TOPOLOGIES BETWEEN COMPACT AND UNIFORM CONVERGENCE ON FUNCTION SPACES

This paper studies two topologies on the set of all continuous real-valued functions
on a Tychonoff space which lie between the topologies of compact convergence and uniform
convergence.

sequences of functions which converge uniformly on compact subsets.One of the distinguishing features of this topology is that whenever x is locally compact the compact-open topology on C(X) is the coarsest topology making the evaluation map e:X C(X)--.continuous (where e(x,])= f(x)).
The compact-open topology and the topology of uniform convergence are equal if and only if x is compact.Because compactness is such a strong condition, there is a considerable gap between these two topologies.This gap was especially felt in [8] while studying the completeness of a normed linear space of continuous linear functionals on C(X) with the compact-open topology.
Because of this, a new class of topologies was introduced in [7] on C*(X) to bridge the gap, where c*(x) is the set of bounded functions in C(x).This also generalized the a-compact-open topology on C*(X) which was studied in [5].The purpose of this present work is to extend these ideas to two natural topologies on C(X), to study the properties of these topologies, and to relate these topologies to the compact-open topology and the topology of uniform convergence.
Let Ck(X and Cu(X denote C'(X) with the compact-open topology and the topology of uniform convergence, respectively.The definitions of the other two topologies that we study here are based on the fact that Ok(x) can be viewed in two different ways.First we can view the compact-open topology as a "set-open topology," where a subbasic open set looks like [K, Vl {I ec(x):l(K)c_ v}, where K is a compact subset of X and V is an opeu subset of the reals.Note that v can always be takeu as a bouuded opeu iuterval.The secoud way that we can view the compact-opeu topology is as a "uniform topology."This approach is developed in the next section.
Throughout the rest of the paper, we use the following conventions.All spaces are Tychonoff spaces.If x and Y are any two spaces with the same underlying set, then we use X Y, X _< I/and X < Y to indicate, respectively, that X and Y have the same topology, that the topology on Y is finer than or equal to the topology on X, and that the topology on Y is strictly finer than the topology on x.The symbols t and 1 denote the spaces of real number and natural numbers, respectively.Finally, the constant zero-function in C(X) is denoted by Jr0.
In this first section we look at "uniform topologies" on C(X) in a general setting.This is done in terms of certain pseudo-seminorms on C(x).By a pseudo-seminorm on a real linear space g is meant a real-valued function p on E such that (1) p(O) 0, (2) p(z) p( z) for all z ( E, and (3) p(= + y) <_ p(z) + p(y) for all =, y ( e. Note that it immediately follows that p(z) >_ 0 and p(.) p(y) <_ p(.y) for all., y g.A pseudo-seminorm p is called a seminorm if the following additional condition holds.
Of course a seminorm p is a norm if p(z) 0 whenever z 0.
Let a be any family of subsets of x satisfying the condition: if A, B a, then there exists a C a such that A o B C_ C.
Then it can be easily verified that the collection {Ae:A c,, > 0} is a base for some uniformity on C(X).We denote the space C(X) with the topology induced by this uniformity be Co, u(X).This topology is called the topology of uniform convergence on .For each Jr C(X),A a and > 0, let < f, A, e > {g {/C(X): If(z)g(z)[ < for all A}.
Then for each Jr ( C(X), the collection < Jr, A, > :A (/a, > O} forms a neighborhood base at jr in C,,,,,(X).Since the topology comes from a uniformity, then C,,,,(X) is completely regular.If a covers x, then C,, (x) is a Tychonoff space.When a= {x}, we get the topology of unilorm convergence, denoted by C,,(x).It is clear that for any a,C,u(X) < C,,(X).VA, {f 15 C(X): PA(I) < }" {VA,:A q t,e > 0}.It can be easily shown that for each 115 C(X),l + 'r {l + v:v '} forms a neighborhood base at I. We say that this topology is generated by the collection of pseudo-seminorms {i0A: A 15 c,}.Note that if we choose 0 < < 1, then for each 115 C(X), we have f +VA, _ <f,A,e> and <I,A,> C_I+VA, for all A 15 .This shows that the topology of uuiform couvergence on a is the same as the topology generated by the collectiou of pseudo-semiuorms {pA:A 15 a}.We see from this poiut of view that C,,,u(x is a topological group with respect to additiou. Observe that for C*(X) we can actually use the seminorms PA defined by PA(I) sup{ 1(.) :* 15 A}.Cousequeutly, C* ,,,,(x) is a locally couvex topological vector space.Oue might wouder wheu C,u(X) is a topological vector space.This is answered by the following theorem.In this theorem the term "bounded" refers to a subset of a space such that each restriction of a continuous realvalued function on the space to this subset is a bounded function.
THEOREM 1.1.The space C,,,,,(X) is a linear topological space if and only if every element of a is bounded.
PROOF.If every element of a is bounded, then as in the case of C* .(x),we can actually use seminorms for PA when A e a.
For the converse, suppose that there is an A e a which is not bounded.Then there is an 115C(X) and {r.:n15N} _c A such that l(r.)_>n for each n.To show that scalar multiplication cannot be continuous in C,,,,,(X), let T be the scalar multiplication operator defined by T(t,g))= tg for 15 and # e C,.(X).We show that T is not continuous at (0,l).Let VA,, be a neighborhood of lo T((O,I)) in C,,,,(X), where A 15 a and 0 < < 1.Then for any neighborhood U of 0 in , there exists an hen such that neU.But xnI(..) > 1>, so that T((n,l))=I qtVa, .
As a consequence of Theorem 1.1, if every element of a is pseudocompact (a space is pseudocompact if every continuous real-valued function on the space is bounded), then C,,,,,(X) is a locally convex topological vector space.Conversely, if every element of a is C-embedded in x (that is, every continuous real-valued function on the subspace has a continuous extension to X) and if C,,,,(X) is a topological vector space, then every element of a is pseudocompact.
For the rest of the paper, we are interested in the particular topology where a is the set of a- compact (countable union of compact) subsets of X.In this case, we denote the space C,,,,(X) by C,.(x).This is called the topology of uniform convergence on a-compact sets.Note that we get the same topology if we take the members of r, to be the closures of the r-compact subsets of X.
For any subset A of X and any open subset V of , define [A, V] {f ( C(X): I(A) C_ V}.This agrees with the usual "set-open" terminology for the compact-open topology because A is compact in this case.Now let ,(X) be the set of ,-compact subsets of X, and let be the set of bounded open intervals in .For the g-compact-open topoloev on C(X), we take as subbase, the family {[A,B]:A G .(X),B }; and we denote this space by Co(X).Note that the same topology is obtained by using [,B], where A 6 ,(X) and B ( 9.This is because for each f .C(X), f(7t c_ f(A); so that f(Tt f(A).The fact that Ca(X) is a Tychonoff space can be proved in a manner similar to the proof of Lemma 5.1 in [9].
It is useful to relate basic open sets in Co(X to basic open sets in C,,,,,(X), which is done in the following two lemmas.
LEMMA 2.1.Let W f'__[ai, v,] be a basic neighborhood of I in C,,(x).Then there exists an > 0 such that < f, A O o A,, > C_ W. PROOF.For each i, since I(A,) is compact, there is an ei > 0 such that the ei-neighborhood of I(A,) is contained in V,.The conclusion now follows by taking LEMMA 2.2.Let A ( ,(X), and let I ( C(X) be such that I(A) is bounded.Then for each > 0, there exists a basic open set W in C,,(X) such that 1" ( W c_ < )',A,e >.PROOF.Since I(A) is bounded, there are open intervals v v, of length such that I(A-----C:_ O ,"__l V,.For each i, let A A r3 f-(V i), and let W be the -neighborhood of V in .Then each Ai r(X), and each W, is an open interval of length e.So define w c,"= [A,,W,], which is basic neighborhood of f in Co(X).It is straightforward to check that w _ < f,A,e >.Lemma 2.1 tells us that the a-compact-open topology is coarser than or equal to the topology of uniform convergence on .-compactsets.On the other hand, it is clear from definitions that the .-compact-opentopology is finer than or equal to the compact-open topology.Therefore we have the following general comparisons.THEOREM 2.3.For every space X, Ct(X <_ Co(X < Co, u(X < C,,(X).
It is well-known that C*(X) is dense in C(X).The same is true for "Co(X), as we see in the next result.THEOREM 2.4.For every space X,C*(X) is dense in Co(X).

PROOF. If n'=
Even though C*(X) is dense in Co(X), the members of C*(X) play a special role in Co(X in the following sense.For each I .C(X), let TI:Co(X)--,Ca(X be the translation operator defined by Tt(g J' + g for all g Co(X).THEOREM 2.5.Let I C(X).Then TI:Co(X)Co(X is continuous if and only if I C*(X).
PROOF.For the sufficiency, suppose that l E C*(X) and that [A, V] is a subbasic neighborhood of Tl(g) in C,,,(X) for some g e C,,(X).Then by Lemma 2.1, there exists an > 0 such that < TI(g),A, > c_ [A, V].Now (f +g)(A)= TI(g)(A is bounded.Since f E C*(X), g(A)=(f +g)(A)-f(A) is bounded.So by Lemma 2.2, there exists a basic open set w in C,,(X) such that g e w c_ < g,A, >.We see that T.(W) C_ Tj( < a,A, > C_ < Tj(a),A, > C_ [A,V], and thus T I must be continuous at g.
When studying function spaces, it is sometimes useful to use induced functions.That is, if 4:XY is a continuous function, define b*:C(Y)C(X) by *(#)=#o for all eC(Y).We can establish the following theorem much like the corresponding theorem for the compact-open topology (cf.[10]).In this theorem, the definition of :X-Y being a or-compact-covering map is that each a- compact subset of is contained in the image of some #-compact subset of x under .THEOREM 2.7.If :X-Y is continuous, then *:C,(Y)-C(X) is continuous.Furthermore, if is a r-compact-covering map, then * is an embedding.
In this section we determine when then inequalities are equalities and give examples to illustrate the differences.
THEOREM 3.1.For every space x, ct:(X C,,(x) if and only if the closure of each a-compact subset of x is compact.
PROOF.The sufficiency is straightforward.For the necessity, suppose that Ct:(x =Ca(X and let A a(X).Then there exist a compact set K in X and an open set V in such that fo e[K, V] _C [A, $\{1}].Suppose, by way of contradiction, that there exists an z \K.Then there would be some f C(X) such that f(z)= and f(y)= 0 for all y K.But then f e which is a contradiction.Therefore c_ K, so that is compact.EXAMPLE 3.12.Let X=\{p}, where peN* (see Example 3.11).
compact and separable, but not compact.Therefore Then X is countably Ck(X) < Co(X) Co, .(x)C.(X).
Note that this shows that the converse of Corollary 3.2 is not true.An example of a space which is not countably compact but which has these same function space relations is the space q in 5I of [4].
Theorem 3.5 can be expanded to include some additional properties of the spaces Co(X and co, .(x).
To prove that (3) implies ( 7), suppose that X does not contain a dense a-compact subset.Now a Gn-set is a subset which can be written as a countable intersection of open subsets.So for each n 6N, let < f0,A,,,e, > be any basic neighborhood of Y0 in Co,,,(X ).We know that there is some z fi X\ tJ= A,.Let f 6 C(X) be such that f(z) and f(y) 0 for all y J.=A.. Then f Nn= < fo, An, 6n >" But since f # f0, we see that {f0} is not a G6-set in Co, u(X).
Since (6) implies (3), it remains only to show that (7) implies (4) and (5).Suppose that X contains a dense a-compact subset D. Let I be any element of Ca(X).We need to demonstrate that I has a countable base in Ca(X).We may take for $ the family of bounded open intervals in l with rational endpoints.Also let C be the closures of the members of 05.Then define .a{.f-(C)oD: e}.
Since D is a-compact, at is a countable subfamily of a(X).Since * is also countable, it now suffices to show that for each subbasic neighborhood [$,v] of ! in Co(X), there is an A 6 at and B 6 * such that/" [A,B] c_ [S, V].
If ! 6 [$, V], then I($) c_ V. Hence since is compact, there exist C 6 e and B 6 such that This means that f .[A,B].To show that [A,B]C_[S,V], let g e; [A,B] and let -'.If W is a neighborhood of t, then g-l(W)n S $.Since S C_ f-l(f(q)) C._ f-l(int C), Now for each A 15 a, define the pseudo-seminorm PA on C(X) by Also for each A t5 a and e > O, let Let PA(f) min{1, sup{ 1(*) :x 15 A}}.
Therefore if we define A y-(C)Cl D, then we havef(A) _ f(f-I(C))CI f(D) C_ CCI f(D) _ B.