . , n, letvi=x+δei. Then,v0, v1, . . . , vn are affinely independent and
x∈intRnv0, v1, . . . , vn ⊂x+δBn⊂D,
hence v0, v1, . . . , vn is a neighborhood of x in D. On the other hand, we have the affine homeomorphism ϕ : ∆n → v0, v1, . . . , vn defined by ϕ(z) = n
i=0z(i+ 1)vi. Since f ϕ(z) = n
i=0z(i+ 1)f(vi), the continuity of the operation above implies that off ϕ, hencef|v0, v1, . . . , vnis continu-ous atx. Then, it follows thatf|D is continuous atx.
SinceD is open inRn, we can apply the same argument as in the proof of Proposition 2.5.1 to prove thatf|D:D →C is open. Consequently,f|D: D→C is a homeomorphism. ⊓⊔
Remark 12.For an arbitrary finite-dimensional convex C, Proposition 2.5.8 does not hold in general. For example, let
C={0} ∪ {(x, y)∈(0,1]2|xy} ⊂R2.
Then, C is a convex set that has a finer topology than usual such that the operation in Proposition 2.5.8 is continuous. Such a topology is generated by open sets in the usual topology and the following sets:
Dr={0} ∪(B((0, r), r)∩C), r >0.
Note that this topology induces the same relative topology onC\{0}as usual.
SinceDε/√2⊂B(0, ε) for eachε >0,{Dr|r >0}is a neighborhood basis at 0∈C with respect to this topology.
We shall show that the operation
C×C×I∋(p, q, t)→(1−t)p+tq∈C
is continuous at (p, q, t) ∈ C×C×I. If (1−t)p+tq = 0, it follows from the continuity with respect to the usual topology. The continuity at (0,0, t) follows from the convexity ofDr,r >0.
To see the continuity at (0, q,0) (q =0), let q = (x, y), where 0< y x 1. Choose s > 0 so that q ∈ Ds (i.e., s > (x2 +y2)/2y). For each 0< r <min{1, s}, let 0tr/2s,p′∈Dr/2, and q′∈Ds. Observe that
2.5 Finite-Dimensionality 109
0 ◦1
Dr
q= (x, y)
r
(1,1)
s Ds
Fig. 2.7.The continuity of the operation at (0, q,0)∈C×C×I
1−t
r/s−t(r/s)p′ ∈ 1−t
r/s−t(r/s)Dr/2⊂ 1
r/2s(r/s)Dr/2=Dr
and (r/s)q′ ∈(r/s)Ds=Dr. SinceDris convex, it follows that (1−t)p′+tq′=
1− t
r/s
1−t
r/s−t(r/s)p′+ t
r/s(r/s)q′ ∈Dr. Thus, the operation is continuous at (0, q,0). The continuity at (p,0,1) (p=0) is the same.
A subsetAof a topological linear spaceE istotally boundedprovided, for each neighborhood U of0∈E, there exists some finite set M ⊂E such that A⊂M+U. In this definition,M can be taken as a subset of A.
Indeed, for each neighborhoodU of0∈E, we have a circled neighborhood V such that V +V ⊂U. Then,A ⊂M+V for some finite set M ⊂E, where it can be assumed that (x+V)∩A=∅for everyx∈M. For each x∈M, chooseax∈Aso thatax∈x+V. Then,x∈ax−V =ax+V. It follows thatA⊂Ëx∈M(x+V)⊂Ëx∈M(ax+V +V)⊂Ëx∈M(ax+U).
IfA⊂E is totally bounded andB ⊂A, thenB is also totally bounded.
It is easy to see that every compact subset ofE is totally bounded and every totally bounded subset ofEis topologically bounded. In other words, we have:
compact ⇒ totally bounded ⇒ top. bounded
For topological linear spaces, the finite-dimensionality can be simply charac-terized as follows:
2.5.9 Theorem LetEbe a topological linear space. The following are equiv-alent:
(a) Eis finite-dimensional;
(b) Eis locally compact;
(c) 0∈E has a totally bounded neighborhood inE.
Proof. Since each n-dimensional topological linear space is linearly homeo-morphic to Rn (Corollary 2.5.2), we have (a) ⇒ (b). Since every compact subset ofE is totally bounded, the implication (b)⇒(c) follows.
(c)⇒(a): LetU be a totally bounded neighborhood of0∈E. By Propo-sition 2.4.1, we have a circled neighborhood V of 0 such that V +V ⊂ U. Then,V is also totally bounded. First, we show the following:
Claim. For each closed linear subspaceF E, there is somex∈U such that (x+V)∩F =∅.
Contrary to the claim, suppose that (x+V)∩F = ∅ for everyx ∈ U. Since V =−V, it follows that U ⊂F +V, so we have V +V ⊂F +V. If (n−1)V ⊂F+V then
nV ⊂(n−1)V +V ⊂F+V +V ⊂F+F+V =F+V.
By induction, we have nV ⊂ F +V for every n ∈ N, which implies that V ⊂
n∈N(F+n−1V).
Takez∈E\F. SinceF is closed inE, we have a circled neighborhoodW of 0∈E such thatW ⊂V and (z+W)∩F =∅. The total boundedness of V implies the topological boundedness, hence V ⊂mW for some m∈N. On the other hand,k−1z∈V for some k∈N. Sincek−1z∈V ⊂F+ (km)−1V, it follows that z ∈ F +m−1V ⊂ F +W. This contradicts the fact that (z+W)∩F=∅.
Now, assume thatEis infinite-dimensional. Letv1∈U\{0}andF1=Rv1. Then,F1 is closed inE (Proposition 2.5.1) andF1=E. Applying the Claim above, we havev2∈U such that (v2+V)∩F1=∅. Note thatv2∈v1+V. Let F2=Rv1+Rv2. Since F2 is closed inE (Proposition 2.5.1) andF2=E, we can again apply the Claim to findv3∈U such that (v3+V)∩F2=∅. Then, note thatv3∈vi+V fori= 1,2. By induction, we havevn ∈U,n∈N, such that vn ∈vi+V for i < n. Then,{vn |n∈N} is not totally bounded. This is a contradiction. Consequently,E is finite-dimensional. ⊓⊔
By Theorem 2.5.9, every infinite-dimensional topological linear space is not locally compact.