Some properties oî the core topologies

R O C Z N IK I P O L S K IE G O TO W A KZY STW A M A TEM A TYCZNEGO Séria I : P E A C E M A TEM A TY CZN E X I X (1977)

Le s z e k Ja n k o w sk i (Poznan)

Some properties oî the core topologies

Abstract. V. L. Klee introduced in [2] the so-called core topologies, Г х and ^{/ 2} in real vector spaces. The main purpose of this paper is to investigate some properties of these “ pathological” topologies, viz. the weight at a point, separability, regularity of Г х and the B aire category. Some results in this paper are slightly stronger than those of Klee’s [3].

Introduction. In this paper we are concerned with two special top­

ologies Г г and Г2 in a real vector space. The topology / \ , called the core topology (see Klee [2], p. 446), is the strongest topology in which both the vector addition and scalar multiplication are seperately continuous.

A vector space with the topology Г х is not a topological vector space except when it is one-dimensional. More familiar is the topology Г 2 ([1], p. 117-119, [4], p. 39, 68, 113) which is the strongest locally convex top­

ology in a real vector space. Using the Г х topology we obtain an example of a Baire space which is not regular. In the sequel we describe these topologies in terms of the so-called core of a set.

First we establish the notation and terminology. We are concerned only with real vector spaces. X will always denote a real vector space.

For x ,y e X , (x , y} will denote the line segment {ax + (1 — a)y : 0 < a < 1}

between x and y, (œ, у) = <(cc, y }\{x ] etc. If {bt}, teT is a Hamel basis in X , then x = £ a tbt denotes the expansion of an element x with respect

teT

to this basis. R n will denote the n-dimensional euclidean space. Let r be a topology in X . If both the vector addition and scalar multiplication in X are separately r-continuous, then {X , r} is called a fi-space (Klee [2], P- 444).

1. The core of a set and the core topologies. Let L be a linear variety.

We recall that for A <= L the core of A with respect to L, denoted A0, is defined to be the set of all points x such that to each y * L \ { x } corre­

sponds ze(x, у) for which (x , z} A. If we consider the cores of a set A

with respect to different variétés, then we shall use different notation for them.

Now we give two examples of sets A such that A 0 Ф (A0)0.

Ex a m p l e 1. Let К с P 2 be a disc centred at 0 and {an} a countable oo / l

( — an, a i \n

Obviously OeM0, and so suppose that a point x Ф 0 belongs to A 0. Then there exists m such that x i ^ —K. Take a point у not a multiple of x in

m

the boundary of the disc. Then the arc defined by the convex angle у Ox contains a subsequence } (%> > m for every k). Then the set

00 / 1 \

u

—

?

п >

\ Пк /

a contradiction with xeA 0. Hence A0 = {0}. But (M0)0 = 0 , because the core of a set containing only one point is empty.

Ex a m p l e 2. Let Q <= P 2 be the set composed with two discs outer tangent at a point x together with the common tangent line at this point.

It is easy to see that oceQ0 but co4(Q0)0.

The family of all sets A <= L for which A = A0 is easily seen to be a topology in L. Following Klee [2] we call it the core topology in L and denote by r ^ L ). We shall also use the notation “ /Vtopology” if there

is no misunderstanding. .

Pr o p o sit io n 1. Let L be a linear variety in X. The topology induced in L by Г г{Х) coincides with Г г(Т).

P roof. Let L = a 4- Y, where Y is a linear subspace of X , and let Z be the complementary subspace to Y in X. Let A a subset of Y. A о and A * will denote the cores of A with respect to X and Y respectively.

It is clear that the core of аф.А in а ф Х is equal to а ф A *. Let P — P 0 for. some P <= X . Then, by definition, we have P r \Y — (P n Y )*.

Conversely, let A = A# <= Y. Then it suffices to check that for Q — A ф Z the equality Q = Q0 holds. Let xeQ and yeX. Let x = д ф и and у = р ф г , where p, qeY and u, veZ. Now we are going to show that there exists z Ф х such that (x, z) <= Q n (x , y}. If q —р-, then this is'obvious, so let p ф q. There exist# se X \{g } such that (q, s} <= A n n <P1 РУ- One can easily check that there exists tv eZ such that for 0 = qфw the following equality holds: <4v,z) = {(q, s') ф Z )n (x , y}. But <#,$> + AZ<^Q, ^so we finally have <a>, s> = (x, y}r\Q. This implies xe'Qo, hence Q = Q0 which completes the proof.

It is easy to see that both the vector addition and scalar multipli­

cation are separately continuous with respect to the Px topology. Hence {X, Pj} is a /bspace. Moreover, it is the strongest topology having this property. This fact follows from the inclusion IntH c AL0 (see Klee [2],

dense subset of its boundary. Let A =

П — Then A 0= {0}.

p. 445) which is satisfied for every subset A of a Д-space X. The above inclusion is equivalent to the equality Int A — (IntJL)0. Namely, if this equality holds, then Int A = (Int A)0 <=■ A 0. Conversely, if Int A <= A0, then Int A — Int (Int A) c (IntX )0 <= Int A. Moreover, Г г is the strongest topology such that the topology induced by it on any straight line coin­

cides with its natural topology. Linear operations are jointy ^-con­

tinuous only in the one-dimensional space. Klee showed in [2], p. 447, that the scalar multiplication is not jointly continuous with respect to the Г х topology if dim X > 2. Now we shall show that also the vector addition is not jointly continuous in this case. For this we need the follow­

ing lemma.

Lemma 1. Suppose ^dim X ^{^ 2} and let & be a family of non-empty Гх-ореп subsets of X such that ^cardâ? = ^{m <} n — ^cardX. Then, given acX, there exists a set M in X such that

(a) a i M;

(b) card-M = m;

(c) 3Ir\B # 0 for every

(d) no line in X contains more than two points from M.

Hence U = X \ M is an open neighbourhood of a in {X , Г х} which contains none of the sets S e J .

P roof. Let p be the least ordinal such that card {a: a is an ordinal

< y] = m. Let us arrange the members of J 1 in a transfinite sequence We are going to define a transfinite sequence {ха}а</л such that the set M = {#a: a < p] has all properties (a) through (d).

Choose any point in B 0\ { a } to be x0. Suppose that for some Д < p we have already defined in such a way that

(*) ccaeBa\ { a } . for each а < Д and

(**) no line in X contains more than two points from {a?a} a</î.

Let 3?p be the family of all the lines which contain exactly two points from {x,p}a<p, and let Lp = • To define Xp we have to treat the cases п = c and n > c separately (c = cardÆ).

1° rt = c. Since card {а: a < Д} < c and card Bp = c, we can find a line l through a intersecting Bp and not containing any of the points жа) а < Д. Further, since InBp is a non-empty open subset of l (in the natural topology of l) and each line in 3?p contains at most one point from l, the set (lr\Bp)'KLp is non-empty, and we define Xp to be any of its points different from a.

2° n > c. In this case cardP^ < c-card {a: a < /?} < n and carding = n, and we take as xpany point in B p\ { L p\j{a}).

In both cases the sequence {xa}a<p+1 satisfies (*) and (**) with /?

replaced by /3+1. This completes the inductive definition of {xfja<li.

It is obvious that the set Ж = {xa: a < y] is as required.

Proposition 2. I f ^{dim X} ^ ^2, then the vector addition in {.X , ^PJ is not jointly continuous.

P roof. If the vector addition is jointly continuous, then for every open neighbourhood V of zero there exists an open neighbourhood V of zero such that V + V <= JJ. If this condition holds in {.X , P J , then so it does by Proposition 1 in each two-dimensional subspace of X endowed with its own -Tj-topology. Hence it suffices to prove the proposition for X = B 2. Now we apply Lemma 1. Let 38 be a basis of the natural topology in B 2 and a = 0. Then the set M from Lemma 1 is dense in B 2 for the natural topology and is / у closed. Then the set B 2\ M is an open /^-neighbourhood of zero. Suppose V is an open /^-neighbourhood of zero such that 7 + F c B 2\M . Then choose any two non-zero points p, q in V, q not a multiple of p, such that <0, p> u <0, #> V. Then the set V + V contains a the parallelogram Q = <0, p } + <0, q}. Since M is dense, we have Qr\M Ф 0 . Hence V + V Ф B 2\ M and we have obtained a contradiction.

The family of all /yopen and convex subsets of X forms a basis for a topology in X called the convex core topology and denoted by /У It is easy to check that {.X , Г 2} is a locally convex topological vector space. Since each subset A of X open in some locally convex topology in X is open in the Pa-topology, we see that Г 2 is the strongest locally con­

vex topology in X .

The topologies Г г and Г 2 are essentially different; nevertheless they possess many common properties. When some property is common for both these topologies we shall write that “/,• possesses this property” . Hence we can say that Г { is a Hausdorff topology, and every linear variety is closed in the /ytopology.

Let L be a linear variety in X. Then one can easily check that the topology induced in L by Г 2(Х) coincides with Г 2(В).

2. Topological properties of the topologies I\. Let us recall that the weight of a topology at a point æ is the smallest cardinal number such that there exists a basis of neighbourhoods of x with this cardinality.

Since the vector addition is separately continuous each translation is a homeomorphism from {X , /7} onto {X , P J. This implies that the weight of P* at any point x is equal to its weight at zero. Let c{X , Г{) denote the weight of P* at zero.

/

Th e o r e m 1. I f X is finite dimensional, then c(X , F 2) is equal to k0.

Otherwise the inequality dim X < c(X , F 2) < 2dlmX holds.

P roof. In the finite dimensional case {X , F J is a topological vector space hence F 2 coincides with the natural topology of X . Hence it suffices to investigate the weight at 0 in X for X infinite dimensional.

Let {bt}teT be a Hamel basis in X and c a r d x 0. Let us suppose that {Bt}teS is a basis of neighbourhoods of zero and 8 <= T. Then for every teS there exists r* > 0 such that < — 2rtbt, 2rtbty cz B t. The set К = {x = £ atbt: for every te8, \at\ < r j is open and convex, and does not contain any of the sets B t. Hence {Bt}us cannot form a basis of teT neighbourhoods of zero. This implies that inequality c(X , F 2) > dim X holds.

Next let TJ be an open convex neighbourhood of zero in ^jT2. For every teT there exists ct > 0 such that ( —ctbt,c tbt}<= V. Then V = conv ( —ctbt, ctbt) is evidently a Fa-neighbourhood of zero and V <=■ TJ.

teT

It follows that the family 38 of all sets of the form conv U ( ~ cthj ct a positive rational, is a basis of neighbourhoods of zero. Since card F teT

= dimX > k0, we have c a r d ^ < 2 dimZ. This completes the proof.

Th e o r e m 2. I f d im X > 2, then the weight of the topology F x at zero is greater than the cardinality of X .

P roof. Let us assume that there exists a basis of neighbourhoods of zero J8 such that its cardinality is not greater than the cardinality of X.

Applying Lemma 1 for this and a = 0 we obtain a IVopen neighbour­

hood of zero which does not contain any set from 38. The proof is completed.

V. L. Klee in [3], p. 29, proved that {.Rn, Fj} is separable. Now we shall proved that a slightly stronger theorem. A proof of this theorem differs a little from the included in [3]. In the proof we shall use a less ' general version of the following lemma.

Lem m a 2. Let r be a topology (or more generally “ a paratopology” ; for a definition see V. L. Klee [2], p. 444) in a group G such that the ad­

dition is separately continuous, in particular such that for G = X , {X , т}

is a fi-space. Let A and В be subsets of G. Then A + B <= A-\-B.

P roof. Since translations are homeomorphisms of {G, r} onto itself, for each aeA we have a-\-B — a-\-B <= A-\-B and also for each beB A-\-b c A -f B. Hence A + B с A + В and A + B <= A + B , and so finally À + B c J T b с A-h В.

Th e o r e m 3. The space {X , F J is separable if and only if X is of at most countable dimension.

Proof. If dim X > x0, then let {bt}UT be a Hamel basis for X . Then the set K — {æ = ^ atbt: \at\ < 1} is open in P2, and {3bt + K }UT is an

U T

uncountable family of pairwise disjoint and open in P2 subsets of X.

Therefore the Suslin condition does not hold, hence the space {X , P2}

is not separable, and neither is {X , P J .

Now we pass to the finite dimensional case. Evidently we may restrict ourselves to the spaces Rn, and we are going to show that for each n > 1

(*) Qn is dense in {Rn, P J ,

where Q denotes the set of rational numbers.

This is obviously true for n — 1, because the Pr topology and the natural topology in R1 coincide.

Suppose (*) is valid for some n. Then since Qn x {0} and {0}№x Q are dense subsets of closed subspaces (Proposition 1) B n x {0} and {0}w x Rl of {Rn+1, P J , in view of Lemma 2 we have

Q **1 э ( ? йх {0} + {0}n x Q = R nx {0 } + {0}nx R 1 = Rn+1.

Thus Qn+ is dense in {Rn+1, P J .

Let in turn X be a countable dimensional space. We may assume

OO 00

that X = [ J Rn. Then the countable set A = U Qn is dense. In fact

n=1 П—1

for every open set G X there exists such m that RmnG ф 0 . Since RmnG is open in {Rm, Pj), we have Gr\QmФ 0 . Hence GnA Ф 0 , and

so A = X . '

Thus лте have proved that (X , Px} is separable of dim X < s 0. To complete the proof it suffices to recall that Г г is final than Г 2.

Cokollaky 1. Every p-space X is separable whenever dim X < s 0- This is an immediate consequence of the fact that I \ is the strongest

topology for which X is a Д-space. \

C^orollary 2. In every uncountable dimensional space {X , P J the Suslin condition does not hold.

This follows from the proof of Theorem 3.

Th e o r e m 4. I f X is a finite dimensional space, then {X , P J is a Baire space. I f X is an infinite dimensional space, then {X , P J is of the Baire first category.

Proof. Let X be an infinite dimensional vector space. Then there existe an increasing sequence of linear subspaces X n such that X n Ф X

OO

for every integer n and U X n= X. Every subspace X n is closed and

n=l

nowhere dense in {X , P J, therefore {X , P J is of the first category.

Since in finite dimensional spaces Г 2 coincides with, the natural topology, it remains to consider the case {B n, Гг}.

We claim that

(*) If {Aj?$ is a sequence of ^-closed subsets of B n such that B n

00

= U 4 ) then there is Tc0 such that Ak contains a non-empty set which

fc=i 0

is open in the natural topology of B n.

For n — 1 assertion (*) is obviously true. So suppose that (*) holds for some n, and consider the space B n+1. Let {A&} be a sequence of Г г- closed set covering B n+1. For each1 xeB n the sets

Ak( œ ) = { U B 1: {oc,t}eAk}, Tc = 1 , 2 , . . . ,

are closed in B 1 and cover B 1, and therefore there exists Tc such that Ak{x) has non-empty interior relative B 1) let Tcx be the first such Tc. Let

~ {Bp. ieN } be a countable base for the topology of B 1. For each xeB n choose ix€N so that B ix <= Akx(cc). .

Then define for each pair {к, tyelSf x Ж CkJ = {xeBn: Tcx = Tc, ix = ï ) .

Since the sets Ck i cover B n, by induction hypothesis there are Tc0, i0 such that Ck()tiQ, the closure of O*0,t-0 in {-®n? A}» contains a non-empty set Q which is open in the natural topology of B n. Let D = Gr\Ckoiit);

D is .Ti-dense in G. Hence

AkQ => D x B io = D x {0} + {0}n x B iQ о D x {0} + {0}пхД . GxB ïq.

Which proves (*) with n-f l in place of n.

Thus {B n, Г i} is of the Baire second category. That this space is actually a Baire space follows easily from that it is a /?-space.

In fact, suppose G is a non-empty IVopen subset of B n, and suppose a sequence {Ak} of subsets of G covers G. On applying a suitable transla-

OO

tion, we may assume that 0 eG. Then \ J jG = B n, and therefore U jA k

j — l j,keN

= B n. By (*) for some j = j 0 and Tc = Tc0, the interior of j 0AkQ = j 0Ak(), and hence of Ak , in the natural topology of B n, is non-empty. This com­

pletes the proof of Theorem.

Assertion (*) and the argument used in the second part of the proof yield immediately the following

Co r o l l a r y. I f G is a non-empty Г г-ореп subset of Bn, then its r x-clos- ure contains a non-empty Г 2-ореп set.

Let G be а Гг ореп set in B n, let H be its />closure and let V be the A-interior of H. Then H W is А -closed, and we claim that it is / >

nowhere dense. Otherwise H W would contain a non-empty P2-open subset, which is impossible. Hence every element x of the set G belongs to the closure of V (in the natural topology).

This corollary is little weaker form of Proposition 2 in ГЗ]. Namely this proposition says that if x e A0 then the /^-closure of a 4 contains non-empty Pa-open set. Now we give an example of a /\-open set A <=

such that there exists xeA and æ is not contained in the P2-interior of the Pj-closure of A.

Ex a m p l e 3. Let Q,QX, P and P x be four different discs such that:

(i) Q c Qx and P <= P 1?

(ii) Qx and P x as well as Q and P outer tangent at the point æ.

Let A the union of {x} and the P2-interior of [R2\(Qx\j Px)]kjQ\jP.

It is easy to see that A is Pj-open set and the Pj-closure of A is equal to its P2-elosure. But the P2-interior of A does not contain x.

A simple proof of the Theorem 5 can be obtained using the corollary of Theorem 4 so as V. L. Klee did it in [3].

But we give here the other simple proof.

Th e o r e m 5. I f dim X > 2, then space {X , Г х} is not regular.

Proof. Proposition 1 implies that it suffices to examine the regularity of the space {R 2, Px}. The separability of the space implies that there exists the Pj-dense and countable subset D of P 2. Then for every Pr open set G we have Gr\D = G. Hence the cardinality of the family {G: G e rx(R2)}

is equal to c. Since the weight of Г х ot zero is greater than c hence there does not exist basis of neighbourhoods of zero consisting of Pr closed sets.

This completes the proof.

Bern ark. Since each group topology must be completely regular Proposition 2 follow from the above theorem.

3. Remarks about the core topologies. We can easily prove that if A is a countable dimensional space then {X , P2} is normal. This follows from the fact in this case P 2 is the strict direct limit of the natural top­

ologies in R n, n = 1 , 2 , . . .

S k e t c h of the proof. If A and В are disjoint closed subsets of the space X , then we form sequences of sets {An} and {Bn} having the following properties:

(i) A n B n c= An and. B n B n a B n, (ii) An <= An+l and B n c B n+1, (iü) Ann B n = 0 ,

(iv) An and B n are open in B n for n — 1,2, ... Nex one can prove

that U An and ( J B n are disjoint open sets in X containing A and В

«,= ! n = 1

respectively.

In paper [3], p. 23, У. L. Klee showed that any sphere in B n is in­

closed and the topology induced by Г г on it is discrete. He used this fact in some proofs. Now we shall mention that the classical inversion on the plane is not continuous with respect to the /\-topoIogy.

Indeed, if P is a straight line which does not contain zero, then its image -with respect to the inversion is equal to a circle passing through zero but with zero excluded from it. Then it is easy to see that the inver­

sion restricted to the line P is not continuous. Hence inversion is not continuous.

I would like to express my gratitude to Dr. L. Drewnowski for his help in preparing this paper.

References

[1] R. E. E d w ard s, Funkcjonalnyj analiz, Tieorija i prilozenija (Functional ana- lysis, Theory and applications), Moskwa 1969.

[2] У. L. K lee, Jr., Convex sets in linear spaces, Duke Math. J . 18 (1951), p. 443-466.

[3] — Some finite-dimensional affine topological spaces, Portugal Math. 14 (1955), p. 27-30.

[4] A. P. R o b e rtso n and W. R o b ertso n , Topologiceskije wektornyje prostranstwa (Topological vector spaces), Moskwa 1967.